SET THEORY AND UNIQUENESS FOR TRIGONOMETRIC SERIES

Alexander S. Kechris* Department of Mathematics Caltech Pasadena, CA 91125

Dedicated to the memory of my friend and colleague Stelios Pichorides

Problems concerning the uniqueness of an expansion of a function in a trigonometric series have a long and fascinating history, starting back in the 19th Century with the work of Riemann, Heine and Cantor. The origins of set theory are closely connected with this subject, as it was Cantor’s research into the nature of exceptional sets for such uniqueness problems that led him to the creation of set theory. And the earliest application of one of Cantor’s fundamental concepts, that of ordinal numbers and transfinite induction, can be glimpsed in his last work on this subject.

The purpose of this paper is to give a basic introduction to the application of set theoretic methods to problems concerning uniqueness for trigonometric series. It is written in the style of informal lecture notes for a course or seminar on this subject and, in particular, contains several exercises. The treatment is as elementary as possible and only assumes some familiarity with the most basic results of general topology, measure theory, functional analysis, and descriptive set theory. Standard references to facts that are used without proof are given in the appropriate places.

The notes are divided into three parts. The first deals with ordinal numbers and transfinite induction, and gives an exposition of Cantor’s work. The second gives an application of Baire category methods, one of the basic set theoretic tools in the arsenal of an analyst. The final part deals with the role of descriptive methods in the study of sets of uniqueness. It closes with an introduction to the modern use of ordinal numbers in descriptive set theory and its applications, and brings us back full circle to the concepts that arose in the beginning of this subject.

There is of course much more to this area than what is discussed in these introductory notes. Some references for further study include: Kechris-Louveau [1989, 1992], Cooke [1993], Kahane-Salem [1994], Lyons [1995]. The material in these notes is mostly drawn from these sources.

Stelios Pichorides always had a strong interest in the problems of uniqueness for trigonometric series, and, in particular, in the problem of characterizing the sets of uniqueness. I remember distinctly a discussion we had, in the early eighties, during a long ride in the Los Angeles freeways, in which he wondered whether lack of further progress on this

* Research partially supported by NSF grant. deep problem perhaps had some logical or foundational explanation. (It should be pointed out that mathematical logic was Stelios’ first love, and he originally went to the University of Chicago to study this field.) His comments are what got me interested in this subject and we have had extensive discussions about this for many years afterwards. Certainly the results proved in the 1980’s (some of which are discussed in Part III below) concerning the Characterization Problem seem to confirm his intuition.

Problems concerning sets of uniqueness have fascinated mathematicians for over 100 years now, in part because of the intrinsic nature of the subject and in part because of its intriguing interactions with other areas of classical analysis, measure theory, functional analysis, number theory, and set theory. Once someone asked Paul Erdös, after he gave a talk about one of his favorite number theory problems, somewhat skeptically, why he was so interested in this problem. Erdös replied that if this problem was good enough for Dirichlet and Gauss, it was good enough for him. To paraphrase Erdös, if the problems of uniqueness for trigonometric series were good enough for Riemann, Cantor, Luzin, Menshov, Bari, Salem, Zygmund, and Pichorides, they are certainly good enough for me.

PART I. ORDINAL NUMBERS AND TRANSFINITE INDUCTION.

§1. The problem of uniqueness for trigonometric series.

A trigonometric series S is an infinite series of the form

S ∼ \frac{a_{0}}{2} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx),

where a_{n} ∈ ℂ, x ∈ ℝ. We view this as a formal expression without any claims about its convergence at a given point x.

The Nth partial sum of this series is the trigonometric polynomial

S_{N}(x) = \frac{a_{0}}{2} + ∑_{n=1}^{N}(a_{n} cos nx + b_{n} sin nx).

If, for some given x, S_{N}(x) → s ∈ ℂ, we write

s = \frac{a_{0}}{2} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx)

and call s the sum of the series at x.

A function f : ℝ → ℂ admits a trigonometric expansion if there is a series S as above so that for every x ∈ ℝ

f(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx).

It is clear that any such function is periodic with period 2π.

It is a very difficult problem to characterize the functions f which admit a trigonometric expansion, but it is a classical result that any “nice” enough 2π-periodic function f, for example a continuously differentiable one, admits a trigonometric expansion

f(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx),

there the coefficients a_{n}, b_{n} can be, in fact, computed by the well-known Fourier formulas

\begin{aligned}a_{n} & = \frac{1}{π} ∫_{0}^{2π} f(t) cos nt dt, \\ b_{n} & = \frac{1}{π} ∫_{0}^{2π} f(t) sin nt dt. \end{aligned}

The following question now arises naturally: Is such an expansion unique?

If f(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx) = \frac{a′_{0}}{2} + ∑_{n=1}^{∞}(a′_{n} cos nx + b′_{n} sin nx), then, by subtracting, we would have a series \frac{c_{0}}{2} + ∑_{n=1}^{∞}(c_{n} cos nx + d_{n} sin nx) with

0 = \frac{c_{0}}{2} + ∑_{n=1}^{∞}(c_{n} cos nx + d_{n} sin nx)

for every x. So the problem is equivalent to the following:

Uniqueness Problem. If \frac{a_{0}}{2} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx) is a trigonometric series and

\frac{a_{0}}{2} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx) = 0

for all x ∈ ℝ, it is true that a_{n} = b_{n} = 0 for all n?

This is the problem (which arose through the work of Riemann and Heine) that Heine proposed, in 1869, to the 24 year old Cantor, who had just accepted a position at the university in Halle, where Heine was a senior colleague.

In the next few sections we will give Cantor’s solution to the uniqueness problem and see how his search for extensions, allowing exceptional points, led him to the creation of set theory, including the concepts of ordinal numbers and the method of transfinite induction. We will also see how this method can be used to prove the first such major extension.

Before we proceed, it would be convenient to also introduce an alternative form for trigonometric series.

Every series

S ∼ \frac{a_{0}}{2} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx)

can be also written as

S ∼ ∑_{n=−∞}^{∞} c_{n}e^{inx},

where, letting b_{0} = 0,

c_{n} = \frac{a_{n} − ib_{n}}{2} , c_{−n} = \frac{a_{n} + ib_{n}}{2} (n ∈ ℕ).

(Thus a_{n} = c_{n} + c_{−n}, b_{n} = i(c_{n} − c_{−n}).) In this notation, the partial sums S_{N}(x) are given by

S_{N}(x) = ∑_{n=−N}^{N} c_{n}e^{inx};

and if they converge as N → ∞ with limit s, we write

s = ∑_{n=−∞}^{+∞} c_{n}e^{inx}.

The standard examples of trigonometric series are the Fourier series of integrable functions. Given a 2π-periodic function f : ℝ → ℂ we say that it is integrable if it is Lebesgue measurable and

\frac{1}{2π} ∫_{0}^{2π} |f(t)|dt < ∞.

In this case we define its Fourier coefficients f̂(n), n ∈ ℤ, by

f̂(n) = \frac{1}{2π} ∫_{0}^{2π} f(t)e^{−int}dt.

We call the trigonometric series

S(f) ∼ ∑_{n=−∞}^{∞} f̂(n)e^{inx}

the Fourier series of f, and write

S_{N}(f, x) = ∑_{n=−N}^{+N} f̂(n)e^{inx}

for its partial sums.

Remark. There are trigonometric series, like ∑_{n=2}^{∞} \frac{sin nx}{log n}, which converge everywhere but are not Fourier series.

§2. The Riemann Theory.

Riemann was the first mathematician to seriously study general trigonometric series (as opposed to Fourier series), in his Habilitationsschrift (1854). We will prove here two of his main results, that were used by Cantor. These results are beautiful applications of elementary calculus.

Let S ∼ ∑ c_{n}e^{inx} be an arbitrary trigonometric series with bounded coefficients,i.e., |c_{n}| ≤ M for some M and all n ∈ ℤ. Then Riemann had the brilliant idea to consider the function obtained by formally integrating ∑ c_{n}e^{inx} twice. This function, called the Riemann function F_{S} of S, is thus defined by

F_{S}(x) = \frac{c_{0}x^{2}}{2} − ∑_{n=−∞}^{+∞_{/}} \frac{1}{n^{2}} c_{n}e^{inx}, x ∈ ℝ,

where ∑′ means n = 0 is excluded.

Clearly, as c_{n} is bounded, F_{S} is a continuous function on ℝ (but is not periodic), since the above series converges absolutely and uniformly, as |\frac{1}{n^{2}} c_{n}e^{inx}| ≤ \frac{M}{n^{2}} .

Now, intuitively, one would hope that F′′_{S}(x) should be the same as ∑ c_{n}e^{inx}, if this sum exists. This may not be quite true, but something close enough to it is.

Given a function F : ℝ → ℂ, let

Δ^{2}F (x, h) = F (x + h) + F (x − h) − 2F (x)

and define the second symmetric derivative or second Schwartz derivative of F at x by

D^{2}F (x) = lim_{h→0} \frac{Δ^{2}F (x, h)}{h^{2}},

provided this limit exists. 2.1. Exercise. If F′′(x) exists, then so does D^{2}F (x) and they are equal, but the converse fails. 2.2. Riemann’s First Lemma. Let S ∼ ∑ c_{n}e^{inx} be a trigonometric series with bounded coefficients. If s = ∑ c_{n}e^{inx} exists, then D^{2}F_{S}(x) exists and D^{2}F_{S}(x) = s. Proof. We have by calculating

\frac{Δ^{2}F_{S}(x, 2h)}{4h^{2}} = ∑_{n=−∞}^{+∞}[\frac{sin nh}{nh}]^{2}c_{n}e^{inx}

(for n = 0, we let \frac{sin nh}{nh} = 1). So it is enough to prove the following:

2.3. Lemma. ∑_{n=0}^{∞} a_{n} = a ⇒ lim_{h→0}[ ∑_{n=0}^{∞}[\frac{sin nh}{nh}]^{2}a_{n}] = a.

Proof. Put A_{N} = ∑_{n=0}^{N} a_{n}. Then

∑_{n=0}^{∞}(\frac{sin nh}{nh})^{2}a_{n} = ∑_{n=0}^{∞}[(\frac{sin nh}{nh})^{2}−(\frac{sin(n + 1)h}{(n + 1)h})^{2}]A_{n}.

Let h_{k} → 0, h_{k} > 0 and put

s_{kn} =(\frac{sin nh_{k}}{nh_{k}} )^{2}−(\frac{sin(n + 1)h_{k}}{(n + 1)h_{k}} )^{2}.

Then we have to show that

A_{n} ⟶^{n} a ⇒ ∑_{n=0}^{∞} A_{n}s_{kn} ⟶^{k} a.

We view the infinite matrix (s_{kn}) as a summability method, i.e., a way of transforming a sequence (x_{n}) into the sequence

(y_{k}) = (s_{kn}) · (x_{n}),

i.e., y_{k} = ∑_{n=0}^{∞} s_{kn}x_{n}.

Example. If s_{kn} = \frac{1}{k+1} for n ≤ k, s_{kn} = 0 for n > k, then y_{k} = \frac{x_{0}+···+x_{k}}{k+1} .

A summability method is called regular if x_{n} ⟶^{n} x ⇒ y_{k} ⟶^{k} x. Toeplitz proved the following result which we leave as an exercise.

2.4. Exercise. (a) If the matrix (s_{kn}) satisfies the following conditions, called Toeplitz conditions:

(i) s_{kn} ⟶^{k} 0, ∀n ∈ ℕ,

(ii) ∑_{n=0}^{∞} |s_{kn}| ≤ C < ∞, ∀k ∈ ℕ,

(iii) ∑_{n=0}^{∞} s_{kn} ⟶^{k} 1,

then (s_{kn}) is regular.

(b) If (s_{kn}) satisfies only (i), (ii) and x_{n} → 0, then y_{k} → 0.

So it is enough to check that (i), (ii), (iii) hold for

s_{kn} =(\frac{sin nh_{k}}{nh_{k}} )^{2}−(\frac{sin(n + 1)h_{k}}{(n + 1)h_{k}} )^{2}.

Clearly (i), (iii) hold. For (ii), let u(x) = (\frac{sin x}{x})^{2}. Then we have

2.5. Exercise. ∫_{0}^{∞}|u′(x)|dx < ∞.

Thus

\begin{aligned}∑_{n=0}^{∞} |s_{kn}| & = ∑_{n=0}^{∞}| ∫_{nh_{k}}^{(n+1)h_{k}} u′(x)dx| \\ & ≤ ∫_{0}^{∞} |u′(x)|dx = C < ∞. \end{aligned}

⊣

2.6. Riemann’s Second Lemma. Let S ∼ ∑ c_{n}e^{inx} be a trigonometric series with c_{n} → 0. Then

\frac{Δ^{2}F_{S}(x, h)}{h} = \frac{F_{S}(x + h) + F_{S}(x − h) − 2F_{S}(x)}{h} → 0,

as h → 0, uniformly on x.

Proof. By direct calculation we have again

\frac{Δ^{2}F_{S}(x, 2h)}{4h} = ∑ \frac{sin^{2}(nh)}{n^{2}h}c_{n}e^{inx}

(where for n = 0, \frac{sin^{2}(nh)}{n^{2}h} is defined to be h). Let as before 0 < h_{k} ≤ 1, h_{k} → 0 and put

t_{kn} = \frac{sin^{2}(nh_{k})}{n^{2}h_{k}}.

We have to show that ∑(c_{n}e^{inx})t_{kn} → 0 as k → ∞, uniformly on x. Since c_{n}e^{inx} → 0, uniformly on x, it is enough to verify that (t_{kn}) satisfies the first two Toeplitz conditions (i), (ii), of 2.4.

Clearly (i) holds. To prove (ii) fix k and choose N > 1 with

N − 1 ≤ h_{k}^{−1} < N.

Then

∑_{n=1}^{∞} |t_{kn}| = ∑_{n=1}^{N−1} \frac{sin^{2}(nh_{k})}{n^{2}h_{k}} + ∑_{n=N}^{∞} \frac{sin^{2}(nh_{k})}{n^{2}h_{k}}

≤ (N − 1) · h_{k} + \frac{1}{h_{k}} ∑_{n=N}^{∞} \frac{1}{n^{2}}

≤ 1 + \frac{1}{h_{k}} ∑_{n=N}^{∞} \frac{1}{n(n − 1)}

= 1 + \frac{1}{h_{k}} · \frac{1}{N − 1} ≤ 3,

and we are done. ⊣

Note that this implies that the graph of F_{S} can have no corners, i.e., if the left- and right-derivatives of F_{S} exist at some point x, then they must be equal.

§3. The Cantor Uniqueness Theorem.

The following result was proved originally by Cantor with “set of positive measure” replaced by “interval”.

3.1. The Cantor-Lebesgue Lemma. If a_{n} cos(nx) + b_{n} sin(nx) → 0, for x in a set of positive (Lebesgue) measure, then a_{n}, b_{n} → 0. So if ∑ c_{n}e^{inx} = 0 for x in a set of positive measure, then c_{n} → 0.

Proof. We can assume that a_{n}, b_{n} ∈ ℝ. Let ρ_{n} = \sqrt{a_{n}^{2} + b_{n}^{2}} and φ_{n} be such that

a_{n} cos(nx) + b_{n} sin(nx) = ρ_{n} cos(nx + φ_{n}).

Thus ρ_{n} cos(nx + φ_{n}) → 0 on E ⊆ [0, 2π), a set of positive measure. Assume ρ_{n} ↛ 0, toward a contradiction. So there is ϵ > 0 and n_{0} < n_{1} < n_{2} < ⋯ such that ρ_{n_{k}} ≥ ϵ. Then cos(n_{k}x + φ_{n_{k}}) → 0, so 2 cos^{2}(n_{k}x + φ_{n_{k}}) → 0, i.e., 1 + cos 2(n_{k}x + φ_{n_{k}}) → 0 for x ∈ E. By Lebesgue Dominated Convergence ∫_{E}(1 + cos 2(n_{k}x + φ_{n_{k}}))dx → 0, i.e., letting χ_{E} be the characteristic function of E in the interval [0, 2π) extended with period 2π over all of ℝ, and μ(E) be the Lebesgue measure of E:

μ(E) + ∫_{0}^{2π} χ_{E}(x) cos 2(n_{k}x + φ_{n_{k}})dx

=μ(E) + 2π[Re χ̂_{E}(−2n_{k}) · cos 2φ_{n_{k}} − Im χ̂_{E}(−2n_{k}) · sin 2φ_{n_{k}}] → 0.

We now have:

3.2. Exercise (Riemann-Lebesgue). If f is an integrable 2π-periodic function, then f̂(n) → 0 as |n| → ∞. [Remark. This is really easy if f is the characteristic function of an interval.]

So it follows that μ(E) = 0, a contradiction. ⊣

We are only one lemma away from the proof of Cantor’s Theorem. This lemma was proved by Schwartz in response to a request by Cantor.

3.3 Lemma (Schwartz). Let F : (a, b) → ℝ be continuous such that D^{2}F (x) ≥ 0, ∀x ∈ (a, b). Then F is convex on (a, b). In particular, if F : (a, b) → ℂ is continuous and D^{2}F (x) = 0, ∀x ∈ (a, b), then F is linear on (a, b).

Proof. By replacing F by F + ϵx^{2}, ϵ > 0, and letting ϵ → 0, we can assume actually that D^{2}F (x) > 0 for all x ∈ (a, b).

Assume F is not convex, toward a contradiction. Then there is a linear function μx + ν and a < c < d < b such that if G(x) = F (x) − (μx + ν), then G(c) = G(d) = 0 and G(x) > 0 for some x ∈ (c, d). Let x_{0} be a point where G achieves its maximum in [c, d]. Then c < x_{0} < d. Now for small enough h, Δ^{2}F (x_{0}, h) ≤ 0, so D^{2}F (x_{0}) ≤ 0, a contradiction. ⊣

We now have:

3.4. Theorem (Cantor, 1870). If ∑ c_{n}e^{inx} = 0 for all x, then c_{n} = 0, ∀n ∈ ℤ.

Proof. Let S ∼ ∑ c_{n}e^{inx}. By the Cantor-Lebesgue Lemma, c_{n} → 0, as |n| → ∞, so, in particular, c_{n} is bounded. By Riemann’s First Lemma, D^{2}F_{S}(x) = 0, ∀x ∈ ℝ, and so by Schwartz’s Lemma F_{S} is linear, i.e.,

c_{0} \frac{x^{2}}{2} − ∑′ \frac{1}{n^{2}} c_{n}e^{inx} = ax + b.

Put x = π, x = −π and subtract to get a = 0. Put x = 0, x = 2π and subtract to get c_{0} = 0. Then ∑′ \frac{1}{n^{2}} c_{n}e^{inx} = b. Since this series converges uniformly, if m ≠ 0 we have

\begin{matrix}0 = & ∫_{0}^{2π} be^{−imx}dx \\ = & ∑′ \frac{1}{n^{2}} c_{n} ∫_{0}^{2π} e^{i(n−m)x}dx \\ = & \frac{c_{m}}{m^{2}} , \end{matrix}

so c_{m} = 0 and we are done. ⊣

Remark. Kronecker (when he was still on speaking terms with Cantor) pointed out that the use of the fact that c_{n} → 0 was unnecessary, i.e., if one can prove that (1) ∑ c_{n}e^{inx} = 0, ∀x ∈ ℝ & c_{n} → 0 ⇒ c_{n} = 0, ∀n ∈ ℤ, then it follows that (2) ∑ c_{n}e^{inx} = 0, ∀x ∈ ℝ ⇒ c_{n} = 0, ∀n ∈ ℤ.

To see this assume (1) and take any series ∑ c_{n}e^{inx} such that ∑ c_{n}e^{inx} = 0, ∀x ∈ ℝ (without any other assumption on the c_{n}). Put for x, x + u and x − u, add and divide by 2 to get

∑ c_{n}e^{inx} cos nu = 0, ∀x, u ∈ ℝ.

or equivalently

c_{0} + ∑_{n=1}^{∞}(a_{n} cos nx + b_{n} sin nx) cos nu = 0, ∀x, u ∈ ℝ.

Considering x fixed now, we note that a_{n} cos nx + b_{n} sin nx → 0 (since c_{0} + ∑(a_{n} cos nx + b_{n} sin nx) = 0) so we can apply (1) to get that a_{n} cos nx + b_{n} sin nx = 0, for all x, n ∈ ℕ, from which easily a_{n} = b_{n} = 0, ∀n ∈ ℕ.

§4. Sets of uniqueness.

Cantor next extended his uniqueness theorem in 1871 by allowing a finite number of exceptional points.

4.1. Theorem (Cantor, 1871). Assume that ∑ c_{n}e^{inx} = 0 for all but finitely many x ∈ ℝ. Then c_{n} = 0, ∀n ∈ ℤ.

Proof. Let S ∼ ∑ c_{n}e^{inx}. Suppose x_{0} = 0 ≤ x_{1} < x_{2} < ⋯ < x_{n} < 2π = x_{n+1} are such that for x ≠ x_{i}, ∑ c_{n}e^{inx} = 0. Then, by Schwartz’s Lemma, F_{S} is linear in each interval (x_{i}, x_{i+1}). Since by Riemann’s Second Lemma the graph of F_{S} has no corners, it follows that F_{S} is linear in [0, 2π). The same holds of course for any interval of length 2π, so F_{S} is linear and thus as in the proof of 3.4, c_{n} = 0, ∀n ∈ ℤ. ⊣

In the sequel it will be convenient to identify the unit circle 𝕋 = {e^{ix} : 0 ≤ x < 2π} with ℝ/2πℤ via the map x ↦ e^{ix}. Often we can think of 𝕋 as being [0, 2π), or [0, 2π] with 0, 2π identified. Functions on 𝕋 can be also thought of as 2π-periodic functions on ℝ. We denote by λ the Lebesgue measure on 𝕋 (i.e., the one induced by the above identification with [0, 2π)) normalized so that λ(𝕋) = 1.

We now introduce the following basic concept.

Definition. Let E ⊆ 𝕋. We say that E is a set of uniqueness if every trigonometric series ∑ c_{n}e^{inx} which converges to 0 off E (i.e., ∑ c_{n}e^{inx} = 0, for e^{ix} ∉ E, which we will simply write “x ∉ E”) is identically 0. Otherwise it is called a set of multiplicity.

So 3.4 says that ∅ is a set of uniqueness and 4.1 says that every finite set is a set of uniqueness.

We denote by U the class of sets of uniqueness and by M the class of sets of multiplicity.

Our next goal is to prove the following extension of Cantor’s Theorem.

4.2. Theorem (Cantor, Lebesgue – see Remarks in §6 below). Every countable closed set is a set of uniqueness.

We will give a proof using the method of transfinite induction.

§5. The Cantor-Bendixson Theorem.

Let E ⊆ 𝕋 be a closed set. We define its Cantor-Bendixson derivative E′ by

E′ = {x ∈ E : x is a limit point of E}.

Note that E′ ⊆ E and E′ is closed as well.

Now define by transfinite induction for each ordinal α a closed set E^{(α)} as follows:

E^{(0)} = E,

E^{(α+1)} = (E^{(α)})′

E^{(λ)} = ⋂_{α<λ} E^{(α)}, λ a limit ordinal.

The E^{(α)} form a decreasing sequence of closed sets contained in E:

E^{(0)} ⊇ E^{(1)} ⊇ E^{(2)} ⊇ ⋯ ⊇ E^{(α)} ⊇ ⋯ ⊇ E^{(β)} ⊇ ⋯, α ≤ β.

5.1. Lemma. If F_{α}, α an ordinal, is a decreasing sequence of closed sets, then for some countable ordinal α_{0} we have that

F_{α_{0}} = F_{α_{0}+1}.

Proof. Fix a basis {U_{n}} for the topology of 𝕋 and let

A_{α} = {n : U_{n} ∩ F_{α} = ∅}.

Then α ≤ β ⇒ A_{α} ⊆ A_{β}. If A_{α} ⫋ A_{α+1} for all countable ordinals α, let f(α) be the least n with n ∈ A_{α+1} \ A_{α}. Then, if ω_{1} is the first uncountable ordinal, which by standard practice we also identify with the set {α : α < ω_{1}} of all countable ordinals, we have that f : ω_{1} → ℕ is injective, which is a contradiction. So for some α_{0} < ω_{1}, A_{α_{0}} = A_{α_{0}+1}, so F_{α_{0}} = F_{α_{0}+1}. ⊣

Thus for each closed set E ⊆ 𝕋, there is a least countable α with E^{(α)} = E^{(α+1)} and thus E^{(α)} = E^{(β)} for all α ≤ β. We denote this ordinal by rk_{CB}(E) and call it the Cantor-Bendixson rank of E. We also put

E^{(∞)} = E^{(rk_{CB}(E))}.

Notice that (E^{(∞)})′ = E^{(∞)}, so E^{(∞)} is perfect, i.e., every point of it is a limit point (it could be ∅ though). We call it the perfect kernel of E.

5.2. Exercise. Show that E^{(∞)} is the largest perfect set contained in E.

5.3. Theorem (Cantor-Bendixson). Let E be closed. Then the set E \ E^{(∞)} is countable. In particular,

E is countable ⇔ E^{(∞)} = ∅.

Proof. Let x ∈ E \ E^{(∞)}, so that for some (unique) α < rk_{CB}(E), x ∈ E^{(α)} \ E^{(α+1)}. Since there are only countably many such α, it is enough to prove the following:

5.4. Lemma. For any closed set F, F \ F′ is countable.

Proof. Fix a countable basis {U_{n}}_{n∈ℕ}. If x ∈ F \ F′, then there is some n with F ∩ U_{n} = {x}. So F \ F′ = ⋃{F ∩ U_{n} : F ∩ U_{n} is a singleton}, which is clearly countable. ⊣

5.5. Exercise. Show that for each closed set E there is a unique decomposition E = P ∪ C, P ∩ C = ∅, P perfect, C countable.

5.6. Exercise. For each countable successor ordinal α find a countable closed set E with rk_{CB}(E) = α.

§6. Sets of uniqueness (cont’d).

We are now ready to give the

Proof of Theorem 4.2.

Let E be a countable closed set. Let S ∼ ∑ c_{n}e^{inx} be such that ∑ c_{n}e^{inx} = 0 off E. Since it is clear that any translate (in 𝕋) of a set of uniqueness is also a set of uniqueness, we can assume that 0 ∉ E. So we can view E as being a closed set contained in (0, 2π). The complement in (0, 2π) of any closed subset F of (0, 2π) is a disjoint union of open intervals with endpoints in F ∪ {0, 2π}, called its contiguous intervals. We will prove by transfinite induction on α, that F_{S} is linear on each contiguous interval of E^{(α)}. Since E^{(α_{0})} = ∅ for some α_{0}, it follows that F_{S} is linear on (0, 2π), so, as before, c_{n} = 0, ∀n.

α = 0: This is clear since ∑ c_{n}e^{inx} = 0 on each contiguous interval of E^{(0)} = E.

α ⇒ α+1: Assume F_{S} is linear in each contiguous interval of E^{(α)}. Let then (a, b) be a contiguous interval of E^{(α+1)}. Then in each closed subinterval [c, d] ⊆ (a, b) there are only finitely many points c ≤ x_{0} < x_{1} < ⋯ < x_{n} ≤ d of E^{(α)}. Then (c, x_{0}), (x_{0}, x_{1}), ⋯, (x_{n}, d) are contained in contiguous intervals of E^{(α)}, so, by induction hypothesis, F_{S} is linear in each one of them, so by the Riemann Second Lemma again, F_{S} is linear on [c, d] and thus on (a, b).

α < λ ⇒ λ (λ a limit ordinal): We use a compactness argument. Fix a contiguous interval (a, b) of E^{(λ)} and a closed subinterval [c, d] ⊆ (a, b). Then

[c, d] ⊆ (0, 2π) \ E^{(λ)}

= (0, 2π) \ ⋂_{α<λ} E^{(α)}

= ⋃_{α<λ}[(0, 2π) \ E^{(α)}].

Since (0, 2π) \ E^{(α)} is open and [c, d] is compact, there are finitely many α_{1}, ⋯, α_{n} < λ with

[c, d] ⊆ ⋃_{α∈{α_{1},···,α_{n}}}[(0, 2π) \ E^{(α)}] ⊆ (0, 2π) \ E^{(β)}

for any α_{1}, ⋯, α_{n} ≤ β < λ. Then [c, d] is contained in a contiguous interval of E^{(β)}, so, by induction hypothesis, F_{S} is linear on [c, d] and thus on (a, b).

This completes the proof. ⊣

Remark. Cantor published in 1872 the proof of Theorem 4.2 only for the case rk_{CB}(E) < ω, i.e., when the Cantor-Bendixson process terminates in finitely many steps. Apparently at this stage he had, at least at some intuitive level, the idea of extending this process into the transfinite at levels ω, ω + 1, · · ·. However this involved conceptual difficulties which led him to re-examine the foundations of the real number system and eventually to create set theory, including, several years later, the rigorous development of the theory of ordinal numbers and transfinite induction. However, after 1872 Cantor never returned to the problem of uniqueness and never published a complete proof of 3.5. This was done, much later, by Lebesgue in 1903.

Theorem 4.1 was further extended by Bernstein (1908) and W. H. Young (1909) to show that an arbitrary countable set is a set of uniqueness. Finally in 1923 Bari showed that the union of countably many closed sets of uniqueness is a set of uniqueness.

6.1. Exercise (Bernstein, Young). Assume these results and the fact that any uncountable Borel set contains a non-∅ perfect subset. Show that every set which contains no perfect non-∅ set is a set of uniqueness.

PART II. BAIRE CATEGORY METHODS.

§7. Sets of uniqueness and Lebesgue measure.

To get some idea about the size of sets of uniqueness, we will first prove the following easy fact.

7.1. Proposition. Let A ⊆ 𝕋 be a (Lebesgue) measurable set of uniqueness. Then A is null, i.e., λ(A) = 0.

Proof. Assume λ(A) > 0, towards a contradiction. Then, by regularity, λ(F) > 0 for some closed subset F ⊆ A. Consider the characteristic function χ_{F} of F and its Fourier series S(χ_{F}). We need the following standard fact.

7.2. Lemma (Localization principle for Fourier series). Let f be an integrable function on 𝕋. Then the Fourier series of f converges to 0 in any open interval in which f vanishes.

Proof. We have

S_{N}(f, x) = ∑_{−N}^{N} f̂(n)e^{inx}

= ∑_{−N}^{N}(\frac{1}{2π} ∫_{0}^{2π} f(t)e^{−int}dt)e^{inx}

= \frac{1}{2π} ∫_{0}^{2π} f(t)( ∑_{−N}^{N} e^{in(x−t)})dt

= \frac{1}{2π} ∫_{0}^{2π} f(t)D_{N}(x − t)dt (= f ∗ D_{n}(x)),

where the Dirichlet kernel D_{n} is defined by

D_{n}(u) = ∑_{−N}^{N} e^{inu} = \frac{sin(n + \frac{1}{2})u}{sin \frac{u}{2}} = cos nu + cot(\frac{u}{2}) sin nu.

So (changing ∫_{0}^{2π} to ∫_{−π}^{π})

S_{N}(f, 0) = \frac{1}{2π} ∫_{−π}^{π} f(t) cos ntdt + \frac{1}{2π} ∫_{−π}^{π} f(t) cot(\frac{t}{2}) sin ntdt.

If now f vanishes in an interval around 0, then clearly f(t) cot(\frac{t}{2}) is integrable, so these two integrals converge to 0 by the Riemann-Lebesgue Lemma (3.2). So S_{N}(f, 0) → 0.

Thus we have shown that if f vanishes in an interval around 0, then its Fourier series converges to 0 at 0. By translation this is true for any other point. ⊣

So S(χ_{F}) converges to 0 in any interval disjoint from F, i.e., S(χ_{F}) converges to 0 off F. So, since F is a set of uniqueness, χ̂_{F} (n) = 0 for all n. But χ̂_{F} (0) = λ(F) > 0, a contradiction. ⊣

7.3. Exercise (Bernstein). Show that there is A ⊆ 𝕋 such that neither A nor 𝕋 \ A contain a non-∅ perfect set (such a set is called a Bernstein set). Conclude that A is not measurable and therefore that A is a set of uniqueness which is not null.

So we have countable ⊆ U ∩ measurable ⊆ null.

In the beginning of the century it was widely believed that U ∩ measurable = null, i.e., if a trigonometric series ∑ c_{n}e^{inx} converges to 0 almost everywhere (a.e.), then it is identically 0. Recall, for example, here the following standard fact (which we will not use later on).

7.4. Theorem. (Fejér-Lebesgue). For every integrable function f on 𝕋, let

σ_{N}(f, x) = \frac{S_{0}(f, x) + S_{1}(f, x) + ⋯ + S_{N}(f, x)}{N + 1}

be the average of the partial sums S_{N}(f, x) of the Fourier series of f. Then σ_{N}(f, x) → f(x) a.e. In particular, if a Fourier series converges a.e. to 0, it is identically 0.

For example, in Luzin’s dissertation Integration and trigonometric series (1915), this problem is discussed and is considered improbable that there is a non-zero trigonometric series that converges to 0 a.e. It thus came as a big surprise when in 1916 Menshov proved that there are indeed trigonometric series which converge to 0 a.e. but are not identically 0. It follows, for example, that any function which admits a trigonometric expansion admits actually more than one, if only convergence a.e. to the function is required.

Remark. Menshov also showed that every measurable function admits an a.e. trigonometric expansion, i.e., for any 2π-periodic measurable f there is a trigonometric series ∑ c_{n}e^{inx} such that f(x) = ∑ c_{n}e^{inx} a.e. By the above this series is not unique.

Menshov actually constructed an example of a closed set of multiplicity of measure 0, by an appropriate modification of the standard construction of the Cantor set. The Cantor set in the interval [0, 2π] is constructed by removing the middle 1/3 open interval, then in each of the remaining two closed intervals remaining the middle 1/3 open interval, etc. Now suppose we modify the construction by removing in the first stage the middle 1/2 interval, at the second stage, the middle 1/3 interval, at the third stage the middle 1/4 interval, etc. Denote the resulting perfect set by E_{M}.

7.5. Exercise. Show that λ(E_{M}) = 0.

Now Menshov showed that E_{M} is a set of multiplicity as follows: This set is in a canonical 1-1 correspondence with the set of infinite binary sequences 2^{ℕ}. Take the usual coin-tossing measure on 2^{ℕ} and transfer it by this correspondence to E_{M}. Call this measure μ_{M}. (Thus μ_{M} gives equal measure 1/2^{n} to each of the 2^{n} closed intervals at the nth stage of this construction.) It is a probability Borel measure on 𝕋, so we can define its Fourier(- Stieltjes) coefficients μ̂_{M}(n), as usual, by

μ̂_{M}(n) = ∫ e^{−int}dμ_{M}(t).

Then it turns out that ∑ μ̂_{M}(n)e^{inx} = 0 for x ∉ E, so E is a set of multiplicity (as μ̂_{M}(0) = μ(E) > 0). This might seem not too surprising, since after all the measure μ “lives” on E_{M}. However by the same token one should also expect that if we consider the usual Cantor set E_{C} and denote the corresponding measure by μ_{C} we should have ∑ μ̂_{C}(n)e^{inx} = 0 off the Cantor set as well, i.e., the Cantor set should also be a set of multiplicity, which is false! This is only one of the many phenomena in this subject which challenge your intuition.

What is the difference between μ_{M}, μ_{C} that accounts for this phenomenon? It turns out that it is the following: μ̂_{M}(n) → 0 but μ̂_{C}(n) ↛ 0 as |n| → ∞. In fact, we have the following result.

7.6. Theorem. Let E ⊆ 𝕋, E ≠ 𝕋 be a closed set and let μ be a probability Borel measure on 𝕋 with μ(E) = 1. Then the following are equivalent:

(i) μ̂(n) → 0;

(ii) ∑ μ̂(n)e^{inx} = 0, ∀x ∉ E.

Clearly (ii) ⇒ (i) by the Cantor-Lebesgue Lemma. The proof of (i) ⇒ (ii) requires some further background in the theory of trigonometric series and we will postpone it for a while (see §§12,13).

Thus Menshov’s proof is based on the fact that μ̂_{M}(n) → 0, which is proved by a delicate calculation.

We will develop in the sequel a totally different approach to Menshov’s Theorem on the existence of null sets of multiplicity, an approach based on the Baire category method.

§8. Baire category.

A set in a topological space is nowhere dense if its closure has empty interior. A set is first category or meager if it is contained in a countable union of nowhere dense sets. Otherwise it is of the second category or non-meager. It is clear that meager sets form a σ-ideal, i.e., are closed under subsets and countable unions. So this concept determines a notion of “topological smallness” analogous to that of null sets in measure theory. Of course, for this to be of any interest it better be that it doesn’t trivialize, i.e., that the whole space is not meager. This is the case in well-behaved spaces, like complete metric spaces. In fact we have an even stronger statement known as the Baire Category Theorem (for such spaces). We call a set comeager if its complement is meager. Notice that a set is comeager iff it contains a countable intersection of dense open sets.

8.1 Theorem (The Baire Category Theorem). Let X be a complete metric space. Every comeager set is dense (in particular non-∅).

This is the basis of a classical method of existence proof in mathematics: Suppose we want to show the existence of a mathematical object x satisfying some property P. The category method consists of finding an appropriate complete metric space X (or other “nice” topological space satisfying the Baire Category Theorem) and showing that {x ∈ X : P (x)} is comeager in X. This not only shows that ∃xP (x), but in fact that in the space X “most” elements of X have property P or as it is often expressed the “generic” element of X satisfies property P. (Similarly, if a property holds a.e. we say that the “random” element satisfies it.)

A standard application of the Baire Category Theorem is Banach’s proof of the existence of continuous nowhere differentiable functions (originally due to Bolzano and Weierstrass in the 19th century). The argument goes as follows:

Let C(𝕋) be the space of real valued continuous 2π-periodic functions with the uniform (or sup) metric

d(f, g) = sup{ |f(x) − g(x)| : x ∈ 𝕋}.

It is well-known that this is a complete metric space. We want to show that

{f ∈ C(𝕋) : ∀x(f′(x) does not exist)}

is comeager in C(𝕋), so non-∅. Consider, for each n, the set

U_{n} = \{f ∈ C(𝕋) : ∀x∃h > 0|\frac{f(x + h) − f(x)}{h}| > n\}.

8.2. Exercise. Show that U_{n} is open in C(𝕋).

It is not hard to show now that U_{n} is also dense in U(𝕋). Simply approximate any f ∈ C(𝕋) by a piecewise linear function g and then approximate g by a piecewise linear function with big slopes, so that it belongs to U_{n}.

Thus ⋂_{n} U_{n} is a countable intersection of dense open sets, so it is comeager. But clearly if f ∈ ⋂_{n} U_{n}, f is nowhere differentiable. (So the “generic” continuous function is nowhere differentiable.)

Before we proceed, let us recall that a set A, in a topological space X, is said to have the Baire property (BP) if there is an open set U such that AΔU is meager. The class of sets with the BP is a σ-algebra (i.e., is closed under countable unions and complements), in fact it is the smallest σ-algebra containing the open sets and the meager sets. The sets with the BP are analogs of the measurable sets.

Although there is some analogy between category and measure it should be emphasized that the concepts are “orthogonal”. This can be expressed by the following important fact:

8.3. Proposition. There is a dense G_{δ} (so comeager) set G ⊆ 𝕋 such that λ(G) = 0 (i.e., G is null).

Proof. Let {d_{n}} be a countable dense set in 𝕋. For each m let I_{n,m} be an open interval around d_{n} with λ(I_{n,m}) ≤ \frac{1}{m} · 2^{−n}. Let U_{m} = ⋃_{n} I_{n,m}, so that U_{m} is dense open with λ(U_{m}) ≤ ∑_{n} λ(I_{n,m}) ≤ 2/m. Let G = ⋂_{m} U_{m}. ⊣

§9. Sets of uniqueness and category.

We have seen in §7 that a (measurable) set of uniqueness is measure theoretically negligible, i.e., null. The question was raised, already in the 1920’s (see, e.g., the memoir of N. Bari in Fundamenta Mathematica, Bari [1927]) whether they are also topologically negligible, i.e., meager (assuming they have the BP). So we have

9.1. The Category Problem. Is every set of uniqueness with the BP of the first category?

This problem was solved affirmatively by Debs and Saint Raymond in 1986. Their original proof used the descriptive set theoretic methods that we will discuss in Part III and was quite sophisticated, making use of machinery established in earlier work of Solovay, Kaufman, Kechris-Louveau-Woodin and Kechris-Louveau. In fact Debs and Saint Raymond established the following stronger result.

9.2. Theorem (Debs-Saint Raymond). Let A ⊆ 𝕋 be a non-meager set with the BP. Then there is a Borel probability measure μ on 𝕋 with μ(A) = 1 and μ̂(n) → 0, as |n| → ∞.

To see that this implies 9.1 we argue as follows:

Let A be a set of uniqueness with the BP. If A is not meager, there is μ, a Borel probability measure, with μ(A) = 1, and μ̂(n) → 0. Since every Borel probability measure is regular, there is closed F ⊆ A with μ(F) > 0. Let ν = μ|F, i.e., ν(X) = μ(X ∩ F).

9.3. Proposition. ν̂(n) → 0, as |n| → ∞.

Proof. We have for any continuous function on 𝕋,

∫ fdν = ∫ fχ_{F} dμ,

so

ν̂(n) = ∫ χ_{F} (t)e^{−int}dμ(t).

For each ϵ > 0, there is a trigonometric polynomial P (x) = ∑_{−N}^{N} c_{k}e^{ikx} such that ∫ |χ_{F} − P |dμ < ϵ. Now if

d_{n} = ∫ P (t)e^{−int}dμ(t),

we have

d_{n} = ∫ ( ∑_{−N}^{N} c_{k}e^{ikx})e^{−int}dμ(t)

= ∑_{k=−N}^{N} c_{k}μ̂(n − k) → 0

as |n| → ∞. Moreover

\begin{aligned}|ν̂(n) − d_{n}| & =| ∫ (χ_{F} (t) − P (t))e^{−int}dμ| \\ & ≤ ∫ |χ_{F} − P |dμ < ϵ, \end{aligned}

so \overline{lim_{n}}|ν̂(n)| ≤ ϵ, and thus ν̂(n) → 0. ⊣

Then we can apply 7.6 to conclude that ∑ ν̂(n)e^{inx} = 0 off E and thus off A. As ν̂(0) ≠ 0, this shows that A is a set of multiplicity, a contradiction.

Starting with the next section we will give a very different than the original and much simpler proof of 9.2, due to Kechris-Louveau, which is based on the category method and employs only elementary functional analysis. Before we do that though we want to show how 9.2 gives also a much different proof of Menshov’s Theorem, which “explains” this result as an instance of the “orthogonality” of the concepts of null and meager sets.

Indeed, by 8.3 fix a dense G_{δ} set G ⊆ 𝕋 with λ(G) = 0. Then G is comeager, so by 9.2, there is a Borel probability measure μ with μ(G) = 1 and μ̂(n) → 0. As before, find F ⊆ G closed with μ(F) > 0. Then ∑ μ̂(n)e^{inx} = 0 off F and thus off G and this shows that ∑ μ̂(n)e^{inx} converges to 0 a.e. without being identically 0 (it also shows that F is a null closed set of multiplicity, as Menshov also showed).

But this method has also many other applications. For example, in the 1960’s Kahane and Salem raised the following question: Recall that a number x ∈ [0, 2π] is called normal in base 2 (say) if for \frac{x}{2π} = 0 · x_{1}x_{2} ⋯, x_{i} ∈ \{0, 1\}, and all (a_{0}, ⋯, a_{k−1}), a_{i} ∈ \{0, 1\}, we have

lim_{n→∞} \frac{1}{n} card \{1 ≤ m ≤ n : x_{m+i} = a_{m+i}, 0 ≤ i ≤ k − 1\} = \frac{1}{2^{k}} .

Denote by N the set of normal numbers and by N′ its complement. A famous theorem of Borel asserts that almost every number is normal, i.e., λ(N) = 1, so λ(N′) = 0. Kahane and Salem asked whether N′ supports any probability Borel measure μ (i.e., μ(N′) = 1) with μ̂(n) → 0. (Measures with μ̂(n) → 0 are somehow considered “thick” - recall that λ̂(n) → 0, since in fact λ̂(n) = 0 if n ≠ 0. It can be also shown that they are continuous, i.e., give every singleton measure 0. In fact a theorem of Wiener asserts that

∑_{x∈𝕋} μ(\{x\})^{2} = lim_{n→∞} \frac{1}{2N+1} ∑_{−N}^{N} |μ̂(n)|^{2}.) Lyons, in 1983, answered this affirmatively, by using

delicate analytical tools. However, a totally different proof can be based on 9.2 by noticing that

9.4. Proposition. N′ is comeager.

Proof. We show that N is meager. Note that

N ⊆ ⋃_{n≥1} F_{n},

where

F_{n} = ⋂_{k≥n} F′_{k},

with

F′_{k} =\{2π ∑_{l=1}^{∞} x_{l}2^{−l} : x_{l} = 0, 1 and |\frac{x_{1} + ⋯ x_{k}}{k} − \frac{1}{2}| ≤ \frac{1}{4}\}.

Now F′_{k} is closed being the continuous image of the closed (thus compact) subset of 2^{ℕ}

\{(x_{1}, x_{2}, ⋯) ∈ 2^{ℕ} : |\frac{x_{1} + ⋯ + x_{k}}{k} − \frac{1}{2}| ≤ \frac{1}{4}\},

by the continuous map

(x_{1}, x_{2}, ⋯) ↦ 2π ∑_{l=1}^{∞} x_{l}2^{−l}.

So F_{n} is closed. 9.5. Exercise. F_{n} contains no open interval.

So each F_{n} is nowhere dense and N is meager. ⊣

§10. Review of duality in Banach spaces.

Good references for the basic results of functional analysis and measure theory that we will use in the sequel are Rudin [1973], [1987].

Let X be a Banach space, i.e., a complete normed linear space, over the complex numbers. Denote by X^{∗} the dual space of X, i.e., the Banach space of all (bounded or, equivalently, continuous) linear functionals x^{∗} : X → ℂ with the norm

\begin{aligned}||x^{∗}|| & = sup\{\frac{||x^{∗}(x)||}{||x||} : x ∈ X, x ≠ 0\} \\ & = sup\{||x^{∗}(x)|| : x ∈ X, ||x|| ≤ 1\}. \end{aligned}

It is often convenient to write

x^{∗}(x) = ⟨x, x^{∗}⟩ = ⟨x^{∗}, x⟩.

A set A ⊆ X is called convex if for every x, y ∈ A and t ∈ [0, 1], tx + (1 − t)y ∈ A.

There is a fundamental collection of results, collectively known as the Hahn-Banach Theorems, which assert the existence of appropriate linear functionals. We will need here the following separation form of Hahn-Banach.

10.1. Theorem (Hahn-Banach). Let X be a Banach space and let A, B ⊆ X be convex, nonempty, A ∩ B = ∅, with A compact and B closed. Then there is x^{∗} ∈ X^{∗}, and α, β ∈ ℝ with

Re ⟨x^{∗}, x⟩ < α < β < Re ⟨x^{∗}, y⟩,

for x ∈ A, y ∈ B.

We now define the weak-topology of X as follows: It is the smallest topology on X for which the maps

x ↦ ⟨x^{∗}, x⟩, for x^{∗} ∈ X^{∗},

are continuous. It is thus contained in the usual or norm-topology of X, i.e., the one induced by its norm, and unless X is finite-dimensional, it is properly contained. By definition, a subbasis of this topology consists of all sets of the form

{x : ⟨x^{∗}, x⟩ ∈ U}

for x^{∗} ∈ X, U ⊆ ℂ open.

10.2. Exercise. Show that X with the weak-topology is a topological vector space (i.e., scalar multiplication and vector addition are continuous) and that a local basis at 0 is given by the sets U_{x_{1}^{∗},···,x_{n}^{∗},ϵ} = {x ∈ X : |⟨x_{1}^{∗}, x⟩|, ⋯, |⟨x_{n}^{∗}, x⟩| < ϵ}, for x_{1}^{∗}, ⋯, x_{n}^{∗} ∈ X^{∗}.

We will denote by \overline{A}^{w} the closure of A ⊆ X in the weak-topology. Clearly \overline{A} (= the closure of A in the norm-topology) ⊆ \overline{A}^{w}, since there are more norm closed sets than weak closed sets. However, for convex sets these closures coincide.

10.3. Theorem (Mazur). Let X be a Banach space. For every convex set A ⊆ X, \overline{A} = \overline{A}^{w}.

Proof. It is enough to show that \overline{A}^{w} ⊆ \overline{A}. Let x_{0} ∉ \overline{A} in order to show that x_{0} ∉ \overline{A}^{w}. By Hahn-Banach applied to \{x_{0}\}, \overline{A} (which is easily convex) there is x^{∗} ∈ X^{∗} and α ∈ ℝ with

Re ⟨x^{∗}, x_{0}⟩ < α < Re ⟨x^{∗}, y⟩

for y ∈ \overline{A}. Then \{x : Re⟨x^{∗}, x⟩ < α\} is a weak-nbhd of x_{0} which is disjoint from A, so x_{0} ∉ \overline{A}^{w}. ⊣

On the dual Banach space X^{∗} we of course have its weak-topology but we can also consider an even weaker (fewer open sets) topology called the weak^{∗}-topology or w^{∗}-topology. This is the smallest topology for which the functions

x^{∗} ↦ ⟨x, x^{∗}⟩

for x ∈ X are continuous. Since every x ∈ X gives rise to a linear functional x^{∗∗} ∈ X^{∗∗}, defined by

⟨x^{∗∗}, x^{∗}⟩ = ⟨x, x^{∗}⟩ for x^{∗} ∈ X^{∗},

this shows that the weak^{∗}-topology of X^{∗} is contained in its weak-topology (in general properly).

10.4. Exercise. Verify that X^{∗} with the weak^{∗}-topology is a topological vector space and that a local basis at 0 is given by the sets V_{x_{1},···,x_{n},ϵ} = {x^{∗} ∈ X : |⟨x_{1}, x^{∗}⟩|, ⋯, |⟨x_{n}, x^{∗}⟩| < ϵ} for x_{1}, ⋯, x_{n} ∈ X.

The crucial property of the weak^{∗}-topology is given by the following.

10.5. Theorem (Banach-Alaoglu). Let X be a Banach space and consider a closed ball

B_{r}(X^{∗}) = {x^{∗} ∈ X^{∗} : ||x^{∗}|| ≤ r}

of X^{∗}. Then B_{r}(X^{∗}) is weak^{∗}-compact, i.e., compact in the weak^{∗}-topology.

Proof. It is enough to consider r = 1. Consider the product space ∏_{x∈X} Δ_{||x||}, where Δ_{r} = {z ∈ ℂ : |z| ≤ r}. This is compact, by Tychonoff’s Theorem. Note that B_{1}(X^{∗}) ⊆ ∏_{x∈X} Δ_{||x||}, since |x^{∗}(x)| ≤ ||x^{∗}|| · ||x|| ≤ ||x||. Moreover, the relative topology that B_{1}(X^{∗}) inherits from ∏_{x∈X} Δ_{||x||} is exactly the weak^{∗}-topology. So it is enough to show that B_{1}(X^{∗}) is closed in ∏_{x∈X} Δ_{||x||}. Clearly B_{1}(X^{∗}) = {f ∈ ∏_{x∈X} Δ_{||x||} : f is linear} = ⋂_{α,β∈ℂ} ⋂_{x,y∈X}{f : f(αx + βy) = αf(x) + βf(y)}. Since the map f ↦ f(x) from ∏_{x∈X} Δ_{||x||} into Δ_{||x||} is continuous, B_{1}(X^{∗}) is the intersection of closed sets, thus closed. ⊣

10.6. Exercise. Show that when X is separable (i.e., has a countable dense set), then B_{r}(X^{∗}) with the weak^{∗}-topology is also metrizable with compatible metric

d(x^{∗}, y^{∗}) = ∑ 2^{−n}|⟨x_{n}, x^{∗}⟩ − ⟨y_{n}, y^{∗}⟩|,

where {x_{n}} is dense in the unit ball B_{1}(X) of X.

Remark. Every element x of X can be identified with the element x^{∗∗} of X^{∗∗} = (X^{∗})^{∗} given by

⟨x^{∗∗}, x^{∗}⟩ = ⟨x, x^{∗}⟩.

So we can view X as a subset of X^{∗∗} (it is in fact a closed subspace of X^{∗}). It is obvious from the definition that the weak-topology on X is exactly the same as the weak^{∗}-topology of X, when it is viewed as a subset of X^{∗∗}(= (X^{∗})^{∗}).

We will now discuss some important, and crucial for our purposes, examples.

A) First, we denote by c_{0} = c_{0}(ℤ) the Banach space of all sequences (x_{n})_{n∈ℤ}, x_{n} ∈ ℂ such that x_{n} → 0 as |n| → ∞, equipped with the sup norm

||(x_{n})||_{∞} = sup_{n∈ℤ} |x_{n}|.

We denote by ℓ^{1} = ℓ^{1}(ℤ) the Banach space of all sequences (x_{n})_{n∈ℤ} such that ∑ |x_{n}| < ∞ with the norm

||(x_{n})||_{1} = ∑_{n∈ℤ} |x_{n}|.

Finally, we denote by ℓ^{∞} = ℓ^{∞}(ℤ) the Banach space of all bounded sequences (x_{n})_{n∈ℤ} (i.e., sup_{n∈ℤ} |x_{n}| < ∞) with the sup norm

||(x_{n})||_{∞} = sup_{n∈ℤ} |x_{n}|

10.7. Exercise. Show that c_{0} is a closed (linear) subspace of ℓ^{∞}. Show that c_{0}, ℓ^{1} are separable, but ℓ^{∞} is not.

We will now identify c_{0}^{∗}, (ℓ^{1})^{∗}. Let Λ ∈ c_{0}^{∗} and put λ_{n} = Λ(e_{−n}), where

e_{n}(i) = \{ \begin{matrix}0, & if i ≠ n \\ 1, & if i = n \end{matrix} , for n, i ∈ ℤ.

(The use of Λ(e_{−n}) instead of Λ(e_{n}) is for technical convenience and assures consistency with the definition of Fourier coefficients μ̂(n) = ∫ e^{−int}dμ(t) later on.)

10.8. Exercise. For any (x_{n}) ∈ c_{0},

Λ((x_{n})) = ∑ λ_{n}x_{−n},

∑ |λ_{n}| < ∞ and ||Λ|| = ||(λ_{n})||_{1}.

From this it immediately follows that the bijection

Λ ↔ (λ_{n})

is a Banach space isomorphism between ℓ^{1} and c_{0}^{∗}, so we simply identify c_{0}^{∗} with ℓ^{1},

c_{0}^{∗} = ℓ^{1},

and view every element of ℓ^{1}, (λ_{n}), as operating on an element (x_{n}) of c_{0}, by

⟨(λ_{n}), (x_{n})⟩ = ⟨(x_{n}), (λ_{n})⟩ = ∑ λ_{n}x_{−n}.

Now consider (ℓ^{1})^{∗} and put, as before, λ_{n} = Λ(e_{−n}).

10.9. Exercise. For any (x_{n}) ∈ ℓ^{1},

Λ((x_{n})) = ∑ λ_{n}x_{−n},

sup |λ_{n}| < ∞ and ||Λ|| = sup |λ_{n}|.

So, as before, we can identify (ℓ^{1})^{∗} with ℓ^{∞},
(ℓ^{1})^{∗} = ℓ^{∞}

and view any element (λ_{n}) ∈ ℓ^{∞} as operating on (x_{n}) ∈ ℓ^{1} by

⟨(λ_{n}), (x_{n})⟩ = ⟨(x_{n}), (λ_{n})⟩ = ∑ λ_{n}x_{−n}.

Note that as c_{0} ⊆ ℓ^{∞} there are two meanings of ⟨(λ_{n}), (x_{n})⟩ for (λ_{n}) ∈ c_{0}, (x_{n}) ∈ ℓ^{1}, but both give, of course, the same value.

10.10. Exercise. Let (x_{n}) be a bounded sequence of elements of c_{0}, i.e., sup ||x_{n}||_{∞} < ∞. Let x ∈ c_{0}. Show that x_{n} → x in the weak-topology iff x_{n}(i) → x(i), ∀i ∈ ℤ. Show that if (x_{n}) is a bounded sequence in ℓ^{1}, and x ∈ ℓ^{1}, then x_{n} → x in the weak^{∗}-topology iff x_{n}(i) → x(i), ∀i ∈ ℤ. Finally, if (x_{n}) is a bounded sequence in ℓ^{∞} and x ∈ ℓ^{∞}, then x_{n} → x in the weak^{∗}-topology (of ℓ^{∞} = (ℓ^{1})^{∗}) iff x_{n}(i) → x(i), ∀i ∈ ℤ.

B) Now consider the Banach space C(𝕋) of all continuous (complex) functions on 𝕋 with the sup norm

||f||_{∞} = sup_{t∈T} |f(t)|.

10.11. Exercise. The trigonometric polynomials with coefficients in ℚ + iℚ are dense in C(𝕋), so C(𝕋) is separable.

The dual of C(𝕋) is identified by the Riesz Representation Theorem.

First recall that a positive Borel measure on 𝕋 is a function μ : {Borel subsets of 𝕋} → [0, ∞] such that (i) μ(∅) = 0 and (ii) μ(⋃_{i∈ℕ} A_{i}) = ∑ μ(A_{i}), if A_{i} are pairwise disjoint Borel sets. It is finite, if μ(𝕋) < ∞, and a probability measure, if μ(𝕋) = 1. A complex Borel measure is a map μ : {Borel subsets of 𝕋} → ℂ which satisfies the above properties (i) and (ii). It turns out that every complex Borel measure μ can be written as μ = μ_{1} − μ_{2} + i(μ_{3} − μ_{4}), where μ_{1}, μ_{2}, μ_{3}, μ_{4} are finite positive Borel measures. It follows that there is C < ∞ such that

∑_{i=1}^{∞} |μ(E_{i})| ≤ C

for any Borel partition {E_{i}}_{i=1}^{∞} of 𝕋. Put

||μ||_{M} = sup ∑_{i=1}^{∞} |μ(E_{i})|,

where the sup is over all these partitions. Notice that ||μ||_{M} = μ(𝕋), if μ is a finite positive Borel measure.

Denote by M(𝕋) the vector space of complex Borel measures on 𝕋 (where we put (αμ + λν)(E) = αμ(E) + λν(E)). Then M(𝕋) with the norm ||μ||_{M} is a Banach space.

10.12. Exercise. For each x ∈ 𝕋, let δ_{x} be the Dirac measure at x, i.e., δ_{x}(E) =

\{ \begin{matrix}1, & if x ∈ E \\ 0, & if x ∉ E \end{matrix}. Show that if x ≠ y, ||δ_{x} − δ_{y}|| = 2. Conclude that M(𝕋) is not separable.

The Riesz Representation Theorem identifies C(𝕋)^{∗} with M(𝕋). Let’s explain this more carefully. First, one can define for each f ∈ C(𝕋) the integral ∫ fdμ and show the properties

(i) ∫ (αf + βg)dμ = α ∫ fdμ + β ∫ gdμ,
(ii) | ∫ fdμ| ≤ ||f||_{∞}||μ||_{M},

so that μ gives rise to the element

f ↦ ∫ fdμ

of C(𝕋)^{∗}. The Riesz Representation Theorem asserts that these are all the elements of C(𝕋)^{∗}. More precisely, to each Λ ∈ C(𝕋)^{∗} one can associate a unique complex Borel measure μ on 𝕋 such that

Λ(f) = ∫ fdμ, for f ∈ C(𝕋).

Moreover Λ ↔ μ is a Banach space isomorphism between C(𝕋)^{∗} and M(𝕋).

So we identify C(𝕋)^{∗} and M(𝕋),
C(𝕋)^{∗} = M(𝕋).

Each μ ∈ M(𝕋) operates on f ∈ C(𝕋) by

⟨f, μ⟩ = ⟨μ, f⟩ = ∫ fdμ.

Now denote by

P (𝕋)

the set of all probability Borel measures on 𝕋. Since

μ ∈ P (𝕋) ⇒ ||μ|| = μ(T) = 1,

it follows that P (𝕋) ⊆ B_{1}(M(𝕋)).

Another part of the Riesz Representation Theorem asserts that in the correspondence Λ ↔ μ, μ is a positive measure iff Λ is positive, i.e., Λ(f) ≥ 0 for f ≥ 0. Thus μ ∈ C(𝕋) is positive, i.e., μ(E) ≥ 0 for Borel E, iff ∫ fdμ ≥ 0 for any f ∈ C(𝕋), with f ≥ 0. (This can be also proved directly by approximation arguments.) Thus P (𝕋) consists exactly of all members of B_{1}(M(𝕋)) which satisfy

∫ 1dμ = 1, ∀f ∈ C(𝕋) (f ≥ 0 ⇒ ∫ fdμ ≥ 0).

It follows that P (𝕋) is a closed subset of B_{1}(M(𝕋)) when the latter is equipped with the weak^{∗}-topology, so it is compact metrizable in this topology. To summarize:

The space P (𝕋) of probability Borel measures on 𝕋 with the weak^{∗}-topology, i.e., the smallest topology for which the maps

μ ↦ ∫ fdμ, f ∈ C(𝕋)

are continuous, is compact metrizable.

Notice again that the sets of the form

V_{f_{1},···,f_{n},ϵ} = {ν :| ∫ f_{i}dμ − ∫ f_{i}dν| < ϵ, i = 1, ⋯, n},

where ϵ > 0 and f_{1}, ⋯, f_{n} ∈ C(𝕋), form a nbhd basis of μ ∈ P (𝕋) in the weak^{∗}-topology.

A finite support probability measure is a measure of the form μ = ∑_{i=1}^{n} α_{i}δ_{x_{i}}, where δ_{x} is the Dirac measure at x, and α_{i} ≥ 0, ∑_{i=1}^{n} α_{i} = 1. We say that μ is supported by {x_{1}, ⋯, x_{n}}.

10.13. Proposition. Let D ⊆ 𝕋 be a dense set in 𝕋. Then the set of finite support probability measures supported by D is dense in P (𝕋) with the weak^{∗}-topology.

Proof. Fix μ ∈ P (𝕋) and an open nbhd {ν : | ∫ f_{i}dμ − ∫ f_{i}dν| < ϵ, i = 1, ⋯, n} (f_{1}, ⋯, f_{n} ∈ C(𝕋), ϵ > 0) of μ in the weak^{∗}-topology. We want to find a finite support probability measure ν supported by D which belongs in this nbhd. Since the functions f_{1}, ⋯, f_{n} are uniformly continuous, we can find δ > 0 such that

|x − y| < δ ⇒ |f_{i}(x) − f_{i}(y)| ≤ ϵ_{1} < ϵ,

for any x, y ∈ 𝕋 and i = 1, ⋯, n. So there is a finite partition I_{1}, ⋯, I_{k} of 𝕋 into half-open intervals such that

x, y ∈ I_{j} ⇒ |f_{i}(x) − f_{i}(y)| ≤ ϵ_{1}, j ≤ k, i ≤ n.

Choose x_{i} ∈ I_{i} ∩ D, let α_{i} = μ(I_{i}) and put ν = ∑ α_{i}δ_{x_{i}}. Then ν is supported by D and for f ∈ {f_{1}, ⋯, f_{n}}

| ∫ fdμ − ∫ fdν|

= | ∫ fdμ − ∑ α_{i}f(x_{i})| (since ∫ fd(δ_{x}) = f(x))

= | ∑_{i=1}^{k}(∫_{I_{i}} fdμ − ∫_{I_{i}} f(x_{i})dμ)| (since μ(I_{i}) = α_{i})

= | ∑_{i=1}^{k} ∫_{I_{i}}(f(x) − f(x_{i}))dμ|

≤ ∑_{i=1}^{k} ∫_{I_{i}} |f(x) − f(x_{i})|dμ

≤ ∑_{i=1}^{k} ∫_{I_{i}} ϵ_{1}dμ

= ϵ_{1} ∫ dμ = ϵ_{1} < ϵ.
⊣

We can generalize this as follows. Fix a closed set E ⊆ 𝕋 and let

P (E) = {μ ∈ P (𝕋) : μ(E) = 1}.

We claim that P (E) is closed in the weak^{∗}-topology of P (𝕋), thus also compact. To see this, let I_{1}, I_{2}, ⋯ enumerate a sequence of open intervals whose union is the complement of E and note that μ ∈ P (E) ⇔ ∀n(μ(I_{n}) = 0). If C_{n} = {f ∈ C(𝕋): f vanishes outside I_{n}}, then clearly χ_{I_{n}} is the pointwise limit of a sequence f_{i} ≤ 1 from C_{n}, so by the Lebesgue Dominated Convergence Theorem, μ(I_{n}) = 0 ⇔ ∫ χ_{I_{n}}dμ = 0 ⇔ ∀f ∈ C_{n}(∫ fdμ = 0), so P (E) is an intersection of closed sets, thus is closed.

10.14. Exercise. Let E ⊆ 𝕋 be closed and let D ⊆ E be dense in E. Then the probability measures supported by finite subsets of D are dense in P (E) with the weak^{∗}-topology.

There is a very interesting connection between M(𝕋) and ℓ^{∞}. To each μ ∈ M(𝕋) associate its Fourier coefficients

μ̂(n) = ∫ e^{−int}dμ.

As |μ̂(n)| ≤ | ∫ e^{−int}dμ| ≤ ||μ||_{M}, we see that μ̂ ∈ ℓ^{∞}. Next note that if μ̂ = ν̂ for two complex measures μ, ν, then ∫ fdμ = ∫ fdν for any trigonometric polynomial f and thus, as these are dense in C(𝕋), ∫ fdμ = ∫ fdν for f ∈ C(𝕋), i.e., μ = ν. So μ ↦ μ̂ is an injection from M(𝕋) into ℓ^{∞}. Denote by { \atop M(𝕋)} its image, and more generally for A ⊆ M(𝕋) let Â = \{μ̂ : μ ∈ A\}. Then { \atop P (𝕋)} ⊆ B_{1}(ℓ^{∞}). We claim that actually μ ↦ μ̂ is a homeomorphism of P (𝕋) with the weak^{∗}-topology and { \atop P (𝕋)} with the weak^{∗}-topology (of B_{1}(ℓ^{∞}), where ℓ^{∞} = (ℓ^{1})^{∗}). Since P (𝕋) is compact, it is enough to show that μ ↦ μ̂ is continuous. So let μ_{n} → μ in the weak^{∗}-topology of P (𝕋). Then ∫ fdμ_{n} → fdμ for any f ∈ C(𝕋), so in particular μ̂_{n}(i) → μ̂(i), thus μ̂_{n} → μ̂ in the weak^{∗}-topology of { \atop P (𝕋)}.

So we can identify, for all practical purposes, P (𝕋) and { \atop P (𝕋)} by identifying μ with μ̂, so we often view P (𝕋) as a subset of ℓ^{∞}.

§11. Rajchman measures and the proof of the Debs-Saint Raymond Theorem.

We say that μ ∈ P (𝕋) is a Rajchman measure if μ̂(n) → 0, as |n| → ∞. (By the way, Rajchman was Zygmund’s teacher.) Denote their class by R. Then Theorem 9.2 says that for every non-meager A ⊆ 𝕋 with the BP, ∃μ ∈ R(μ(A) = 1).

We will now give the promised proof of 9.2, due to Kechris-Louveau, which is based on the Baire category method.

The streamlined presentation below is due to Lyons.

First, since A ⊆ 𝕋 has the property of Baire, there is an open set U ⊆ 𝕋, with AΔU meager. As A is not meager, U ≠ ∅, so there is a closed interval I = [a, b], a ≠ b, and a sequence U_{n} ⊆ I of open sets, dense in I, with ⋂_{n} U_{n} ⊆ A.

Let for E ⊆ 𝕋
R(E) = {μ ∈ R : μ(E) = 1}.

We want to show that R(A) ≠ ∅. If we try to apply the category method in P (𝕋) with the weak^{∗}-topology, we run into a problem since R(A) ⊆ R and R is unfortunately meager in P (𝕋) with the weak^{∗}-topology. (See 11.1 below). The trick is to work instead with R̂ ⊆ c_{0} and the norm-topology.

Claim 1. R̂ is a norm-closed subset of c_{0}. More generally, if E ⊆ 𝕋 is closed, { \atop R(E)} is a norm-closed subset of c_{0}.

Proof. Take μ_{n} ∈ R(E) and assume μ̂_{n} → x ∈ c_{0} in norm, i.e., ||μ̂_{n} − x||_{∞} → 0. Now this implies immediately that μ̂_{n}(i) → x(i), ∀i ∈ ℤ. But recall that P (𝕋) is compact, so there is a subsequence n_{0} < n_{1} < ⋯ with μ_{n_{j}} → μ ∈ P (𝕋) (for some μ ∈ P (𝕋)), with respect to the weak^{∗}-topology, so μ̂_{n_{j}}(i) → μ̂(i), ∀i ∈ ℤ, i.e., μ̂(i) = x(i), so x = μ̂ ∈ R̂. We now want to show that μ ∈ P (E), i.e., μ(E) = 1, so that μ̂ ∈ { \atop R(E)}. But this is clear as P (E) is closed in the weak^{∗}-topology of P (𝕋). ⊣

So { \atop R(I)} is in particular a complete metric space and we can apply the Baire Category Theorem to it. It will be clearly enough to show that each { \atop R(U_{n})} is dense, G_{δ} in { \atop R(I)} in the norm-topology of { \atop R(I)}. Because then ⋂_{n} { \atop R(U_{n})} is dense, G_{δ} in { \atop R(I)} and so nonempty, i.e., there is μ ∈ R such that for each n, μ(U_{n}) = 1, thus μ(⋂_{n} U_{n}) = 1 and since ⋂_{n} U_{n} ⊆ A, μ(A) = 1, and the proof is complete.

Thus our final claim is:

Claim 2. If I is a closed non-trivial interval in 𝕋 and U ⊆ I is open and dense in I, then { \atop R(U)} is dense G_{δ} in { \atop R(I)}, in the norm-topology of { \atop R(I)}.

Proof. First we check that { \atop R(U)} is G_{δ} in { \atop R(I)} in the norm-topology. We have

{ \atop R(U)} = ⋂_{n≥1} ⋃_{{f∈C(𝕋) \atop 0≤f≤χ_{U}}} \{μ̂ ∈ { \atop R(I)} : ∫ fdμ > 1 − \frac{1}{n}\}.

(To see this recall that U is a disjoint union of open intervals.) It is then enough to check that for each f ∈ C(𝕋),

\{μ̂ ∈ { \atop R(I)} : ∫ fdμ > 1 − \frac{1}{n}\}

is open in { \atop R(I)} in the norm-topology of c_{0}. In fact, we can easily see that it is open in { \atop R(I)} in the weak-topology of c_{0}. This is because

μ ↦ ∫ fdμ

is continuous in the weak^{∗}-topology of R(I) and thus in the weak^{∗}-topology of { \atop R(I)}. But since { \atop R(I)} ⊆ c_{0}, the weak^{∗}-topology of { \atop R(I)} is the same as the weak-topology of { \atop R(I)}.

It remains to prove that { \atop R(U)} is dense in { \atop R(I)} for the norm-topology. Since clearly { \atop R(U)} is a convex subset of c_{0}, it is enough, by Mazur’s Theorem 10.3, to show that { \atop R(U)} is weakly dense in { \atop R(I)}. But again, as { \atop R(I)} ⊆ c_{0}, this is the same thing as saying that { \atop R(U)} is weak^{∗}-dense in { \atop R(I)}, where we now view these as subsets of ℓ^{∞}. But this again means the same thing as R(U) being weak^{∗}-dense in R(I), where we work in P (𝕋) now.

We will in fact show that \overline{R(U)}^{w^{∗}} (= the weak^{∗}-closure of R(U) in P (𝕋)) = P (I), which of course completes the proof. Since the probability measures with finite support contained in U are dense in P (I), and R(U) is convex, it is enough to show that every Dirac measure δ_{x}, with x ∈ U, is the limit of a sequence in R(U) in the weak^{∗}-topology. But this is easy. Let I_{n} ⊆ U be a decreasing sequence of open intervals with λ(I_{n}) < \frac{1}{n} and \{x\} = ⋂_{n} I_{n}. Let μ_{n} = (λ|I_{n})/λ(I_{n}). Then, by direct calculation, μ̂_{n}(i) → 0, as |i| → ∞, so μ_{n} ∈ R(U). Now μ_{n} → δ_{x} in the weak^{∗}-topology. This is because for any f ∈ C(𝕋),

\begin{aligned}∫ fdμ_{n} − ∫ fd(δ_{x}) & = \frac{1}{λ(I_{n})} ∫_{I_{n}} f(t)dt − f(x) \\ & = \frac{1}{λ(I_{n})} ∫_{I_{n}}(f(t) − f(x))dt, \end{aligned}

so that if ϵ is given and n is large enough, so that |f(t) − f(x)| < ϵ for t ∈ I_{n}, we have

| ∫ fdμ_{n} − ∫ fd(δ_{x})| < ϵ.

This completes the proof.

11.1. Exercise. Show that R is meager in P (𝕋) with the weak^{∗}-topology.

Although R is meager, it still has an interesting largeness property. Note that for a set A in a complete metric space, if A has the BP, then A is comeager iff for each open non-∅ set U ⊆ X and each sequence of open dense in U sets, U_{n}, we have A ∩ (⋂_{n})U_{n} ≠ ∅. Now R, although meager, is convexly comeager, in the following sense.

11.2. Theorem (Kechris-Louveau). For every non-empty open set U ⊆ P (𝕋) (in the weak^{∗}-topology) and any sequence U_{n} of open dense in U convex sets, we have R ∩ (⋂_{n} U_{n}) ≠ ∅.

I will omit the proof (see Kechris-Louveau [1989], VIII. 3.6]).

11.3. Exercise. Use this to give another proof of the Debs-Saint Raymond Theorem.

§12. Paying a debt: Proof of 7.6.

To bring this chapter into conclusion I will give the proof that (i) ⇒ (ii) in 7.6, which we omitted earlier. The proof will be based on a classical result of Riemann, known as the localization principle.

12.1. Riemann localization principle. Let S ∼ ∑ c_{n}e^{inx} be a trigonometric series with c_{n} → 0. If the Riemann function F_{S} is linear in some open interval (a, b), then ∑ c_{n}e^{inx} = 0 on (a, b) (and uniformly on closed subintervals).

Note that the hypothesis is equivalent (by Schwartz’s Lemma 3.3) to saying that D^{2}F_{S}(x) = 0, ∀x ∈ (a, b). Recall also Riemann’s First Lemma 2.2 which implies that if ∑ c_{n}e^{inx} = 0, then D^{2}F_{S}(x) = 0. So 12.1 asserts a converse, but only under the hypothesis that D^{2}F_{S}(x) vanishes in a whole interval.

Before proving 12.1, let me first show how it can be used to prove (i) ⇒ (ii) in 7.6. To start with, note the following fact.

12.2. Exercise. Let f ∈ C(𝕋) be such that ∑ |f̂(n)| < ∞. Then f(x) = ∑ f̂(n)e^{inx} uniformly on x.

Now fix a ∈ ℝ, 0 < h < π and let ψ_{a,h} be the 2π-periodic function defined in the period [a − π, a + π] as follows: ψ_{a,h}(a) = 2π/h, ψ_{a,h}(x) = 0 off [a − h, a + h], and ψ_{a,h} in linear in [a − h, a], [a, a + h]. Then one can easily check that the Fourier series of ψ_{a,h} is

S(ψ_{a,h}) ∼ ∑_{n=−∞}^{∞} e^{−ina}(\frac{sin(nh/2)}{nh/2})^{2}e^{inx},

so that ∑ |ψ̂_{a,h}(n)| < ∞ and

ψ_{a,h}(x) = ∑_{n=−∞}^{∞} e^{−ina}(\frac{sin(nh/2)}{nh/2})^{2}e^{inx}.

Let us next note another fact.

12.3. Exercise. Let f ∈ C(𝕋) be such that ∑ |f̂(n)| < ∞. Then for any μ ∈ M(𝕋),

∫ fdμ = ∑ μ̂(n)f̂(−n).

So if E ⊆ 𝕋 is a closed set and μ(E) = 1, we have for any a ∉ E and h small enough that

\begin{matrix}0 = ∫ ψ_{a,h}dμ = & ∑_{n=−∞}^{∞} μ̂(n)ψ̂_{a,h}(−n) \\ = & ∑_{n=−∞}^{∞} μ̂(n)(\frac{sin(nh/2)}{nh/2})^{2}e^{ina} \\ = & \frac{Δ^{2}F_{S}(a, h)}{h^{2}} \end{matrix}

where S ∼ ∑ μ̂(n)e^{inx}. So F_{S} is linear on each open interval disjoint from E and thus, by the Riemann Localization Principle, ∑ μ̂(n)e^{inx} = 0 off E.

So it only remains to prove 12.1.

§13. The Rajchman Multiplication Theory.

We will first develop a theory, due to Rajchman, concerning the formal multiplication of trigonometric series by “nice” functions. Beyond being useful in proving the Riemann localization principle, it has many other applications, some of which we will see later on.

Let S ∼ ∑ c_{n}e^{inx} have bounded coefficients |c_{n}| ≤ M < ∞. Let f ∈ C(𝕋) have absolutely convergent Fourier coefficients ∑ |f̂(n)| < ∞, so that f(x) = ∑ f̂(n)e^{inx} uniformly. Define the formal product S(f) · S (another trigonometric series) by

S(f) · S ∼ ∑ C_{n}e^{inx},

where C_{n} = ∑_{k} c_{k}f̂(n − k). Clearly ∑_{k} c_{k}f̂(n − k) is convergent and |C_{n}| ≤ sup_{k} |c_{k}| ·

∑_{n∈ℤ} |f̂(n)|.

13.1. Exercise. c_{n} → 0 ⇒ C_{n} → 0. 13.2. Lemma. If ∑_{ℓ=0}^{∞} ∑_{|n|≥ℓ} |f̂(n)| < ∞ (e.g., if f̂(n) = O(\frac{1}{|n|^{3}} )) and c_{n} → 0, then

∑_{−N}^{N} C_{n}e^{inx} − f(x) ∑_{−N}^{N} c_{n}e^{inx} → 0 uniformly on x (i.e., ∑_{−∞}^{∞}(C_{n}e^{inx} − f(x)c_{n}e^{inx}) = 0,

uniformly on x). Proof. First we prove that if f(x) = 0 for x ∈ P ⊆ 𝕋, then ∑_{−N}^{N} C_{n}e^{inx} = 0 uniformly on x ∈ P. To see this note that

∑_{−N}^{N} C_{n}e^{inx} = ∑_{n=−N}^{N}( ∑_{k=−∞}^{∞} c_{k}e^{ikx}f̂(n − k)e^{i(n−k)x})

= ∑_{k=−∞}^{∞} c_{k}e^{ikx}( ∑_{n=−N}^{N} f̂(n − k)e^{i(n−k)x})

= ∑_{k=−∞}^{∞} c_{k}e^{ikx}( ∑_{m=−N−k}^{N−k} f̂(m)e^{imx})

= I_{1} + I_{2},

where

\begin{aligned}I_{1} & = ∑_{|k|≤ \frac{1}{2} N} ⋯, \\ I_{2} & = ∑_{|k|> \frac{1}{2} N} · · · . \end{aligned}

Since ∑_{m=−∞}^{∞} f̂(m)e^{imx} = 0, ∑_{m=−N−k}^{N−k} f̂(m)e^{imx} = −( ∑_{m=−∞}^{−N−k−1} f̂(m)e^{imx} + ∑_{m=N−k+1}^{∞}

f̂(m)e^{imx}), so

|I_{1}| ≤ sup_{k} |c_{k}| · 2( ∑_{ℓ≥N/2} ∑_{|m|≥ℓ} |f̂(m)|) → 0

as N → ∞, uniformly on x, since ∑_{ℓ=0}^{∞} ∑_{|m|≥ℓ} |f̂(m)| < ∞. Also

\begin{aligned}|I_{2}| & ≤ sup_{|k|> \frac{N}{2}} |c_{k}| · ∑_{k∈ℤ}( ∑_{m=−∞}^{−N−k−1} |f̂(m)| + ∑_{m=−∞}^{N−k+1} |f̂(m)|) \\ & ≤ sup_{|k|> \frac{N}{2}} |c_{k}| · 3( ∑_{ℓ=0}^{∞} ∑_{|m|≥ℓ} |f̂|(m)|) → 0 \end{aligned}

uniformly on x (consider cases as k ≤ −N, k ∈ (−N, N), k ≥ N.)

Note that this argument applies as well to any series of the form

∑ f̃(n, x)e^{inx}

provided that ∑_{|n|≥ℓ} |f̃(n, x)| ≤ M_{ℓ}, with ∑_{ℓ≥0} M_{ℓ} < ∞. In this case C_{n} = ∑_{k} c_{k}f̃(n − k, x) and the hypothesis f(x) = 0 is replaced by ∑ f̃(n, x)e^{inx} = 0.

Now consider the general situation. We have

\begin{aligned}& ∑_{−N}^{N} C_{n}e^{inx} − f(x) ∑_{−N}^{N} c_{n}e^{inx} \\ = & ∑_{−N}^{N}(C_{n} − f(x)c_{n})e^{inx}. \end{aligned}

But

\begin{aligned}C_{n} − f(x)c_{n} & = ∑_{−∞}^{∞} c_{k}f̂(n − k) − f(x)c_{n} \\ & = ∑_{−∞}^{∞} c_{k}f̃(n − k, x), \end{aligned}

where f̃(m, x) = \{ \begin{matrix}f̂(m), & if m ≠ 0 \\ f̂(0) − f(x), & if m = 0 \end{matrix}. Now ∑ f̃(n, x)e^{inx} = ∑ f̂(n)e^{inx} − f(x) = 0

for all x. Also if M_{k} = ∑_{|n|≥k} |f̂(n)|, for k > 0, and M_{0} = ∑ |f̂(n)| + ||f||_{∞}, we have ∑_{|n|≥k} |f̃(n, x)| = M_{k} if k > 0 and ∑_{n} |f̃(n, x)| = M_{0}. Since ∑_{k≥0} M_{k} < ∞, it follows from the preceding remarks that

∑_{−N}^{N}(C_{n} − f(x)c_{n})e^{inx} → 0

uniformly on x, i.e.,

∑_{−N}^{N} C_{n}e^{inx} − f(x) ∑_{−N}^{N} c_{n}e^{inx} → 0

uniformly on x. ⊣ We are now ready to prove the Riemann Localization Principle. Let S ∼ ∑ c_{n}e^{inx} be a trigonometric series with c_{n} → 0 and assume F_{S} is linear on

(a, b). Let [c, d] ⊆ (a, b). Take a nice function f ∈ C(𝕋), say with continuous derivatives of all orders, such that f = 1 on [c, d] and 0 off (a, b). Integration by parts shows that f̂(n) = O(\frac{1}{|n|^{k}} ) for all k ≥ 0, so in particular ∑_{k} ∑_{|n|≥k} |f̂(n)| < ∞. So

∑_{−∞}^{∞}(C_{n}e^{inx} − f(x)c_{n}e^{inx}) = 0,

uniformly on x. Then by 2.3

lim_{h→0} ∑_{−∞}^{∞}(C_{n}e^{inx} − f(x)c_{n}e^{inx})(\frac{sin nh}{nh})^{2} = 0.

Now F_{S} is linear on (a, b), so for x ∈ (a, b) and small enough h,

∑_{−∞}^{∞} c_{n}e^{inx}(\frac{sin nh}{nh})^{2} = \frac{Δ^{2}F_{S}(x, 2h)}{4h^{2}} = 0,

so for x ∈ (a, b)

lim_{h→0} ∑_{−∞}^{∞} f(x)c_{n}e^{inx}(\frac{sin nh}{nh})^{2} = 0,

thus this is true for all x since f(x) = 0 off (a, b). So

lim_{h→0} ∑_{−∞}^{∞} C_{n}e^{inx}(\frac{sin nh}{nh})^{2} = 0.

But if T ∼ ∑ C_{n}e^{inx}, then this limit is simply D^{2}F_{T} (x), so

D^{2}F_{T} (x) = 0

for all x, thus F_{T} is linear, and so C_{n} = 0 for all n, thus

∑_{−∞}^{∞} f(x)c_{n}e^{inx} = 0

uniformly for all x, thus as f(x) = 1 on [c, d], ∑ c_{n}e^{inx} = 0 uniformly for x ∈ [c, d]. ⊣ PART III. DESCRIPTIVE METHODS

§14. Perfect sets of uniqueness.

Until now the only examples of sets of uniqueness that we have seen are the countable ones. So it is conceivable that U = countable. This turned out to be false since in the period 1921–23 Rajchman and Bari came up independently with examples of perfect sets of uniqueness. We will give here Rajchman’s approach which makes use of his multiplication theory.

For x ∈ [0, 2π] and m ∈ ℤ we let mx = mx(mod 2π). (If we identify x with e^{ix} ∈ 𝕋, then mx = e^{imx}.) For A ⊆ 𝕋 we let mA = {mx : x ∈ A}. The next definition is due to Rajchman.

Definition. (Rajchman). A set E ⊆ 𝕋 is called an H-set if for some nonempty open interval I ⊆ 𝕋 and some sequence 0 ≤ n_{0} < n_{1} < n_{2} < ⋯, we have (n_{k}E) ∩ I = ∅ for all k.

Examples. (i) Every finite set is an H-set (but not every countable set).

(ii) The Cantor 1/3-set in [0, 2π], i.e., the set E of numbers of the form 2π ∑_{n=1}^{∞} ϵ_{n}/3^{n},

with ϵ_{n} = 0, 2, is an H-set. Indeed, 3^{n}E avoids the middle 1/3 interval.

14.1. Theorem (Rajchman). Every H-set is a set of uniqueness. So the Cantor 1/3-set is a set of uniqueness.

Proof. Notice that the closure of an H-set is an H-set, so we will work with a closed H-set E. Let I ≠ ∅ be an open interval and let 0 ≤ n_{0} < n_{1} < ⋯ be such that (n_{k}E) ∩ I = ∅. Let S ∼ ∑ c_{n}e^{inx} be a trigonometric series with ∑ c_{n}e^{inx} = 0 off E. We will show that c_{n} = 0. Clearly c_{n} → 0, by the Cantor-Lebesgue Lemma.

Choose a f ∈ C(𝕋) which has derivatives of all orders, f̂(0) = 1 and supp(f) = \overline{\{x : f(x) ≠ 0\}} ⊆ I. Put

f_{k}(x) = f(n_{k}x).

Then f_{k} = 0 on E. Let

S(f_{k}) · S ∼ ∑ C_{n}^{k}e^{inx}.

Claim. C_{n}^{k} → c_{n}, as k → ∞

Since by 13.2 we have that

∑_{−N}^{N} C_{n}^{k}e^{inx} − f_{k}(x) ∑_{−N}^{N} c_{n}e^{inx} → 0

for all x, it follows that

∑_{−∞}^{∞} C_{n}^{k}e^{inx} = 0

for all x (as f_{k}(x) = 0 on E and ∑_{−∞}^{∞} c_{n}e^{inx} = 0 off E). So C_{n}^{k} = 0 and thus by the claim,

c_{n} = 0.

Proof of the claim. Note that since

f(x) = ∑ f̂(n)e^{inx},

we have that

f_{k}(x) = f(n_{k}x) = ∑ f̂(n)e^{in·n_{k}x},

thus

f̂_{k}(i) = \{ \begin{matrix}f̂(n), & if i = n · n_{k} \\ 0, & otherwise \end{matrix} ,

so that ∑_{i∈ℤ} |f̂_{k}(i)| ≤ C < ∞ for all k, f̂_{k}(0) = 1, and lim_{k→∞} f̂_{k}(i) = 0, for i ≠ 0.

Now we have

\begin{aligned}C_{n}^{k} & = ∑_{m} c_{n−m}f̂_{k}(m) \\ & = ∑_{|m|≤N} ⋯ + ∑_{|m|>N} ⋯ \end{aligned}

for any N > |n|. The first sum converges to c_{n} as k → ∞ and the second is bounded by sup{|c_{k}| : |k| ≥ N − |n|} · C, which goes to 0 as N → ∞, so C_{n}^{k} → c_{n} as k → ∞. ⊣

§15. The Characterization Problem and the Salem-Zygmund Theorem.

We have now seen that (for measurable sets)

countable ⫋ U ⫋ null,

so an attempt to identify the sets of uniqueness with other types of “thin” sets like countable or null has failed. This raises the more general question of whether it is possible to characterize in some sense the sets of uniqueness. This problem was already prominent in the 1920’s and in fact even in the simplest case, that of closed sets or even perfect sets. For example, in Bari’s memoir on the problems of uniqueness in Fundamenta Mathematicae, Bari [1927], the following problem is explicitly stated.

The Characterization Problem. Find a necessary and sufficient condition for a perfect set to be a set of uniqueness.

As usual with such characterization problems this is a somewhat vague question. It appears though that the intended meaning was to find a characterization which can be expressed fairly explicitly in terms of a standard description of a given perfect set E, e.g., its sequence of contiguous intervals. Many attempts have been made to obtain such a characterization (see, e.g., Bari’s monograph, Bari [1964]) without success in the general case of an arbitrary perfect set. However, in the 1950’s Salem and Zygmund, following earlier work of Bari and Piatetski-Shapiro, proved a remarkable theorem which characterized when a perfect symmetric set of constant ratio of dissection is a set of uniqueness. We will next state the Salem-Zygmund Theorem.

Fix a sequence of numbers ξ_{1}, ξ_{2}, ⋯ with 0 < ξ_{i} < 1/2. The symmetric perfect set with dissection ratios ξ_{1}, ξ_{2}, ⋯, in symbols E_{ξ_{1},ξ_{2},···}, is defined as follows: For each interval [a, b], and 0 < ξ < 1/2 consider the middle open interval (a + ξℓ, b − ξℓ), where ℓ = b − a, and let E = [0, a + ξℓ] ∪ [b − ξℓ, b] be the remaining closed intervals. We say that E is obtained from [a, b] by a dissection of ratio ξ. Starting with [0, 2π] define E_{1} ⊇ E_{2} ⊇ ⋯, where E_{k} is a union of 2^{k} closed intervals in [0, 2π], by letting E_{1} be obtained from [0, 2π] by a dissection of ratio ξ_{1} and E_{k+1} be obtained form E_{k} by applying a dissection of ratio ξ_{k+1} to each interval of E_{k}. Let

E_{ξ_{1},ξ_{2},···} = ⋂_{k} E_{k}.

Then E_{ξ_{1},ξ_{2},···} is a perfect nowhere dense set and λ(E_{ξ_{1},ξ_{2},···}) = 0 iff 2^{k}ξ_{1}ξ_{2} ⋯ ξ_{k} → 0.

If ξ = ξ_{1} = ξ_{2} = ⋯, we write E_{ξ} instead of E_{ξ_{1},ξ_{2},···} and call E_{ξ} the symmetric perfect set of constant ratio of dissection ξ. Clearly (as ξ < 1/2) λ(E_{ξ}) = 0. The classical Cantor set is the set E_{1/3}.

The Salem-Zygmund Theorem characterizes when E_{ξ} is a set of uniqueness. Remarkably this depends on a subtle number theoretic property of ξ. We need the following definition.

Definition. An algebraic number θ is called an algebraic integer if θ is the root of a polynomial P (x) ∈ ℤ[x] with leading coefficient 1. Then there is a unique polynomial P (x) of least degree with leading coefficient 1 for which P (θ) = 0, called the minimal polynomial by θ. Say it has degree n ≥ 1. Write θ = θ^{(1)}, θ^{(2)}, ⋯, θ^{(n)} for its roots. We call θ^{(2)}, ⋯, θ^{(n)} the conjugates of θ. We say that θ is a Pisot number if θ > 1 and all its conjugates have absolute value < 1. (So θ must be real.)

Examples. (i) Every n ∈ ℕ, n ≥ 2 is Pisot, since it satisfies x − n = 0 which has only one root.

(ii) \frac{1+\sqrt{5}}{2} is Pisot, since it satisfies x^{2} − x − 1 = 0 and its conjugate is \frac{1−\sqrt{5}}{2} with |\frac{1−\sqrt{5}}{2}| < 1.

(iii) A rational p/q is Pisot iff it is an integer > 1 (otherwise it is not an algebraic integer).

(iv) \sqrt{2} is not Pisot.

Intuitively, a Pisot number is a number θ > 1, whose powers θ^{m} are “almost” integers. To see this first let θ^{(2)}, ⋯ θ^{(n)} be the conjugates of the Pisot number θ. Then θ^{m} + (θ^{(2)})^{m} + ⋯ + (θ^{(n)})^{m} is a symmetric polynomial of the roots of the minimal polynomial P (x) of θ, so it is an integer. But |θ^{(i)}| < 1, so (θ^{(i)})^{m} → 0 as m → ∞, for i > 1, so θ^{(m)} is closer and closer to an integer as m → ∞. Conversely, it can be shown that if we let

{x} = distance of x to the nearest integer, and θ > 1 is such that ∑_{n=0}^{∞}{θ^{n}}^{2} < ∞, then θ

is a Pisot number, and if θ is already algebraic, then {θ^{n}} → 0 is enough.

One remarkable fact about Pisot numbers is the following:

(Pisot) The set of Pisot numbers is closed (and of course countable).

It turns out that it has Cantor-Bendixson rank exactly ω.

We now have:

15.1. The Salem-Zygmund Theorem. Let 0 < ξ < \frac{1}{2} and let E_{ξ} be the symmetric perfect set of constant ratio dissection ξ Then

E_{ξ} is a set of uniqueness ⇔ θ = \frac{1}{ξ} is P isot.

Thus it appears that number theoretic issues enter into the arena of the characterization problem.

Salem and Zygmund extended somewhat their theorem to a wider class of perfect sets. We will state this generalization for further reference.

Fix η_{0} = 0 < η_{1} < ⋯ < η_{k} < η_{k+1} = 1 and put ξ = 1 − η_{k}. Assume that ξ < η_{i+1} − η_{i} for i < k. The so-called homogeneous perfect set associated to (ξ; η_{1}, ⋯, η_{k}), E(ξ; η_{1}, ⋯ η_{k}) is defined as follows:

For each closed interval [a, b] with length ℓ = [a, b], consider the disjoint intervals [a + ℓη_{i}, a + ℓη_{i} + ℓξ], i = 0, ⋯, k and let E be their union. We say then that E results from [a, b] by a dissection of type (ξ; η_{1}, ⋯, η_{k}). Starting from [0, 2π] define closed sets of E_{1} ⊇ E_{2} ⊇ ⋯ by performing to each interval of E_{n} a dissection of type (ξ; η_{1}, ⋯, η_{k}) to get E_{n+1}, and let

E(ξ; η_{1}, ⋯, η_{k}) = ⋂_{n} E_{n}.

Clearly E_{n} is made up of (k + 1)^{n} intervals of length ξ^{n}, so, as (k + 1)ξ < 1 we have that (k + 1)^{n}ξ^{n} → 0, thus λ(E(ξ; η_{1} ⋯ η_{k})) = 0. Note that E(ξ; η_{1}) = E(ξ; 1 − ξ) = E_{ξ}.

We now have:

15.2. The General Salem-Zygmund Theorem. The set E(ξ; η_{1}, ⋯, η_{k}) is a set of uniqueness iff

(i) θ = \frac{1}{ξ} is Pisot

and

(ii) η_{1}, ⋯, η_{k} ∈ ℚ(θ).

The proof of 15.2 can be found in Kahane-Salem [1994].

This is essentially the best known positive result concerning the Characterization Problem. For example, there is no known characterization of when E_{ξ_{1},ξ_{2},···} is a set of uniqueness. Any such potential characterization would have to look quite different since Meyer has shown that if ∑ ξ_{n}^{2} < ∞, then E_{ξ_{1},ξ_{2},···} is a set of uniqueness.

In the preface of Zygmund’s classic treatise, Zygmund [1979], he states: “Two other major problems of the theory also await their solution. These are the structure of the sets of uniqueness and the structure of the functions with absolutely convergent Fourier series .... in a search for solutions we shall probably have to go beyond the domains of the theory of functions, in the direction of the theory of numbers and Diophantine approximation.” (This was of course written after the proof of the Salem-Zygmund Theorem, which was proved in 1955.)

In the rest of this chapter we will develop another approach to the Characterization Problem based on the concepts and methods of descriptive set theory. This approach has led also to other significant dividends, as, for example, the original solution of the Category Problem.

This approach, based on the idea of studying the global structure of the class of closed sets of uniqueness from a descriptive standpoint, has led to interesting conclusions concerning the Characterization Problem for arbitrary perfect sets, by providing sharp limitations on the possibility of a positive solution. Whether these results actually provide a negative solution to the Characterization Problem is a matter of interpretation of the original question, which is rather vague. It certainly rules out characterizations of the type that researchers in the field have tried to establish over the years. Independently, of this, the point of view and the techniques that will be explained in the sequel should be useful in general in attacking similar characterization problems in analysis or other areas of mathematics.

§16. The hyperspace K(𝕋) of closed subsets of the circle.

Descriptive set theory is the study of “definable” sets in Polish, i.e., complete separable metric spaces (like ℝ, ℂ, ℝ^{n}, etc.). In this theory, sets in such spaces are classified in hierarchies according to the complexity of their definitions, and the structure of the sets in each level of these hierarchies is studied in detail.

We want to apply this theory to the study of the global structure of closed sets of uniqueness which we will denote by U:

U = {E ⊆ 𝕋 : E is a closed set of uniqueness}.

It is also important to consider a wider class, the so-called sets of extended uniqueness which are those sets E ⊆ 𝕋 which satisfy uniqueness for series of the form ∑ μ̂(n)e^{inx} with μ ∈ M(𝕋). We denote the class by U_{0}, so that

E ∈ U_{0} ⇔ for every μ ∈ M(𝕋), if ∑ μ̂(n)e^{inx} = 0 off E, then μ̂(n) = 0, ∀n ∈ ℤ.

We also let U_{0} = {E ⊆ 𝕋 : E is a closed set of extended uniqueness}.

Clearly U ⊆ U_{0} (and Piatetski-Shapiro actually showed that this inclusion is proper).

Both the sets U, U_{0} are “definable” subsets (and we will see later on at what level of complexity) of the so-called hyperspace of 𝕋, i.e., the space of all closed subsets of the circle equipped with an appropriate topology, which we will now describe.

Let us start more generally with a compact metric space (X, d), with d ≤ 1 (for normalization purposes), like, e.g., the circle 𝕋 with the usual metric (normalized arclength). Denote by K(X) the set of all closed (= compact) subsets of X. Define on K(X) the following metric, called the Hausdorff metric (associated to d), d_{H}:

d_{H}(K, L) = 0, if K = L = ∅,
= 1, if K ≠ L, K = ∅ or L = ∅,
= max{δ(K, L), δ(L, K) : K, L ≠ ∅},

where

δ(K, L) = max_{x∈K} d(x, L).

16.1. Exercise. If

B(E, ϵ) = {x : d(x, E) < ϵ},

show that

d_{H}(K, L) < ϵ ⇔ K ⊆ B(L, ϵ) & L ⊆ B(K, ϵ).

16.2. Exercise. (i) Show that (K(X), d_{H}) is complete. Hint. If {K_{n}} is Cauchy with K_{n} ≠ ∅, then K = ⋂_{n}(\overline{⋃_{i=n}^{∞} K_{i}}) is the limit of \{K_{n}\}.

(ii) Show that (K(X), d_{H}) is compact. Hint. If F ⊆ X is finite, with ∀x ∈ X∃y ∈ F (d(x, y) < ϵ), then K(F) = {K ∈ K(𝕋) : K ⊆ F } is finite and ∀K ∈ K(X)∃L ∈ K(F)(d_{H}(K, L) < 2ϵ).

(iii) Show that if D ⊆ X is dense, then K_{f}(D) = {K ∈ K(𝕋) : K ⊆ D, K finite} is dense in K(X).

Thus (K(X), d_{H}) ≡ K(X) is a compact metric space, so it is separable, and thus a Polish space.

Although the metric on K(X) depends on the chosen metric on X, the topology of K(X) depends only on the topology of X.

For any topological space X, we let K(X) be the space of compact subsets of X. We give K(X) the so-called Vietoris topology which is the one generated by the sets

\begin{matrix}\{K ∈ K(X) : K ⊆ U\}, \\ \{K ∈ K(X) : K ∩ U ≠ ∅\}, \end{matrix}

where U ⊆ X is open. So a basis of this topology is given by the sets

{K ∈ K(X) : K ⊆ U_{0} & K ∩ U_{1} ≠ ∅ & ⋯ & K ∩ U_{n} ≠ ∅}

for U_{0}, U_{1}, ⋯, U_{n} ⊆ X open. 16.3. Exercise. Show that the topology of (K(X), d_{H}) is exactly the Vietoris topology on K(X).

The following facts are not hard to prove. (Sometimes the best method is to use 16.3.) 16.4. Exercise. (i) x ↦ {x} is an isometry of X into K(X).

(ii) {(x, K) : x ∈ K}, {(K, L) : K ⊆ L}, {(K, L) : K ∩ L ≠ ∅}, are closed in X × K(X), K(X) × K(X), resp.

(iii) (K, L) ↦ K ∪ L is continuous (from K(X) × K(X) into K(X)) but, in general (K, L) ↦ K ∩ L is not.

(iv) If f : X → Y is continuous, so is f′′ : K(X) → K(Y) given by f′′(K) = f[K].

(v) The operation ⋃ : K(K(X)) → K(X) given by ⋃ K = ⋃{K : K ∈ K} for any closed K ⊆ K(X) is continuous.

(vi) K_{f}(X) = {K ∈ K(X) : K is finite} is F_{σ} in K(X), K_{p}(X) = {K ∈ K(X) : K is perfect} is G_{δ} in K(X).

(vii) If A ⊆ X, let

K(A) = {K ∈ K(X) : K ⊆ A}.

If A is closed, open, G_{δ}, then K(A) is closed, open, G_{δ}, resp. §17. Review of descriptive set theory.

Our reference for the concepts and results of descriptive set theory that we will use here is Kechris [1995].

Let X be a Polish space. A set A ⊆ X is Borel if it belongs to the smallest σ-algebra containing the open sets. So all open, closed, F_{σ}, G_{δ}, ⋯ sets are Borel. We ramify Borel sets in a transfinite hierarchy of ω_{1} (= the first uncountable ordinal) stages, called the Borel hierarchy. We let

Σ_{1}^{0} = open, Π_{1}^{0} = closed,

and for α < ω_{1}, we inductively define Σ_{α}^{0}, Π_{α}^{0} by

Σ_{α}^{0} = {⋃_{n} A_{n} : A_{n} is Π_{α_{n}}^{0} for some α_{n} < α},

Π_{α}^{0} = {X \ A : A ∈ Σ_{α}^{0}}.

So Σ_{2}^{0} = F_{σ}, Π_{2}^{0} = G_{δ}, Σ_{3}^{0} = countable unions of G_{δ} sets = G_{δσ}, Π_{3}^{0} = complements of Σ_{3}^{0} sets = countable intersections of F_{σ} sets = F_{σδ}, etc.

Let also

Δ_{α}^{0} = Σ_{α}^{0} ∩ Π_{α}^{0},

so that Δ_{1}^{0} = clopen, for example. To emphasize that we work in the space X, we denote this also by Σ_{α}^{0}(X), Π_{α}^{0}(X), Δ_{α}^{0}(X) if necessary. We also let B = B(X) be the class of Borel sets in X.

We have Σ_{α}^{0} ∪ Π_{α}^{0} ⊆ Σ_{α+1}^{0} ∩ Π_{α+1}^{0} = Δ_{α+1}^{0}, so that we have an increasing hierarchy of sets and

B = ⋃_{α<ω_{1}} Σ_{α}^{0}( = ⋃_{α<ω_{1}} Π_{α}^{0} = ⋃_{α<ω_{1}} Δ_{α}^{0}).

We call {Σ_{α}^{0}, Π_{α}^{0}, Δ_{α}^{0}}_{α<ω_{1}} the Borel hierarchy. It is proper, i.e., Σ_{α}^{0} ≠ Π_{α}^{0}, Σ_{α}^{0} ∪ Π_{α}^{0} ⫋ Δ_{α+1}^{0}, if X is uncountable.

A subset A ⊆ X is Σ_{1}^{1} or analytic if for some Polish space Y, Borel B ⊆ Y, and continuous f : Y → X we have f[B] = A. A set A ⊆ X is Π_{1}^{1} or co-analytic if X \ A is Σ_{1}^{1}. Inductively define

Σ_{n+1}^{1} = the class of continuous images of Π_{n}^{1} sets,

Π_{n+1}^{1} = the complements of Σ_{n+1}^{1} sets.

Also put Δ_{n}^{1} = Σ_{n}^{1} ∩ Π_{n}^{1}. Again we write Σ_{n}^{1}(X), Π_{n}^{1}(X), Δ_{n}^{1}(X) to emphasize that we look at subsets of X. It turns out that Σ_{n}^{1} ∪ Π_{n}^{1} ⊆ Δ_{n+1}^{1}. The projective subsets of X are defined by

P = P(X) = ⋃_{n} Σ_{n}^{1}(X)(= ⋃_{n} Π_{n}^{1}(X) = ⋃_{n} Δ_{n}^{1}(X)).

We call {Σ_{n}^{1}, Π_{n}^{1}, Δ_{n}^{1}} the projective hierarchy. It is proper, i.e., Σ_{n}^{1} ≠ Π_{n}^{1}, Σ_{n}^{1}∪Π_{n}^{1} ⫋ Δ_{n+1}^{1}, if X is uncountable.

In descriptive set theory one studies the structure of sets in these hierarchies (and even more extended ones). We will only need to consider in these lectures Borel sets and sets in the 1st level of the projective hierarchy, i.e., analytic (Σ_{1}^{1}) and co-analytic (Π_{1}^{1}) sets.

These are tied up by the well-known Souslin Theorem

B = Δ_{1}^{1}.

(It is not hard to check that B ⊆ Δ_{1}^{1}; it is the inclusion that Δ_{1}^{1} ⊆ B that is the main point here.) So if a set is both analytic and co-analytic, it is Borel. However, there are, in every uncountable Polish space X, analytic (and so co-analytic sets) which are not Borel. One of the early examples is due to Hurewicz: If X is an uncountable compact metric space, then K_{ω}(X) = {K ∈ K(X): K is countable} is Π_{1}^{1} but not Borel. Another early example is due to Mazurkiewicz: In C(𝕋), the set {f ∈ C(𝕋): f is differentiable} is Π_{1}^{1} but not Borel.

If a set A, in a given space X, is not Borel, then this implies that one cannot give a necessary and sufficient criterion for membership in A, i.e., a characterization of membership in A, which is simple enough to be expressible in terms of countable operations starting from the basic information describing the members of X. So such a fact about the descriptive complexity of A gives important information about possible characterizations of membership in A. We want to apply this descriptive approach to the (closed) sets of uniqueness.

For that purpose it will be useful to first discuss another example of a co-analytic non-Borel set. First we recall a standard fact from the theory of analytic sets.

17.1. Theorem. For every Polish space X and Σ_{1}^{1} set A ⊆ X, there is a G_{δ} set G ⊆ X × C, where C = 2^{ℕ} is the Cantor space, such that A = proj_{X}[G], i.e.,

x ∈ A ⇔ ∃y ∈ 2^{ℕ}(x, y) ∈ G.

Proof. This is clear if A = ∅, so we assume that A ≠ ∅. The nonempty analytic sets can be also characterized as the continuous images of Polish spaces and since every Polish space is the continuous image of the Baire space N = ℕ^{ℕ} (which is homeomorphic to the irrationals), it follows that there is continuous g : N ⟶ X with g[N] = A. Let G ⊆ X ×N be defined by

(x, y) ∈ G ⇔ f(y) = x.

Then A = proj_{X}[G]. So A is the projection of a closed subset of X × N. But it is easy to see that N is (homeomorphic to) a G_{δ} subset of C. (View C as 2^{ℕ×ℕ} and identify x ∈ ℕ^{ℕ} with its graph which is a subset of ℕ × ℕ.) So G, viewed as a subset of X × C, is G_{δ} in X × C and we are done. ⊣

We now have:

17.2. Theorem (Hurewicz). Let ℚ′ = ℚ ∩ [0, 1]. Then K(ℚ′) is Π_{1}^{1} but not Borel in K([0, 1]).

Proof. Note first that K(ℚ′) is Π_{1}^{1}: Put N = [0, 1] \ ℚ′. Then

∼ K(ℚ′) = {K ∈ K([0, 1]) : K ∩ N ≠ ∅}
= {K ∈ K([0, 1]) : ∃x(x ∈ K & x ∈ N)}
= proj_{K([0,1])}[G],

where G ⊆ K([0, 1]) × [0, 1] is defined by

(K, x) ∈ G ⇔ x ∈ K & x ∈ N.

So G is G_{δ} and thus ∼ K(ℚ′) is Σ_{1}^{1}, and K(ℚ′) is Π_{1}^{1}.

To show that K(ℚ′) is not Borel, we will first work with K(Q), where Q ⊆ C = 2^{ℕ} is the countable dense consisting of the eventually periodic sequences. We will show that K(Q) is not Borel (in K(C)). Granting this, we complete the proof as follows: Let f : C →

[0, 1] be defined by f(x) = ∑_{n=0}^{∞} x(n)2^{−n−1}. Then f is continuous and x ∈ Q ⇔ f(x) ∈ ℚ′.

Let F : K(C) → K([0, 1]) be defined by F (K) = f′′(K) = f[K]. Then F is continuous and

K ∈ K(Q) ⇔ F (K) ∈ K(ℚ′),

so K(Q) = F^{−1}(K(ℚ′)). If K(ℚ′) was Borel, then K(Q) would be Borel too, being a continuous preimage of a Borel set, a contradiction.

So it is enough to show K(Q) is not Borel in K(C). We will, in fact, prove that this holds for any countable dense set Q ⊆ C. This is based on the following lemma.

Lemma. Let F ⊆ 2^{ℕ} be F_{σ}. Then there is a continuous function g : 2^{ℕ} → 2^{ℕ} such that F = g^{−1}(Q).

Assuming this, we complete the proof as follows: Fix a Π_{1}^{1} not Borel set P ⊆ C. Then, by 17.1, let F be F_{σ} in C × C such that

x ∉ P ⇔ ∃y(x, y) ∉ F

or

x ∈ P ⇔ ∀y(x, y) ∈ F.

Now C×C is homeomorphic to C, so that there is continuous g : C×C → C with g^{−1}[Q] = F. Let G : C → K(C) be defined by

G(x) = g({x} × C).

Then G is continuous and

x ∈ P ⇔ ∀y(x, y) ∈ F
⇔ ∀y(g(x, y) ∈ Q)
⇔ G(x) ⊆ Q
⇔ G(x) ∈ K(Q).

So, as before, if K(Q) was Borel, so would be P, a contradiction.

Proof of the Lemma. Consider the following game: In a run of the game, Players I,II take turns (I starting first) choosing successively x(0), y(0), x(1), y(1), ⋯; x(i), y(i) ∈ {0, 1}. II wins iff x ∈ F ⇔ y ∈ Q. A strategy for II is a map σ : {0, 1}^{<ℕ} → {0, 1}. II follows this strategy in a run of this game if for each n, y(n) = σ(x|n) (where x|n = (x(0), ⋯, x(n − 1))). It is a winning strategy if II wins every run of the game in which he follows σ. (Strategies for I are similarly defined.) Put σ^{∗}(x) = y iff ∀n(y(n) = σ(x|n)), i.e., σ^{∗}(x) is what II plays following σ, when I plays x in a given run of the game. Thus if σ is a winning strategy for II, clearly x ∈ F ⇔ σ^{∗}(x) ∈ Q. Since easily σ^{∗} : C → C is a continuous function, it is enough to show that II has a winning strategy in this game. We will define such a strategy below.

Let F = ⋃_{n} F_{n}, F_{n} closed. Let for each closed set H ⊆ C, T_{H} = {x|n : x ∈ H}. Then T_{H} is a tree on {0, 1}, i.e., a subset of {0, 1}^{<ℕ} closed under initial segments (i.e., s = (s_{0}, ⋯, s_{n−1}) ∈ T_{H} and m < n implies that (s_{0}, ⋯, s_{m−1}) ∈ T_{H}). Moreover

[T_{H}] = {x : ∀n(x|n ∈ T_{H})} = H.

Let T_{F_{n}} = T_{n} and Q = {q_{n}}. Here is then the strategy for II:

As I plays x(0), x(1), ⋯ II plays y(0), y(1), ⋯ as follows: As long as I stays within T_{0}, i.e., x|n ∈ T_{0}, II plays y(0) = q_{0}(0), y(1) = q_{0}(1), ⋯, i.e., follows q_{0} ≡ q′_{0}. If x ever gets out of T_{0} let n_{0} + 1 be least with x|(n_{0} + 1) ∉ T_{0}. Then II plays y(n_{0}) ≠ q_{0}(n_{0}) and chooses q′_{1} ∈ Q with y|(n_{0} + 1) an initial segment of q′_{1}. This can be done as Q is dense. From then on, if x stays within T_{1}, II follows q′_{1}. If x ever gets out of T_{1}, let n_{1} > n_{0} be least with x|(n_{1} + 1) ∉ T_{1}. Then II plays y(n_{1}) ≠ q_{1}(n_{1}) and chooses q′_{2} ∈ Q with y|(n_{1} + 1) an initial segment of q′_{2}, and so on ad infinitum. ⊣

§18. The theorem of Kaufman and Solovay.

I will prove here that the set U of closed sets of uniqueness is not Borel in the space K(𝕋). This result is due to Kaufman and Solovay independently. The proof that I will give is a simplification of Solovay’s argument and is based on two facts about U: (1) Bari’s Theorem that the countable union of closed sets of uniqueness is also a set of uniqueness and (2) the general form of the Salem-Zygmund. I will give the proof of Bari’s Theorem after giving the proof of the Kaufman-Solovay result.

18.1. Theorem (Kaufman,Solovay). The set U of closed sets of uniqueness is not Borel (in K(𝕋)). Proof. In the notation of 15.2, let, for each x ∈ [0, 1],

f(x) = E(1/4; \frac{3}{8} + \frac{x}{9}, 3/4).

Then f(x) is a perfect set in 𝕋, so f : [0, 1] → K(𝕋). 18.2. Exercise. f is continuous.

From 15.2 it now follows that

x ∈ ℚ ⇔ f(x) ∈ U.

Let F : K([0, 1]) → K(𝕋) be defined by

\begin{aligned}F (K) & = ⋃ f′′(K) \\ & = ⋃\{f(x) : x ∈ K\}. \end{aligned}

Then F is continuous and

K ⊆ ℚ ⇔ F (K) ∈ U,

since the union of countably many closed sets of uniqueness is a set of uniqueness. Thus, in the notation of 17.2,

K(ℚ′) = F^{−1}(U),

so U cannot be Borel, since K(ℚ′) is not Borel. ⊣

In 15.2 it is actually proved that if 1/ξ is not a Pisot or else one of η_{1}, ⋯, η_{k} is not in ℚ(ξ), then not only E(ξ; η_{1}, ⋯, η_{k}) ∉ U but also E(ξ; η_{1}, ⋯, η_{k}) ∉ U_{0}. So it follows, in the notation of the preceding proof, that

K(ℚ′) = F^{−1}(U_{0}),

so we also have: 18.3. Corollary (of the proof). U_{0} is not Borel. Remark. Another proof of (much stronger versions of) 18.2 and 18.3, which is selfcontained and independent of the Salem-Zygmund theorem, will be given in §27 below.

Since the sets F (K) are also perfect it finally follows that: 18.4. Corollary. The class of perfect sets of uniqueness is not Borel (in K(𝕋)).

This result has obvious implications for the Characterization Problem: It is impossible to characterize, given a standard description of a perfect set (e.g., in terms of the sequence of its contiguous intervals), whether it is a set of uniqueness, by conditions which are explicit enough to be expressed by countable operations involving this description. This is because any such description would give a Borel definition of the set U in K(𝕋). It should be noted that all positive results obtained so far, including the Salem-Zygmund Theorem and its generalizations, are of this nature. Later on we will see some even stronger conclusions ruling out even more general types of characterization (see §24).

§19. Descriptive classification as a method of existence proof.

We describe here a method of existence proof based on the concept of descriptive classification of sets in Polish spaces. Suppose we have two properties R, S and every object satisfying R satisfies also S. Our problem is to find an object satisfying S but not R. The descriptive method consists of finding an appropriate Polish space X and calculating the descriptive complexity of {x ∈ X : R(x)} = R^{∗}, {x ∈ X : S(x)} = S^{∗}. Clearly R^{∗} ⊆ S^{∗}. If for instance R^{∗} turns out to be of descriptive complexity different than S^{∗}, e.g., if R^{∗} is non-Borel but S^{∗} is Borel, or R^{∗} is not Σ_{1}^{1} but S^{∗} is Σ_{1}^{1}, then clearly R^{∗} ⫋ S^{∗}, thus ∃x ∈ X(S(x) & ¬R(x)), so we have shown the existence of an object satisfying S but not R.

Here is an example (due to Bourgain) of the application of this method: Given a class S of separable Banach spaces, a separable Banach space X is universal for S if every Y ∈ S is isomorphic to a closed subspace of X, i.e., can be embedded into X. An old problem in the theory of Banach spaces asked whether there is a separable Banach space with separable dual which is universal for the class of separable Banach spaces with separable dual (Problem 49 in the Scottish book, Mauldin [1981]). This was answered negatively by Wojtaszczyk. Bourgain then showed that any separable Banach space universal for the above class must be universal for the class of all Banach spaces (so it cannot have separable dual). The method of proof is the following: Let X_{0} be universal for the class of separable Banach spaces with separable dual. Then one can calculate that

S = {K ∈ K(C) : C(K) is isomorphic to a closed subspace of X_{0}}

is Σ_{1}^{1}. Let

R = {K ∈ K(e) : K is countable }.

Then, by a result of Hurewicz (see §22 below), R is Π_{1}^{1} but not Borel, so not Σ_{1}^{1} (by Souslin’s Theorem). Now

R ⊆ S

(as the dual of C(K) is the space M(K), which, as K is countable, is easily separable). So R ≠ S and there is K ∈ K(C), K uncountable with C(K) isomorphic to a closed subspace of X_{0}. But, as K is uncountable, C(K) is universal for all separable Banach spaces, and thus so is X_{0}.

19.1. Exercise. Show that {K ∈ K(𝕋) : λ(K) = 0} is G_{δ} in K(𝕋). Use this and just the statement of 18.1 to deduce Menshov’s Theorem (that there is a closed null set of multiplicity).

§20. Bari’s Theorem.

I will now prove the following important result of Bari that was used in the proof of 18.1.

20.1. Bari’s Theorem. The union of countably many closed sets of uniqueness is a set of uniqueness.

Proof. We will need the following result of de la Vallée-Poussin (1912), which should be contrasted with Menshov’s Theorem that a trigonometric series can converge a.e. without being identically 0.

20.2. Theorem (de la Vallée-Poussin). Let S ∼ ∑ c_{n}e^{inx} be a trigonometric series such

that for each x, there is M_{x} < ∞ with | ∑_{n=−N}^{N} c_{n}e^{inx}| ≤ M_{x} for all N ≥ 0 (i.e., ∑ c_{n}e^{inx}

has bounded partial sums). Then if ∑ c_{n}e^{inx} = 0 a.e., we have that c_{n} = 0, ∀n.

I will postpone for a while the proof of 20.2 and use it to prove Bari’s Theorem. So assume E_{n} ⊆ 𝕋 are closed and E_{n} ∈ U. Put E = ⋃_{n} E_{n}. Let ∑ c_{n}e^{inx} = 0 off E, in order to show that c_{n} = 0, ∀n. Clearly λ(E_{n}) = 0, so λ(E) = 0, thus c_{n} → 0. Assume c_{n} is not identically 0, towards a contradiction. Let

G = {x : {S_{N}(x)} is unbounded},

where

S_{N}(x) = ∑_{n=−N}^{N} c_{n}e^{inx}.

Then G ⊆ E, G is G_{δ} and G ≠ ∅ by de la Vallé-Poussin’s Theorem. So G is Polish in the relative topology and if E_{n} ∩ G = G_{n}, clearly G_{n} is closed in G and ⋃_{n} G_{n} = G, so, by the Baire Category Theorem, there is an open interval I_{0} and some n_{0} with G ∩ I_{0} = G_{n_{0}} ∩ I_{0} ≠ ∅. We will show that ∑ c_{n}e^{inx} = 0 on I_{0}, thus I_{0} ∩ G = ∅, so we have a contradiction.

We use Rajchman multiplication. Choose f ∈ C(𝕋) infinitely differentiable with f > 0 on I_{0} and f = 0 off I_{0}. Let T = S(f) · S, T ∼ ∑ C_{n}e^{inx}. Recall that, by 13.2,

∑_{−∞}^{∞}(C_{n} − f(x)c_{n})e^{inx} = 0, ∀x. If we can show that C_{n} = 0, ∀n, then we are done. As E_{n_{0}}

is a set of uniqueness, it is enough to show that ∑ C_{n}e^{inx} = 0 for x ∉ E_{n_{0}}. So let x ∉ E_{n_{0}}. We can assume that x ∈ I_{0} ∩ E, since ∑ C_{n}e^{inx} = 0 off I_{0} ∩ E. So let J_{0} ⊆ I_{0} be an interval containing x such that \overline{J} ∩ E_{n_{0}} = ∅.

Choose again an infinitely differentiable g ∈ C(𝕋) with g(x) = 1 and supp(g) ⊆ \overline{J}. Again if

R ∼ S(g) · T, R ∼ ∑ D_{n}e^{inx},

∑ D_{n}e^{inx} = 0 a.e., because ∑ C_{n}e^{inx} = 0 a.e. (as ∑ c_{n}e^{inx} = 0 a.e.), and has bounded partial sums outside \overline{J_{0}} ∩ G = \overline{J_{0}} ∩ G_{n_{0}} = ∅ (since ∑ c_{n}e^{inx}, and thus ∑ C_{n}e^{inx}, has the same property outside G), i.e., ∑ D_{n}e^{inx} has bounded partial sums everywhere, so by de la Vallé-Poussin again, D_{n} = 0, ∀n, and, by 13.2, ∑ C_{n}e^{inx} = 0.

Proof of 20.2.

Lemma. Let G ⊆ [0, 2π] be G_{δ} and null. Then there is g ≥ 0 continuous nondecreasing in [0, 2π] with g′(x) = +∞ for x ∈ G.

Proof. Let G = ⋂_{n} G_{n}, G_{n} open, λ(G_{n}) < 2^{−n}. Let g_{n}(x) = \frac{1}{2π} ∫_{0}^{x} χ_{G_{n}}(t)dt. Then 0 ≤ g_{n} ≤ 2^{−n}. Let g = ∑ g_{n}. It is enough to show that g′(x) = ∞ for x ∈ G. Fix K > 0. For any n_{0} > K, let ϵ > 0 be such that (x − ϵ, x + ϵ) ⊆ G_{0} ∩ ⋯ ∩ G_{n_{0}}. Then if 0 < |h| < ϵ,

\frac{g(x + h) − g(x)}{h} ≥ \frac{(n_{0} + 1)h}{2π · h} > \frac{k}{2π} ,

so g′(x) = +∞. ⊣

Now assume ∑ c_{n}e^{inx} = 0 a.e., and ∑ c_{n}e^{inx} has bounded partial sums at each point x. Let G be a null G_{δ} set with

x ∉ G ⇒ ∑ c_{n}e^{inx} = 0.

Let g be as in the preceding lemma. Put

f(x) = ∫_{0}^{x} g(t)dt + C,

where C < 0 is chosen so that f(2π) = 0. So f(x) is convex and f′(x) = g(x).

Let F_{S} be the Riemann function of S and choose a, b so that if

F (x) = F_{S}(x) + ax + b,

then F (0) = F (2π) = 0. If we can show that F = 0 on [0, 2π], then F_{S} is linear on [0, 2π], so c_{n} = 0, ∀n.

We will show that F ≥ 0, F ≤ 0 on [0, 2π]:

F ≤ 0: For ϵ > 0, x ∈ [0, 2π] let

F_{ϵ}(x) = F (x) − ϵx(2π − x) + ϵf(x).

Then F_{ϵ}(0) < 0, F_{ϵ}(2π) = 0, so if F ≤ 0 fails, towards a contradiction, there is ϵ > 0 and x_{0} ∈ (0, 2π) at which F_{ϵ} achieves a maximum which is positive. Then for small enough h

0 ≥ \frac{Δ^{2}F_{ϵ}(x_{0}, h)}{h^{2}} = \frac{Δ^{2}F (x_{0}, h)}{h} + 2ϵ + \frac{ϵΔ^{2}f(x_{0}, h)}{h^{2}}.

Now consider 2 cases:

Case 1. x_{0} ∉ G. Then ∑ c_{n}e^{inx_{0}} = 0, so D^{2}F_{S}(x_{0}) = D^{2}F (x_{0}) = 0, thus \frac{Δ^{2}F_{S}(x_{0},h)}{h^{2}} → 0 as h → 0. But, f being convex, Δ^{2}f(x_{0}, h) ≥ 0, so we have a contradiction. Case 2. x_{0} ∈ G. Since {S_{N}(x_{0})} is bounded, we claim that

|\frac{Δ^{2}F (x_{0}, h)}{h^{2}} | < K < ∞  (∗)

But D^{2}f(x_{0}) = g′(x_{0}) = +∞, so again we have a contradiction.

We can see (∗) as follows: we have

\frac{Δ^{2}F (x_{0}, 2h)}{4h^{2}} = \frac{Δ^{2}F_{S}(x_{0}, 2h)}{4h^{2}}

= ∑_{n∈ℤ}(\frac{sin nh}{nh})^{2}c_{n}e^{inx_{0}}

= ∑_{n=0}^{∞}(\frac{sin nh}{nh})^{2}(a_{n} cos nx_{0} + b_{n} sin nx_{0}).

We know that the partial sums

\begin{aligned}S_{N}(x_{0}) & = ∑_{n=0}^{N}(a_{n} cos nx_{0} + b_{n} sin nx_{0}) \\ & = ∑_{n=−N}^{N} c_{n}e^{inx_{0}} \end{aligned}

are bounded, say in absolute value by M. But we have

∑_{n=0}^{∞}(\frac{sin nh}{nh})^{2}(a_{n} cos nx_{0} + b_{n} sin nx_{0})

= ∑_{n=0}^{∞} S_{N}(x_{0})((\frac{sin nh}{nh})^{2}−(\frac{sin(n + 1)h}{(n + 1)h})^{2}),

so

\begin{aligned}|\frac{Δ^{2}F (x_{0}, 2h)}{4h^{2}} | & ≤ M · ∑_{n=0}^{∞}|(\frac{sin nh}{nh})^{2}−(\frac{sin(n + 1)h}{(n + 1)h})^{2}| \\ & ≤ M · C = K < ∞, \end{aligned}

where

C = ∫_{0}^{∞} |u′(x)|dx,

with u(x) = (sin x/x)^{2} (see 2.5).

F ≥ 0: Replace ϵ by −ϵ in the above argument. ⊣

It follows from Bari’s Theorem that the union of countably many F_{σ} sets of uniqueness is a set of uniqueness. It is not known if the union of even two G_{δ} sets of uniqueness is a set of uniqueness.

§21. Computing the exact descriptive complexity of U, U_{0}.

We have seen that U, U_{0} are not Borel sets. We will now compute that they are both Π_{1}^{1}, and since then they are not Σ_{1}^{1}, this determines the exact descriptive complexity of U, U_{0}. If one looks at the definition of U, U_{0}, a rather straightforward calculation shows that U, U_{0} must be Π_{2}^{1}. However, this is a very crude estimate and with some more work, which is based on some appropriate generalizations of the nontrivial Theorem 7.6, we can bring the complexity down to Π_{1}^{1}, which is the exact level.

Given a closed set K ∈ K(𝕋), and a trigonometric series S ∼ ∑ c_{n}e^{inx}, with |c_{n}| bounded, we reformulate the condition “∑ c_{n}e^{inx} = 0 off K” in functional analytic terms. The key is to identify S with the element of ℓ^{∞} = (ℓ^{1})^{∗} given by {c_{n}}. So from now on we will view trigonometric series S ∼ ∑ c_{n}e^{inx} with bounded coefficients as elements of ℓ^{∞} and simply write c_{n} = S(n).

Next we view an element λ_{n} of ℓ^{1} as identified with the function f(x) = ∑ λ_{n}e^{inx}. Note that f̂(n) = λ_{n}. These are of course exactly the functions with absolutely convergent Fourier series, and their class is traditionally denoted by A(𝕋) = A. Thus we let A ≡ ℓ^{1}. Under this identification the element e_{n} of ℓ^{1} (where e_{n}(j) = 1, if j = n, e_{n}(j) = 0, if j ≠ n) is identified with e^{inx}.

Now each S ∈ ℓ^{∞} operates on f = ∑ c_{n}e^{inx} ∈ A by
⟨f, S⟩ = ∑ f̂(n)S(−n).

Thus S(n) = ⟨e^{inx}, S⟩. In particular, if μ ∈ M(𝕋) and S = μ̂, then S ∈ ℓ^{∞} (as |μ̂(n)| ≤ ||μ||_{M} < ∞) and

⟨f, μ̂⟩ = ∑ f̂(n)S(−n) = ∫ fdμ

for f ∈ A. So we can view ⟨f, S⟩ as some kind of generalized integral and S as some kind of generalized measure (operating though only on functions in A) and thus it is customary to call elements of ℓ^{∞} pseudomeasures and write PM instead of ℓ^{∞},

PM = ℓ^{∞}.

Finally, every f ∈ L^{1}(𝕋) gives rise to f̂ ∈ c_{0}, so that it is customary to think of the elements of c_{0} as generalized functions and thus call them pseudofunctions and write PF instead of c_{0},

PF = c_{0}.

So we have PF^{∗} = A, A^{∗} = PM (and also PF is a closed subspace of PM).

Using this idea, we can define what it means to say that a closed set K ⊆ 𝕋 supports a pseudomeasure S.

Definition. Let S ∈ PM and K ∈ K(𝕋). Then K supports S iff for any open interval I ∩ E = ∅ and infinitely differentiable φ ∈ C(𝕋) supported by I (i.e., sup_{P} (φ) ⊆ I), we have ⟨φ, S⟩ = 0. (Note that φ ∈ A too.)

21.1. Exercise. Show that if μ ∈ P (𝕋), E ∈ K(𝕋), then E supports μ iff μ(E) = 1.

Remark. One can easily see that if S is supported by K, then actually for any interval I disjoint from K and any f ∈ A supported by I we also have ⟨f, S⟩ = 0. To see this fix ϵ > 0 and let φ be infinitely differentiable with φ = 1 on sup_{P} (f) and sup_{P} (φ) ⊆ I. Letting for f ∈ A

||f||_{A} = ||f̂||_{ℓ^{1}} = ∑ |f̂(n)|,

let P be a trigonometric polynomial with ||f − P ||_{A} < ϵ/||φ||_{A}. Then noting that A is actually a Banach algebra under pointwise multiplication, i.e., for f, g ∈ A, fg ∈ A and ||fg||_{A} ≤ ||f||_{A}||g||_{A}, we have that

\begin{aligned}||f − P φ||_{A} & = ||fφ − P φ||_{A} \\ & ≤ ||f − P ||_{A} · ||φ||_{A} < ϵ. \end{aligned}

Thus f can be approximated in the norm of A by infinitely differentiable functions ψ with support contained in I, so ⟨f, S⟩ = 0, as ⟨ψ, S⟩ = 0 for any such ψ.

We now have the following generalization of 7.6.

21.2. Theorem. Let K ∈ K(𝕋) be a closed set and let S ∈ PF. Then the following are equivalent:

(i) S is supported by K,

(ii) ∑ S(n)e^{inx} = 0 off K.

We will postpone the proof of this for a while. From 21.2 it immediately follows that, letting

M = K(𝕋) \ U, M_{0} = K(𝕋) − U_{0}

be the classes of closed sets of multiplicity and restricted multiplicity, resp., we have

21.3 Corollary. Let E ∈ K(𝕋). Then

(i) E ∈ M ⇔ ∃S ∈ P M(||S||_{∞} ≤ 1 & S ∈ P F & S ≠ 0 & E supports S),

(ii) E ∈ M_{0} ⇔ ∃μ ∈ M(𝕋)(||μ||_{M} ≤ 1 & μ ∈ P F & E supports μ).

Using this one can easily prove the following. 21.4. Theorem. The sets U, U_{0} are Π_{1}^{1}. Proof. It is clearly enough to show that M, M_{0} are Σ_{1}^{1}.

First consider M. By 21.3 (i) the set M ⊆ K(𝕋) is the projection of the following set in K(𝕋)×B_{1} (PM), where B_{1} (PM) is the unit ball of PM (= ℓ^{∞}) with the weak^{∗}-topology:

P = {(K, S) ∈ K(𝕋) × B_{1}(PM) : lim_{|n|→∞} |S(n)| = 0 & S ≠ 0 & K supports S}.

Since for each n, the map S ↦ S(n) = ⟨e^{−inx}, S⟩ is continuous in B_{1} (PM), the condition lim_{|n|→∞} |S(n)| = 0 & S ≠ 0 is clearly Borel. We next claim that

Q = {(K, S) ∈ K(𝕋) × B_{1}(PM) : K supports S}

is closed, which shows that P is Borel, so M is Σ_{1}^{1}.

To see that S is closed, let K_{i}, S_{i} ∈ Q and K_{i} → K (in the Hausdorff metric or, equivalently, the Vietoris topology) and S_{i} → S (in the weak^{∗}-topology, i.e., ⟨f, S_{i}⟩ → ⟨f, S⟩ for each f ∈ A). Now take an interval I disjoint from K and infinitely differentiable φ supported by I, in order to show that ⟨φ, S⟩ = 0.

Let J ⊆ I be a closed interval containing supp(φ). Then if U = 𝕋 \ J we have K ⊆ U, so, by the definition of the Vietoris topology, K_{n} ⊆ U for all large enough n. Thus φ is supported by an interval disjoint from K_{n}, and so ⟨φ, S⟩ = 0 and we are done.

The proof for M_{0} is similar, with B_{1}(M(𝕋)) instead of B_{1} (PM). ⊣ We finally give the proof of 21.2:

Proof of 21.2. (i) ⇒ (ii) By the Riemann Localization Principle it is enough to show that F_{S} is linear in each open interval I disjoint from K. Let a ∈ I and choose h small enough so that ψ_{a,h}, as defined in §12, is supported by I. Then

\begin{aligned}0 & = ⟨ψ_{a,h}, S⟩ = ∑ ψ̂_{a,h}(−n)S(n) \\ & = \frac{Δ^{2}F_{S}(a, h)}{h^{2}}, \end{aligned}

so F_{S} is linear on I.

(ii) ⇒ (i). Let I be an open interval disjoint from K and let φ ∈ C(𝕋) be infinitely differentiable supported by I. We want to show that

⟨φ, S⟩ = ∑ φ̂(n)S(−n) = 0.

Consider the formal product

S(φ) · S = T,

where T (m) = ∑ φ̂(n)S(m − n). Then T (0) = ∑ φ̂(n)S(−n) = ⟨φ, S⟩, so it is enough to show that T = 0. But by the Rajchman multiplication theory

∑(T (n) − φ(x)S(n))e^{inx} = 0,

so, as ∑ S(n)e^{inx} = 0 on I and φ(x) = 0 off I, ∑ T (n)e^{inx} = 0 for all x, so T = 0. ⊣

§22. Co-analytic σ-ideals of compact sets.

We summarize two key properties of the classes U, U_{0}.

(1) U, U_{0} are Π_{1}^{1} (and not Borel).

(2) U, U_{0} are σ-ideals in the following sense: Definition. Let X be a compact metric space and I ⊆ K(X) a class of closed sets in X. We say that I is a σ-ideal of closed sets if it satisfies the following two properties:

(i) K ∈ I, L ⊆ K, L ∈ K(X) ⇒ L ∈ I;

(ii) K_{n} ∈ I, K ∈ K(X), K = ⋃_{n} K_{n} ⇒ K ∈ I.

Kechris, Louveau and Woodin undertook the general study of Π_{1}^{1} σ-ideals of closed sets in compact metric spaces. This theory turned out to have interesting applications to several problems concerning uniqueness sets. I will next discuss some of the main results of this theory and some of its applications.

Before getting into these though, I want to give some examples of σ-ideals. Examples. (i) For A ⊆ X, let

K(A) = {K ∈ K(X) : K ⊆ A},
K_{ω}(A) = {K ∈ K(X) : K ⊆ A, K countable}.

If A is Π_{1}^{1}, then K(A) and K_{ω}(A) are Π_{1}^{1}. The first is easy and the second follows from: 22.1. Exercise. The set K_{ω}(X) is Π_{1}^{1}. Hint. Use the Cantor-Bendixson Theorem.

(ii) I_{meager} = {K ∈ K(X): K is meager (e.g. nowhere dense)} is a σ-ideal, which is in fact G_{δ}.

(iii) If μ is a Borel probability Borel measure on X, then I_{μ} = {K ∈ K(X) : μ(K) = 0} is a σ-ideal, which is also G_{δ}.

(iv) More generally, if M is a class of Borel probability measures on X and M ⊆ P (X) is Σ_{1}^{1} in P (X) with the weak^{∗}-topology (P (X) is defined as in §10, where we dealt with the special case X = 𝕋), then

I_{M} = {K ∈ K(𝕋) : ∀μ ∈ M(μ(K) = 0)} is a Π_{1}^{1} σ-ideal.

(v) U, U_{0} are Π_{1}^{1} σ-ideals. Note that U_{0} = I_{R}, where R = {μ ∈ P (𝕋): μ is a Rajchman measure}.

The first result I will discuss is a surprising dichotomy which limits sharply the possible descriptive complexities of Π_{1}^{1} σ-ideals of closed sets.

22.2. The Dichotomy Theorem (Kechris-Louveau-Woodin). Let I be a Π_{1}^{1} σ-ideal of closed sets in a compact metric space X. Then either I is G_{δ} or else it is not Borel (and thus Π_{1}^{1} but not Σ_{1}^{1}).

We can prove this theorem by applying the following result of Hurewicz. Below, if Y is a topological space and C ⊆ Y, we say that C is a Cantor set if C is homeomorphic to C = 2^{ℕ}.

22.3. Theorem (Hurewicz). Let Y be a Polish space and A ⊆ Y a Π_{1}^{1} set. Then either A is G_{δ} or else there is a Cantor set C ⊆ Y such that C ∩ A is countable dense in C.

Note that exactly one of these possibilities must occur, since if A is G_{δ}, C ∩ A is G_{δ} in C and so cannot be countable dense in C.

Proof of 22.2. Let Y = K(X) and assume I ⊆ Y is not G_{δ}. Then by 22.3 there is a Cantor set C ⊆ Y with C ∩ I countable dense in C. This means that there is a φ : C → C and a countable dense set Q ⊆ C, with φ^{−1}[C ∩ I) = Q. Let f : K(C) → K(X) be defined by f(K) = ⋃ φ′′(K) = ⋃ φ[K]. Clearly f is continuous. Moreover we claim that

K ∈ K(Q) ⇔ f(K) ∈ I,

because if K ⊆ Q, then φ[K] ⊆ I and since φ[K] is countable, ⋃ φ[K] = f(K) ∈ I. Conversely if K ⊈ Q, and x ∈ K \ Q, then φ(x) ∈ φ[K] \ φ[Q], so φ(x) ∈ φ[K] \ I, and, since ⋃ φ[K] ⊇ φ(x), f(K) = ⋃ φ[K] ∉ I.

We have seen in the proof of 17.2 that K(Q) = f^{−1}[I] is not Borel, so I can’t be Borel either. ⊣

This result distinguishes all Π_{1}^{1} σ-ideals of closed sets into two main categories according to descriptive complexity:

(1) The simple ones, which are G_{δ}. Examples include:

(a) K(A), for A ∈ G_{δ};

(b) I_{meager};

(c) I_{μ}.

(2) The complicated ones, which are Π_{1}^{1} but not Borel. Examples include:

(a) K(A), if A is not G_{δ}.

(Proof. Since x ∈ A ⇔ {x} ∈ K(A), if A is not G_{δ}, K(A) cannot be G_{δ}, so it must be complicated.)

(b) (Hurewicz) K_{ω}(A), if A contains a Cantor set (e.g., if A = X and X is uncountable).

(Proof. Let C ⊆ A be a Cantor set. Since K_{ω}(C) = K(C) ∩ K_{ω}(A), it is enough to show K_{ω}(C) is not Borel. If it was, then it would be G_{δ}. But notice that K_{ω}(C) is dense in K(C), since it contains all the finite sets. But K_{p}(C) = {K ∈ K(C): K is perfect} is also dense in G_{δ} in K(C), so it must intersect K_{ω}(C) by the Baire category theorem, which is a contradiction.)

(c) U, U_{0}.

22.4. Exercise. Show that K_{ω}(A) is complicated for any uncountable A ⊆ X, A ∈ Π_{1}^{1}. (If A is countable, clearly K_{ω}(A) = K(A).) [Hint. Consider cases as A is Borel or not. Use the fact that every uncountable Borel set contains a Cantor set.]

Before I proceed I will say a few things about the proof of 22.3. A nice proof of 22.3 can be given using games. I will consider only the case when Y is actually compact metrizable.

First we can reduce the problem to C. For this we use the fact that, since Y is compact metrizable, there is a continuous surjection φ : C → Y. Consider the φ^{−1}[A] = A′. Then A′ is Π_{1}^{1}. Assume the result has been proved for the space C. Then either A′ is G_{δ} or else there is a Cantor set C′ ⊆ C with C′ ∩ A′ countable dense in C′.

In the first case B′ = C \ A′ is F_{σ}, so φ[B′] = X \ A is F_{σ} (since we are working in compact spaces), so A is G_{δ}.

In the second case, let C′′ = φ[C′]. Then C′′ is closed in Y, and A ∩ C′′ is countable dense in C′′. Note also that (Y \ A) ∩ C′′ is dense in C′′, so C′′ is perfect. It is now easy, by a Cantor-type construction, to find a Cantor set C ⊆ C′′ such that C ∩ A is countable dense in C. So it is enough to assume that Y = C.

Fix a countable dense subset Q ⊆ C and consider the following game as in the proof of the lemma in 17.2: I plays x ∈ C, II plays y ∈ C and II wins iff x ∈ Q ⇔ y ∈ A.

If I has a winning strategy, this gives a continuous function f : C → C such that y ∈ A ⇔ f(y) ∉ Q, so A = f^{−1}[C \ Q], and since C \ Q is G_{δ}, so is A. If on the other hand II has a winning strategy, then this gives a continuous function g such that

x ∈ Q ⇔ g(x) ∈ A.

Let g[C] = K. Then notice that K is closed and g[Q], g[C \ Q] are disjoint and dense in K, so K is perfect. Moreover A ∩ K = g[Q], so A ∩ K is countable dense in K. So again, by a simple Cantor-type construction, we can find a Cantor set C ⊆ K with A ∩ C countable dense in C.

So if this game is determined, i.e., one of the players has a winning strategy, the proof is complete. The payoff set of this game, i.e., the set

{(x, y) : x ∈ Q ⇔ y ∈ A}

is a Boolean combination of Π_{1}^{1} sets (in C ×C), so by a theorem of Martin, the determinacy of this game follows from appropriate large cardinal axioms in set theory. However, we cannot prove the determinacy of such complex games in classical set theory (ZFC), since the best result provable in it is the determinacy of all Borel games (Martin). This problem can be handled by considering an appropriate modification of this game, which still does the job, and turns out to be Borel, in fact a Boolean combination of F_{σ} sets, so its determinacy can be established in classical set theory (see Kechris [1995], §21].

§23. Bases for σ-ideals.

Definition. Let I be a σ-ideal of closed sets in a compact metric space X. A basis for I is a subset B ⊆ I such that B is hereditary, i.e., K ∈ B, L ⊆ K, L ∈ K(X) ⇒ L ∈ B, and B generates I, i.e., for any K ∈ I, ∃{K_{n}} ⊆ B with K = ⋃_{n} K_{n}.

We will consider here the question of whether a given σ-ideal admits a Borel basis. The motivation comes again from the Characterization Problem for U.

Although one cannot hope to find a very explicit characterization of when a closed set K ⊆ 𝕋 is in U or not, it may still be possible to find a simple subclass B of U, like e.g., the H-sets that we considered in §14, so that every U-set can be written on a countable union of sets in B. Such questions have been raised in this subject periodically. For example, it was indeed considered whether every U set can be written as a countable union of Hsets or more generally a countable union of so-called H^{(n)}-sets, a generalization of H-sets (H = H^{(1)})). The answer turned out to be negative in this case (Piatetski-Shapiro). The general philosophy is the following: Is it possible to understand U-sets as countable unions of some explicitly characterizable subclass? This can then be formalized as follows:

The Basis Problem. Does the σ-ideal U of closed uniqueness sets admit a Borel basis?

A negative answer would provide a much stronger limitative result concerning the characterization problem. But it would also be a powerful existence theorem (again a use of the descriptive method): Given any simply definable (i.e., Borel) hereditary collection of closed uniqueness sets B, there exists a K ∈ U which cannot be written as a countable union of sets in B. For example, since the H-sets (and the H^{(n)}-sets) can be easily shown to form a Borel class, this would immediately imply the result of Piatetski-Shapiro quoted earlier. But instead of relying on ad-hoc constructions to deal with existence of such examples for any potentially proposed class B, a negative answer to the Basis Problem would once and for all deal with all such (reasonable) possibilities without such constructions.

A similar basis problem can of course be raised for the σ-ideal U_{0}.

Our main goal here is to develop a method for demonstrating non-existence of Borel bases. In fact, the main result that I will prove below establishes (under certain conditions) an important and quite strong property that all σ-ideals with Borel bases must necessarily have. This can be used to prove non-existence of Borel bases by showing that a given σ-ideal fails to have this strong property. This is how one shows that U has no Borel basis. But on the other hand, if it happens to be the case that one deals with an ideal that has a Borel basis (and, as it turns out, U_{0} is such an example) this establishes this very strong property. And this is how the original solution of the Category Problem came about.

Before I proceed I would like to mention a couple of examples. Examples. (i) Every G_{δ} σ-ideal has a Borel basis (namely itself).

(ii) K_{ω}(X) has a Borel basis, namely {∅} ∪ {{x} : x ∈ X}.

(iii) K(A) has no Borel basis, if A is Π_{1}^{1} but not Borel (since for any basis B ⊆ K(A), x ∈ A ⇔ {x} ∈ B). In fact it turns out K(A) has a Borel basis iff A is the difference of two G_{δ} sets (Kechris-Louveau-Woodin).

For any hereditary B ⊆ K(X) let

B_{σ} = {K ∈ K(X) : K = ⋃_{n} K_{n}, K_{n} ∈ B},

be the σ-ideal generated by I. Thus B is a basis for I iff I = B_{σ}. 23.1. Exercise. For B ⊆ K(X) hereditary and any K ∈ K(X) define the B-derivative K′_{B} of F by

K′_{B} = \{x ∈ K : ∀ open V (x ∈ V ⇒ \overline{K ∩ V} ∉ B)\}.

(Notice that for B = {∅} ∪ {{x} : x ∈ B}, K′_{B} = K′). Then by transfinite induction define K_{B}^{(0)} = K, K_{B}^{(α+1)} = (K_{B}^{(α)})′, K_{B}^{(λ)} = ⋂_{α<λ} K_{B}^{(α)}. There is again a countable ordinal α_{0} such that K_{B}^{(α_{0})} = K_{B}^{(β)}, ∀β ≥ α_{0}. The least one is denoted by rk_{B}(K) and called the Cantor-Bendixson rank of K associated to B. Put K_{B}^{(∞)} = K_{B}^{(rk_{B}(K))}. Show that

K ∈ B_{σ} ⇔ K_{B}^{(∞)} = ∅.

Call K ∈ K(X) B-perfect if K′_{B} = K, i.e., ∀ open V (K ∩ V ≠ ∅ ⇒ \overline{K ∩ V} ∉ B). Show the analog of Cantor-Bendixson, namely that any K can be uniquely written as

K = P ∪ C

with P B-perfect and C contained in a countable union of sets in B. Show that P = K \ ⋃\{V open: \overline{K ∩ V} ∈ B_{σ}\} = K_{B}^{(∞)}.

Show that if B is Π_{1}^{1}, then

{K ∈ K(X) : K is B−perfect}

in Σ_{1}^{1}. [Hint. Show that f : Y → K(X) is Borel iff for any open W ⊆ Y, {y : f(y)∩W ≠ ∅} is Borel. Use this to show that K ↦ \overline{K ∩ V} is Borel, for any open V ⊆ X.]

Conclude that if B ⊆ K(X) is hereditary Π_{1}^{1}, then B_{σ} is also Π_{1}^{1}.

(In particular, this shows that only Π_{1}^{1} σ-ideals can have Borel (in fact even Π_{1}^{1}) bases.)

If B is Borel show that the map

K ↦ K′_{B}

is Borel. Thus one has a “semi-Borel” test for membership in B_{σ} : K ∈ B_{σ} iff the transfinite iteration of the Borel operation K ↦ K′_{B} terminates after countably many steps (depending on K) with the empty set. Remark. It can be also shown that the following are equivalent for any Π_{1}^{1} σ-ideal I:

(i) I admits a Borel basis, (ii) I admits a Σ_{1}^{1} basis. (iii) There is B ⊆ I (not necessarily hereditary), B ∈ Σ_{1}^{1}, such that I = {K ∈ K(X):

∃{K_{n}} ⊆ B(K ⊆ ⋃_{n} K_{n})}.

Before I state the main result I need one more definition. Definition. A σ-ideal I of closed sets is calibrated if for any closed set F, and any sequence F_{n} ∈ I if K(F \ ⋃_{n} F_{n}) ⊆ I, then F ∈ I. Examples. (i) I_{meager} (in any perfect X) is not calibrated. Because if {x_{n}} is dense, F_{n} = {x_{n}} ∈ I_{meager}, K(X \ ⋃_{n} F_{n}) ⊆ I_{meager}, but X ∉ I_{meager}.

(ii) K_{ω}(X) is calibrated (since every uncountable G_{δ} set contains an uncountable closed set).

(iii) If M ⊆ P (X), then I_{M} is calibrated (since for every μ ∈ P (X), every Borel set of positive μ-measure contains a closed set of positive μ-measure).

Thus calibration can be thought of as a (weak) generalization of the idea of inner regularity of measures.

We now have the following result. 23.2. The Basis Theorem (Kechris-Louveau-Woodin) Let I be a calibrated σ-ideal of closed subsets of X. Assume I admits a non-trivial basis, in the local sense that there is a basis B ⊆ I such that for every open ∅ ≠ V ⊆ X, there is K ⊆ V, K ∈ I \ B. Then if A ⊆ X has the Baire property and K(A) ⊆ I (i.e., every closed subset of A is in I), then A is meager. Proof. Assume A is not meager. Then, as A has the property of Baire, there is an open set U ≠ ∅ on which A is comeager, so A contains a G_{δ} set G which is dense in U. We will derive a contradiction by showing that there is K ∉ I, K ∈ K(X), K ⊆ G.

To simplify the notation, we will also assume that U = X (otherwise we can do the construction below within U). So the context is the following:

G ⊆ X is dense G_{δ} and we want to construct K ∈ K(X), K ∉ I, K ⊆ G.

Notice first that every K ∈ I is meager: Otherwise K = ⋃_{n} K_{n}, K_{n} ∈ B, and so for some K_{n} ∈ B, K_{n} is non-meager, so there is non-∅ open V with V ⊆ K_{n}. Thus

I ∩ K(V) = B ∩ K(V)(= K(V)), a contradiction.

Notice also that if K ∈ K(X) is meager and V is open with K ⊆ V, then there is a countable set of points, say D(K, V), with no point of D(K, V) a limit point of D(K, V), D(K, V) ⊆ (G ∩ V) \ K, and \overline{D(K, V)} = K ∪ D(K, V). To see this, let {d_{1}, d_{2}, ⋯} ⊆ K be dense in K and let for each n, x_{1}^{(n)}, ⋯, x_{n}^{(n)} be points of (G ∩ V) \ K which have distance < \frac{1}{n} from d_{1}, ⋯, d_{n}, resp. (We are using here that no open ball can be contained in K, and that G is dense in V.) Let D(K, V) = {x_{i}^{(n)} : n ≥ 1, i ≤ n}.

Finally, from our hypothesis, for each nonempty open U ⊆ X there is compact K_{U} ⊆ U, K_{U} ∈ I \ B.

Let now G = ⋂_{n} W_{n}, W_{n} ⊇ W_{n+1}, W_{n} dense open. We will construct for each s ∈ 2^{<ℕ}, by induction on |s| = length(s), a compact set K_{s} and an open set U_{s} satisfying the following:

(i) U_{s} ≠ ∅, \overline{U}_{s} ⊆ W_{|s|}, K_{s} = K_{U_{s}} (so K_{s} ∈ I \ B);
(ii) \overline{U_{sˆ n}} ⊆ U_{s}, \overline{U_{sˆ n}} ∩ \overline{⋃_{m≠n} U_{sˆ m}} = ∅;
(iii) K_{s} ∩ \overline{U_{sˆ n}} = ∅;
(iv) diam(U_{sˆ n}) ≤ min\{2^{−|s|}, \frac{1}{n}\};
(v) \overline{⋃_{s} U_{sˆ n}} = ⋃_{n} \overline{U_{sˆ n}} ∪ K_{s};
(vi) K_{s} ⊆ \overline{⋃_{n} K_{sˆ n}}.

Step 1. U_{∅} = W_{0}, K_{∅} = K_{U_{∅}}.

Step k + 1. Suppose we have constructed U_{s}, K_{s} for |s| ≤ k, satisfying (i)–(vi). Let for s ∈ ℕ^{k}, D(K_{s}, U_{s}) = {x_{sˆ n} : n ∈ ℕ} and let U_{sˆ n} ⊆ W_{k+1} be a small enough open set containing x_{sˆ n} so that diam(U_{sˆ n}) < min\{\frac{1}{n}, 2^{−|s|}\} and (ii), (iii), (iv) are satisfied. This can be done as no point of D(K_{s}, U_{s}) in a limit point of D(K_{s}, U_{s}), D(K_{s}, U_{s}) ⊆ (G ∩ U_{s}) \ K ⊆ (W_{n+1} ∩ U_{s}) \ K_{s}, and \overline{D(K_{s}, U_{s})} = D(K_{s}, U_{s}) ∪ K_{s}. It also follows that \overline{⋃_{n} U_{sˆ n}} = ⋃_{n} \overline{U_{sˆ n}}∪K_{s}. Also if K_{sˆ n} = K_{U_{sˆ n}} clearly K_{s} ⊆ \overline{⋃_{n} K_{sˆ n}}, as K_{s} ⊆ \overline{D(K_{s}, U_{s})}.

Let H = ⋂_{n} ⋃_{s∈ℕ^{n}} U_{s}, K = H ∪⋃_{s} K_{s}. Clearly H is G_{δ} and as ⋃_{s∈ℕ^{n}} U_{s} ⊆ W_{n}, H ⊆ G.

Claim. K is closed.

Proof. It is enough to show that if L = ⋂_{n} \overline{⋃_{s∈ℕ^{n}} U_{s}}, then K = L. Clearly H ⊆ ⋂_{n} \overline{⋃_{s∈ℕ^{n}} U_{s}}. Since, by (vi), K_{s} ⊆ \overline{⋃_{n} K_{sˆ n}}, for all s we have that K_{s} ⊆ \overline{⋃_{n} K_{sˆ n}}, K_{s} ⊆ \overline{⋃_{n,m} K_{sˆ nˆ m}}, ⋯ so if |s| = k, K_{s} ⊆ \overline{⋃_{s∈ℕ^{k+1}} U_{s}}, K_{s} ⊆ \overline{⋃_{s∈ℕ^{k+2}} U_{s}}, ⋯, i.e., K_{s} ⊆ L and so ⋃_{s} K_{s} ⊆ L. Thus K ⊆ ⋂_{n} \overline{⋃_{s∈ℕ^{n}} U_{s}}. Let now x ∈ ⋂_{n} \overline{⋃_{s∈ℕ^{n}} U_{s}}, in order to show that x ∈ K. If x ∈ ⋃_{s} K_{s} we are done. So assume x ∉ ⋃_{s} K_{s}, in order to show that x ∈ H. Then, since x ∈ \overline{⋃_{n} U_{(n)}}, we have, by (v), that x ∈ ⋃_{n} \overline{U_{n}}, so x ∈ \overline{U_{(n_{0})}} for some n_{0} ∈ ℕ. Again x ∈ \overline{⋃_{s∈ℕ^{2}} U_{s}}, so, as x ∈ \overline{U_{(n_{0})}}, by (ii), x ∈ \overline{⋃_{n} U_{(n_{0},n)}}, and thus by (v) again x ∈ ⋃_{n} \overline{U_{(n_{0},n)}}, so x ∈ \overline{U_{(n_{0},n_{1})}} for some n_{1} ∈ ℕ, etc. Thus ∀k(x ∈ \overline{U_{(n_{0},n_{1},···,n_{k},n_{k+1})}} ⊆ U_{(n_{0},n_{1}···,n_{k})}), so x ∈ ⋂_{k} ⋃_{s∈ℕ^{k}} U_{s}, i.e., x ∈ H.

Claim. K ∉ I: Otherwise, K = ⋃_{n} K_{n}, K_{n} ∈ B, so by the Baire Category Theorem there is open U_{0}, and n_{0} with ∅ ≠ U_{0} ∩ K ⊆ K_{n_{0}}, so \overline{U_{0} ∩ K} ∈ B. Let x ∈ U_{0} ∩ K. If x ∈ H, then there is unique α ∈ ℕ^{ℕ} with x ∈ U_{α|n} for each n. Since diam(U_{α|n}) → 0, there is n with U_{α|n} ⊆ U_{0}, so K_{α|n} ⊆ U_{0} and thus K_{α|n} ⊆ U_{0} ∩ K, so, as K_{α|n} ∉ B, \overline{U_{0} ∩ K} ∉ B, a contradiction. If now x ∈ K_{s} for some s, then, by (v) and (iv), K_{sˆ n} ⊆ U_{0} for some n, so K_{sˆ n} ⊆ U_{0} ∩ K, and again \overline{U_{0} ∩ K} ∉ B, a contradiction.

Since K = H ∪ ⋃_{s} K_{s} and K_{s} ∈ I, while K ∉ I, it follows, by calibration, that K(K\⋃_{s} K_{s}) ⊈ I, so as K\⋃_{s} K_{s} ⊆ H, K(H) ⊈ I, i.e., there is K ⊆ H, K ∈ K(X), K ∉ I. But H ⊆ G, so ∃K ∈ K(X), K ⊆ H, K ∉ I and the proof is complete. ⊣

The following is an important application of the Basis Theorem.

23.3 The Covering Theorem (Debs-Saint Raymond). Let I be a σ-ideal of closed sets in X. Assume

(i) I is calibrated,

(ii) I admits a basis B such that for any L ∈ K(X)\I, there is K ∈ K(L), K ∈ I \B.

Then for any Σ_{1}^{1} set A ⊆ X, if K(A) ⊆ I, then there is a sequence {K_{n}} ⊆ I with A ⊆ ⋃_{n} K_{n}.

Proof. Assume A ⊆ X is Σ_{1}^{1}, and A cannot be covered by countably many sets in I. We will show that there is K ∈ K(X), K ∉ I with K ⊆ A.

By 17.1, let G ⊆ X × C be G_{δ} such that A = proj_{X}[G]. Let G′ = G \ ⋃{V open in X × C: proj_{X}[V ∩ G] can be covered by countably many sets from I}. Since A cannot be covered by countably many sets from I, G′ ≠ ∅. Let F = \overline{G′}, so that F is compact. We define the following σ-ideal J ⊆ K(F):

K ∈ J ⇔ proj_{X}[K] ∈ I.

We claim that J satisfies the hypothesis of the Basis Theorem 23.2, i.e., is calibrated and has a non-trivial basis. It will follow then (as G′ is dense in G_{δ} in F) that there is compact L ⊆ F with L ⊆ G′ and L ∉ J, i.e., proj_{X}[L] = K ∈ K(A) \ I, so we are done.

J is calibrated: Let K ∈ K(F), {K_{n}} ⊆ J, K(K \ ⋃_{n} K_{n}) ⊆ J. Consider proj_{X}[K] \ ⋃_{n} proj_{X}[K_{n}]. We have proj_{X}[K_{n}] ∈ I. If K ∉ J, towards a contradiction, then proj_{X}[K] ∉ I, so by the calibration of I, proj_{X}[K] \ ⋃_{n} proj_{X}[K_{n}] contains a compact set L with L ∉ I. Then K ∩ (L × C) ⊆ K \ ⋃_{n} K_{n} is not in J (as proj_{X}[K ∩ (L × C)] = L), a contradiction.

J has a non-trivial basis: Let D = {K ∈ K(F): proj_{X}[K] ∈ B}. Then D is a basis for J. If V is an open set intersecting F, then, by the definition of G′, proj_{X}[V ∩ G′] cannot be covered by countably many sets in I (otherwise proj_{X}[V ∩G] can be so covered, so V ∩G′ = ∅, contradicting the density of G′ in F). So L = \overline{proj_{X}[V ∩ F]} = \overline{proj_{X}[V ∩ G′]} ∉ I. Then there is K ⊆ L, K ∈ I \ B. Now L = \overline{proj_{X}[V ∩ F]} = proj_{X}[\overline{V ∩ F}] (by compactness), so by looking at (K × C) ∩ [\overline{V ∩ F}] we conclude that J ∩ K(\overline{V ∩ F}) ≠ D ∩ K(\overline{V ∩ F}). Since this is true for all open V that intersect F, it follows that J ∩ K(U) ≠ D ∩ K(U) for all non-∅ U open relative to F and the proof is complete. ⊣

The following corollary gives a definability context under which 23.3 can be applied.

23.4. Corollary. Let I be a σ-ideal of closed sets in X. Assume

(i) I is calibrated;

(ii) if L ∈ K(X) \ I, then I ∩ K(L) is not Borel;

(iii) I admits a Borel basis.

Then for any Σ_{1}^{1} set A ⊆ X, if K(A) ⊆ I, then there is a sequence {K_{n}} ⊆ I with A ⊆ ⋃_{n} K_{n}.

§24. Non-existence of Borel bases for U.

We will now apply the methods of §23 to the σ-ideal U of closed sets of uniqueness.

Debs and Saint Raymond used a deep result from harmonic analysis due to Körner (the existence of so-called Helson sets of multiplicity), to show that there is a closed set E such that for every open W with W ∩ E ≠ ∅, \overline{W ∩ E} ∉ U and a G_{δ} set G ⊆ E which is dense in E such that K(G) ⊆ U. This is one of the ingredients needed to apply 23.2. The other two ingredients are

(i) If E ∈ M = K(𝕋) \ U, then U ∩ K(E) is not Borel.

This is a local version of the Kaufman-Solovay Theorem 18.1 and has been proved independently by Debs-Saint Raymond, Kaufman and Kechris-Louveau.

(ii) U is calibrated.

This was proved independently by Debs-Saint Raymond and Kechris-Louveau. The proofs of all these results are given in Kechris-Louveau [1989].

Putting all them together we have:

24.1. Theorem (Debs-Saint Raymond). The σ-ideal U of closed sets of uniqueness has no Borel basis.

Proof. Assume U had a Borel basis B ⊆ U. Let E ∈ K(𝕋) be such that for every open W with W ∩ E ≠ ∅ we have \overline{W ∩ E} ∉ U, but there is a G_{δ} set G ⊆ E, dense in E, with K(G) ⊆ U. Now consider

I = U ∩ K(E).

It is a Π_{1}^{1} σ-ideal of closed subsets of K(E). It is (i) calibrated, (ii) for every L ∈ K(E), L ∉ I, I ∩ K(L) is not Borel and (iii) I admits a Borel basis, namely B ∩ K(E). So by 23.4 applied to A = G we must have that there is a sequence K_{n} ∈ I with G ⊆ ⋃_{n} K_{n}. But each K_{n} is meager in E, since otherwise there would be W open with W ∩ E ≠ ∅ and W ∩ E ⊆ K_{n}, so \overline{W ∩ E} ⊆ K_{n}, a contradiction since \overline{W ∩ E} ∈ M and K_{n} ∈ U. So G is meager, a contradiction. ⊣

This result, as we explained earlier, has very strong implications concerning the Characterization Problem for U-sets. One cannot characterize U-sets as countable unions of any reasonably explicitly characterizable subclass, e.g., H-sets, H^{(n)}-sets, etc. Or, one can use this as an existence theorem: Given any reasonable explicitly characterizable sublcass of U-sets, say B, there is a closed set E ∈ U which is not a countable union of sets in B. Thus this gives a new proof that for each n there are U-sets which are not countable unions of H^{(n)}-sets (a result originally due to Piatetski-Shapiro) or that there are U-sets which are not countable unions of H^{(n)}-sets for varying n (a new result), etc.

§25. Existence of a Borel basis for U_{0}.

Recall that U_{0} is the class of all closed sets of extended uniqueness, i.e., K ∈ U_{0} if for every measure μ ∈ M(𝕋), if ∑ μ̂(n)e^{inx} = 0 off K, then μ̂(n) = 0, ∀n ∈ ℤ (i.e., μ = 0). By 21.3, this is equivalent to saying that there is no μ ∈ M(𝕋), μ ≠ 0, with μ̂(n) → 0 which is supported by K. By using a bit of additional measure theory we can see that in this characterization we can restrict ourselves to probability Borel measures, i.e., μ ∈ P (𝕋). To see this, it is enough to check that if μ ≠ 0, μ ∈ M(𝕋), μ̂(n) → 0 and μ is supported by K, then there is μ ∈ P (𝕋) with μ̂(n) → 0 also supported by K.

For this let |μ| be the so-called total variation of μ. This is the finite positive Borel measure defined by

|μ|(A) = sup ∑_{i=1}^{∞} |μ(E_{i})|,

where the sup varies over all Borel partitions of A. Thus

||μ||_{M} = |μ|(𝕋).

Moreover, by the Radon-Nikodym Theorem, there is a Borel function h : 𝕋 → ℂ with |h| = 1 such that

μ(E) = ∫ hd|μ|,

and so

∫ fdμ = ∫ fhd|μ|

for any bounded Borel f. Thus, in particular,

∫ fh^{−1}dμ = ∫ fd|μ|.

Note also that | ∫ fdμ| ≤ ∫ |f|d|μ|.

It follows that |μ| is supported by K. It is thus enough to show that ˆ|μ|(n) → 0. (Although |μ| might not be a probability measure, ν = |μ|/|μ|(𝕋) will be, it will have the same support as |μ|, and ν̂(n) → 0 as well.)

Fix ϵ > 0. As h^{−1} ∈ L_{1}(|μ|) (since ∫ |h^{−1}|d|μ| = ∫ d|μ| = |μ|(𝕋) < ∞), there is a trigonometric polynomial P (x) = ∑_{k=−N}^{N} c_{n}e^{ikx} with

||h^{−1} − P ||_{L^{1}(|μ|)} = ∫ |h^{−1} − P |d|μ| < ϵ.

Then

∫ e^{−inx}d|μ|(x) = ∫ e^{−inx}h^{−1}dμ(x)
= ∫ e^{−inx}P (x)dμ(x)−
∫ e^{−inx}(P (x) − h^{−1}(x))dμ(x)
= ∑_{k=−N}^{N} c_{k}μ̂(n − k)−
∫ e^{−inx}(P (x) − h^{−1}(x))dμ(x).

But

| ∫ e^{−inx}(P (x) − h^{−1}(x))dμ(x)| ≤ ∫ |h^{−1} − P |d|μ| < ϵ

and, since μ̂(n) → 0 as |n| → ∞, it follows that there is N_{ϵ} such that for |n| ≥ N_{ϵ},

| ∑_{k=−N}^{N} c_{n}μ̂(n − k)| < ϵ,

so for |n| ≥ N_{ϵ}, | ∫ e^{inx}d|μ|(x)| < 2ϵ and thus ˆ|μ|(n) → 0 as |n| → ∞.

Thus we have (recalling that R = {μ ∈ P (𝕋) : μ̂(n) → 0}): 25.1. Proposition. For any K ∈ K(𝕋),

K ∈ U_{0} ⇔ ∀μ ∈ P (𝕋)(K supports μ ⇒ μ ∉ R).

For μ ∈ P (𝕋) let

R(μ) = \overline{lim}|μ̂(n)|.

Thus 0 ≤ R(μ) ≤ 1 and

μ ∈ R ⇔ R(μ) = 0.

Put

U′_{0} = {K ∈ K(𝕋) : ∃ϵ > 0∀μ ∈ P (𝕋)(K supports μ ⇒ R(μ) ≥ ϵ).

Thus clearly U′_{0} ⊆ U_{0}. (Note that U_{0} = {K ∈ K(𝕋) : ∀μ ∈ P (𝕋) (K supports μ ⇒ R(μ) > 0)}.)

For example, it turns out that the Cantor set E_{1/3} is in U′_{0}. In fact we have:

25.2. Proposition. Every closed H-set is in U′_{0}.

Proof. Let E be a closed H-set and let 0 < n_{0} < n_{1} < ⋯ be a sequence and I an open interval with n_{i}x ∉ I for any x ∈ E. Let φ ∈ A be a function supported by some closed interval contained in I and φ̂(0) = 1. Put f_{k}(x) = φ(n_{k}x). Then f_{k}(x) = 0 if x ∈ E, so for any μ ∈ P (E), ∫ f_{k}dμ = ⟨f_{k}, μ⟩ = ∑ f̂_{k}(n)μ̂(−n) = 0. Note that f̂_{k}(0) = 0 and f̂_{k}(n) → 0 as k → ∞, for any n ≠ 0.

Take ϵ > 0 such that ϵ · ||φ||_{A} < 1. We claim that if μ ∈ P (E), then R(μ) ≥ ϵ. Otherwise let μ ∈ P (E) be such that R(μ) < ϵ. Then fix N ∈ ℕ so that |μ̂(n)| < ϵ for any |n| > N. We have

\begin{aligned}0 & = ∑ f̂_{k}(n)μ̂(−n) \\ & = 1 + ∑_{n≠0_{n=−N}}^{N} f̂_{k}(n)μ̂(−n) + ∑_{|n|>N} f̂_{k}(n)μ̂(−n), \end{aligned}

thus

1 ≤| ∑_{n≠0_{n=−N}}^{N} f̂_{k}(n)| + ∑_{|n|>N} |f̂_{k}(n)| · ϵ.

The first summand can be made arbitrarily small by letting k → ∞ and the second is bounded by ||φ||_{A} · ϵ < 1, so we have a contradiction. ⊣

25.3. Proposition. U′_{0} is Borel and hereditary.

Proof. We have for K ∈ K(𝕋)

K ∉ U′_{0} ⇔ ∀ϵ∃μ ∈ P (𝕋)(K supports μ & R(μ) < ϵ)
⇔ ∀ϵ∃μ ∈ P (𝕋)(K supports μ & ∃n∀m(|m| > n ⇒ |μ̂(m)| ≤ ϵ)).

Now P = {(K, μ) : K supports μ & ∃n∀m(|m| > n ⇒ |μ̂(m)| ≤ ϵ)} is F_{σ} in K(𝕋) × P (𝕋), so K(𝕋) \ U′_{0} is Π_{3}^{0}, and thus U′_{0} is Σ_{3}^{0}.

That U′_{0} in hereditary is obvious. ⊣

25.4. Theorem (Kechris-Louveau). U′_{0} is a basis for U_{0}, so U_{0} admits a Borel basis.

Proof. Let E ∈ K(𝕋), E ∉ (U′_{0})_{σ}. We will show that E ∉ U_{0}. Since K ∉ (U′_{0})_{σ}, by 23.1, there is ∅ ≠ F ⊆ E, F ∈ K(𝕋) which is U′_{0}-perfect, i.e., for any open V.

F ∩ V ≠ ∅ ⇒ \overline{F ∩ V} ∉ U′_{0},

i.e., F ∩ V ≠ ∅ ⇒ ∀ϵ∃μ ∈ P (𝕋) (\overline{F ∩ V} supports μ and R(μ) < ϵ). We will find μ ∈ R supported by F, so F ∉ U_{0} and thus E ∉ U_{0}.

Consider first P (F), the set of probability measures supported by F. This is a closed subset of P (𝕋) (always equipped with the weak^{∗}-topology).

Let

R_{ϵ} = {μ ∈ P (𝕋) : R(μ) < ϵ}

and

R_{ϵ}(F) = P (F) ∩ R_{ϵ}.

We claim that R_{ϵ}(F) is dense in P (F). Since R_{ϵ}(F) is convex, it suffices (by 10.14) to show that every Dirac measure δ_{x}, x ∈ F, is in the closure of R_{ϵ}(F). Let (by 10.4) V_{n} be a sequence of nbhds of x with diam(V_{n}) → 0. Then \overline{V_{n} ∩ F} ∉ U′_{0}, so there is μ_{n} ∈ P (\overline{V_{n} ∩ F}) with μ_{n} ∈ R_{ϵ}(F). But μ_{n} ⟶^{w^{∗}} δ_{x} (see §11) and so we are done.

We will now construct, by induction on n, a sequence μ_{n} ∈ P (F), and 0 < N_{0} < N_{1} < N_{2} < ⋯ < N_{n} < ⋯ such that for each n:

(∗)  i ≤ n, N_{i} ≤ |k| < N_{i+1} ⇒ |μ̂_{n}(k)| < 2^{−i−1}.

Then, by the compactness of P (F), let μ be a w^{∗}-limit of a subsequence {μ_{n_{j}}} of {μ_{n}}. We clearly have (as μ̂_{n_{j}}(k) → μ̂(k) for every k) that |μ̂(k)| ≤ 2^{−i−1} if |k| ≥ N_{i}, so μ ∈ R and μ ∈ P (F), thus we are done.

To construct {μ_{n}}, {N_{n}} we will actually choose μ_{n} to satisfy (∗) for i ≤ n − 1 and

(∗∗)  ∀|k| ≥ N_{n}(|μ̂_{n}(k)| < 2^{−n−2}).

(Then (∗) will be satisfied for i = n no matter what N_{n+1} is).

n = 0 . Find μ_{0} ∈ P (F) so that μ_{0} ∈ R_{2^{−2}}(F) and then choose N_{0} so that |μ̂_{0}(k)| < 2^{−2} for |k| ≥ N_{0}.

n → n + 1 . Suppose μ_{0}, ⋯ μ_{n}, N_{0}, ⋯, N_{n} have been defined satisfying (∗) for i ≤ n − 1 and (∗∗). Let m ≥ N_{n}. Then there is μ^{m} ∈ P (F) and φ(m) > m such that

(i) μ^{m} satisfies (∗) for i ≤ n − 1,
(ii) |μ^{m}(k)| < 2^{−n−2} for N_{n} ≤ |k| < m,
(iii) |μ^{m}(k)| < 2^{−n−3} for |k| ≥ φ(m).

This is because μ_{n} satisfies (i), (ii) and so, by the density of R_{2^{−n−3}}(F), there is μ^{m} ∈ R_{2^{−n−3}}(F) satisfying (i), (ii). Then choose φ(m) to make (iii) true.

Now define a sequence ν_{j} ∈ P (F) and m_{j} by

\begin{aligned}ν_{0} & = μ^{N_{n}}, m_{0} = φ(N_{n}), \\ ν_{j+1} & = μ^{m_{j}}, m_{j+1} = φ(m_{j}). \end{aligned}

Let for each k

θ_{k} = \frac{1}{k + 1} ∑_{j=0}^{k} ν_{j}.

Then θ_{k} satisfies (∗) for i ≤ n − 1 and |θ̂_{k}(m)| < 2^{−n−3} for |m| ≥ m_{k}. If

N_{n} ≤ |m| < m_{k},

there is at most one j, namely the one such that m_{j} ≤ |m| < m_{j+1} for which |ν_{j}(m)| ≥ 2^{−n−2}. So (as always |{ \atop ν_{j}(p)}| ≤ 1),

|θ̂_{k}(m)| ≤ \frac{k · 2^{−n−2} + 1}{k + 1}.

Choose then k large enough so that \frac{k·2^{−n−2}+1}{k+1} < 2^{−n−1}. Put μ_{n+1} = θ_{k}, N_{n+1} = m_{k}. Then clearly μ_{n+1} satisfies (∗) for i ≤ n − 1 (as each ν_{j} does). Also for N_{n} ≤ |k| < N_{n+1}, |μ̂_{n+1}(k)| < 2^{−n−1}, so μ_{n+1} satisfies (∗) for i ≤ n, and finally it clearly satisfies (∗∗), i.e., ∀|k| ≥ N_{n+1}(|μ̂_{n+1}(k)| < 2^{−n−3}). ⊣

We are now in a position to apply 23.3. It is clear that U_{0} is calibrated. Kaufman had showed that for every L ∈ M_{0}, there is K ⊆ L, K ∈ U_{0} \ U′_{0} (see also §27 below). So by 23.3 we have:

25.5. Theorem (Debs-Saint Raymond). For each Σ_{1}^{1} set A ⊆ 𝕋, if A ∈ U_{0} there is a sequence K_{n} ∈ U_{0} with A ⊆ ⋃_{n} K_{n}. In particular, A is of the first category.

Proof. Recall that by definition

A ∈ U_{0} ⇔ ∀μ ∈ M(𝕋)(∑ μ̂(n)e^{inx} = 0 off E ⇒ μ̂(n) = 0, ∀n ∈ ℤ).

Then we must have for every Rajchman measure μ ∈ R, μ(A) = 0. (Otherwise, μ(A) > 0, since every Σ_{1}^{1} set is μ-measurable (see Kechris [1995, 29.7]. Then there is F ∈ K(𝕋) with μ(F) > 0. Let ν = (μ|F)/μ(F). Then by 9.3, ν̂(n) → 0 and so, by 7.6, ∑ ν̂(n)e^{inx} = 0, ∀x ∉ F, and so ∀x ∉ A, a contradiction.) So by 23.3 there are K_{n} ∈ U_{0}, with A ⊆ ⋃_{n} K_{n}. ⊣

25.6. Corollary. Let A ⊆ 𝕋 have the BP and be in U_{0}. Then A is of the first category.

Proof. If A is not of the first category, then A is comeager in some open set U, so A contains a G_{δ} set G which is dense in U. Then G is obviously Σ_{1}^{1} and in U_{0} but not of the first category, a contradiction. ⊣

This was the original solution of the Category Problem 9.1 (which in fact established the stronger version about sets of extended uniqueness with the BP being of the first category). In §11 we gave a different proof based on Baire Category methods. Such a technique can be also used to give a proof of 25.5 as well.

In conclusion we can summarize as follows the key structural and definability properties of U, U_{0}:

For I either U or U_{0} we have

(i) I is a calibrated σ-ideal,

(ii) I is Π_{1}^{1}, but for any F ∈ K(𝕋), F ∉ I, I ∩ K(F) is not Borel.

These are properties that both U and U_{0} share. However they differ in one important respect.

(iii) U_{0} has a Borel basis, but U does not have a Borel basis.

It should be noted that the three properties (i)–(iii) encapsulate a large part of the theory of sets of uniqueness (and extended uniqueness). For example, they imply Menshov’s Theorem (existence of null closed sets of multiplicity); the Debs-Saint Raymond Theorem (sets of extended uniqueness with the BP are of the first category), which in turn has several consequences, like Lyons’ theorem that there are Rajchman measures supported by the non-normal numbers; Piatetski-Shapiro’s Theorem that U ≠ U_{0}, etc.

§26. Co-analytic ranks.

The last topic in these lectures will bring us back full circle to the first method we introduced here, ordinals and transfinite induction. Ordinals play a crucial role in classical as well as modern descriptive set theory, through various concepts of rank. For our purposes here, the crucial concept is that of a co-analytic or Π_{1}^{1}-rank. A general reference for descriptive set theoretic results used in this section is Kechris [1995].

A rank on a set A is simply a function φ : A → Ordinals, assigning to each element of A an ordinal number. It is a fundamental property of all co-analytic sets that they admit ranks φ : A → ω_{1} (= the set of all countable ordinals) with very nice definability properties. Roughly speaking, such φ exist for which the initial segments

{x ∈ A : φ(x) ≤ φ(y)},

for y ∈ A, are “uniformly” Δ_{1}^{1}. Let me be more precise.

Definition. Let X be a Polish space and A ⊆ X a Π_{1}^{1}-set. A Π_{1}^{1}-rank on A is a map φ : A → ω_{1} which has the following property:

There are P, S ⊆ X^{2}, in Π_{1}^{1}, Σ_{1}^{1}, resp., such that
y ∈ A ⇒ [x ∈ A & φ(x) ≤ φ(y) ⇔ (x, y) ∈ P ⇔ (x, y) ∈ S].

Thus for y ∈ A,

{x ∈ A : φ(x) ≤ φ(y)} = P^{y} = S^{y}

(where R^{y} = {x : (x, y) ∈ R}), so that the initial segment determined by y is both Σ_{1}^{1} and Π_{1}^{1}, i.e., Δ_{1}^{1}, but in fact in a uniform way.

It is a basic fact of the theory of Π_{1}^{1} sets that they admit Π_{1}^{1}-ranks.

26.1. Theorem. Every Π_{1}^{1} set A admits a Π_{1}^{1}-rank.

Note that if φ : A → ω_{1} is a Π_{1}^{1}-rank on the Π_{1}^{1} set A, and for each countable ordinal ξ we let

A_{ξ} = {x ∈ A : φ(x) ≤ ξ},

then A_{ξ} is Borel. (Proof. Recall that Δ_{1}^{1} = Borel. This is now easily proved by induction on ξ. If ξ = 0, then either A_{0} = ∅, or else if y ∈ A_{0}, clearly A_{0} = {x ∈ A : φ(x) ≤ φ(y)} is Borel. Assume it holds for all ξ < η, and consider A_{η}. If there is no y ∈ A with φ(y) = η, clearly A_{η} = ⋃_{ξ<η} A_{ξ} is Borel, as this is a countable union of Borel sets. If on the other hand there is y ∈ A with φ(y) = η, then clearly A_{η} = {x ∈ A : φ(x) ≤ φ(y)}, so A is again Borel.)

Since

A = ⋃_{ξ<ω_{1}} A_{ξ},

this gives a decomposition of A as a union of ω_{1} many Borel sets. So, although A may not be Borel, it can be “approximated” by Borel sets, in some sense.

An important application of the concept of Π_{1}^{1}-rank is the following:

26.2 Boundedness Theorem. Let A be a Π_{1}^{1} set and φ : A → ω_{1} a Π_{1}^{1}-rank. If B ⊆ A is Σ_{1}^{1}, then there is a countable ordinal ξ such that

φ(x) ≤ ξ

for all x ∈ B.

This result suggests the following technique for establishing the non-Borelness of a given Π_{1}^{1} set, called the rank method: Given a Π_{1}^{1} set A, for which we want to show that it is not Borel (equivalently: not analytic), find a Π_{1}^{1}-rank φ : A → ω_{1} which is unbounded, i.e., for each countable ordinal ξ there is some x ∈ A with φ(x) > ξ.

For example, one can use this method to give another proof that the set K_{ω}(X) of countable closed subsets of an uncountable compact metric space X is not Borel (see 22.4). One lets φ(K) = the Cantor-Bendixson rank of K. Then it turns out that φ is a Π_{1}^{1}-rank. Since, as in 5.6, one can see that φ is unbounded, this shows that K_{ω}(X) is not Borel.

For this and other reasons it is important, when studying the descriptive properties of a given Π_{1}^{1} set, like U or U_{0}, to find canonical Π_{1}^{1}-ranks that reflect the structural properties of the sets (and don’t just come from applying the abstract Theorem 26.1). There is indeed such a canonical Π_{1}^{1}-rank for U, called the Piatetski-Shapiro rank (see Kechris-Louveau [1989]), but I will not discuss this here. There is also a canonical rank for U_{0}, which I will now discuss, since it has a particularly simple description using the basis U′_{0} for U_{0} and the generalized Cantor-Bendixson procedure described in 23.1.

Let X be a compact metric space, and B ⊆ K(X) a hereditary Borel set. Let I = B_{σ} be the σ-ideal generated by B. By 23.1, I is Π_{1}^{1}. Define, as in 23.1 again, the following rank on I:

φ(K) = rk_{B}(K).

Let us call this the B-rank of I. The following can be then proved.

26.3. Theorem. For any hereditary Borel B ⊆ K(X), the B-rank on I = B_{σ} is a Π_{1}^{1}-rank.

This is of course a generalization of the corresponding fact for the Cantor-Bendixson rank, which we used in the example above. It also shows that the U′_{0}-rank on U_{0} is a Π_{1}^{1}-rank on U_{0}.

One then can use the rank method to establish that I = B_{σ} is not Borel: It is enough for that to show that for each countable ordinal ξ there is K ∈ B_{σ} with rk_{B}(K) > ξ.

For the following exercise note that if B is an ideal (i.e., it is also closed under finite unions), then rk_{B}(K) ≤ 1 ⇔ K ∈ B.

26.4. Exercise. Assume B ⊆ K(X) is a Borel ideal consisting of nowhere dense sets and J ⊆ B_{σ} is a σ-ideal such that for every open non-∅ V ⊆ X there is K ∈ J \ B, K ⊆ V. Show that for every countable ordinal ξ, there is a K ∈ J with rk_{B}(K) > ξ. Conclude that J is not analytic. [Hint. Show by transfinite induction that for every countable ordinal ξ, and for every open non-∅ set V ⊆ X, there is K ∈ J, K ⊆ V with rk_{B}(K) > ξ + 1.)

We will apply this method to show that U_{0} is not Borel. In fact, by applying appropriately 26.4 we will see a much stronger property of U_{0}. In a sense, U_{0} is “hereditarily” non Borel.

§27. A hereditary property of U_{0}.

I will prove here the following:

27.1. Theorem (Kechris). There is no simple (i.e., Borel, or, equivalently, G_{δ}) σ-ideal I such that

K_{ω}(𝕋) ⊆ I ⊆ U_{0}.

Before giving the proof, I want to make some comments.

(i) Since K_{ω}(𝕋) ⊆ U ⊆ U_{0}, this shows that U, U_{0} (as well as K_{ω}(𝕋)) are not Borel, so it gives another proof of these results.

(ii) The proof also shows that the hypothesis can be relaxed to the following: For any E ∈ K(𝕋) which is U_{0}-perfect, there is no Borel ideal with K_{ω}(E) ⊆ I ⊆ U_{0}. In particular, this implies that for any E ∈ K(𝕋), E ∉ U_{0}, U_{0} ∩ K(E) is not Borel (which was one of the ingredients needed to apply 23.4 to U_{0}).

(iii) This result has also implications concerning characterizations of subclasses of closed uniqueness sets. For example, A. Olevskii proposed, in a private conversation, a specific explicit characterization of the following class of uniqueness sets:

U_{diff} = {E ∈ U : for every diffeomorphism h of 𝕋 h[E] ∈ U}.

Since clearly U_{diff} is a σ-ideal and

K_{ω}(𝕋) ⊆ U_{diff} ⊆ U_{0},

it follows that U_{diff} cannot be Borel. However, Olevskii’s proposed characterization would easily lead to a Borel definition of U_{diff}, which shows that this proposed characterization does not work.

(iv) Kechris-Louveau-Woodin have shown that any Σ_{1}^{1} σ-ideal of closed sets is actually simple (i.e., Borel). So 27.1 also implies that there is no Σ_{1}^{1} ideal between K_{ω}(𝕋) and U_{0}.

Proof of 27.1. We will apply 26.4 to X = 𝕋, B = U′_{0}, J = I. First we will verify that U′_{0} is an ideal, i.e., is closed under finite unions. This is due to Lyons and Host-Parreau. We need a couple of lemmas first.

Lemma 1. Let μ, ν be finite, positive Borel measures on 𝕋 so that ν(𝕋) ≤ 1 and ν ≤ μ (i.e., ν(A) ≤ μ(A) for all Borel A or, equivalently, ∫ fdν ≤ fdμ for all Borel f ≥ 0). Let n_{1}, ⋯, n_{k} ∈ ℤ and let w_{j} ∈ ℂ be such that |ν̂(n_{j})| = w_{j}ν̂(n_{j}) (we agree that w_{j} = 1, if ν̂(n_{j}) = 0). Then

(\frac{1}{k} ∑_{j=1}^{k} |ν̂(n_{j})|)^{2} ≤ \frac{μ̂(0)}{k} + \frac{2}{k^{2}} ∑_{1≤a<b≤k} Re w_{b} w̄_{a} μ̂(n_{b} − n_{a}).

Proof. We have

\begin{aligned}(\frac{1}{k} ∑_{j} |ν̂(n_{j})|)^{2} & = \frac{1}{k^{2}} ( ∑_{j} w_{j} ∫ e^{−in_{j}x}dν)^{2} \\ & ≤ \frac{1}{k^{2}} ( ∫ | ∑_{j} w_{a}e^{−in_{j}x}|dν)^{2}. \end{aligned}

Letting f(x) = ∑_{j} w_{a}e^{−in_{j}x} we have, by the Cauchy-Schwartz inequality, that

( ∫ |f|dν)^{2} ≤( ∫ |f|^{2}dν)·( ∫ 1dν) ≤ ∫ |f|^{2}dν,

as ν(𝕋) ≤ 1. So, since ν ≤ μ,

\begin{aligned}(\frac{1}{k} ∑_{j} |ν̂(n_{j})|)^{2} & ≤ \frac{1}{k^{2}} ∫ ( ∑_{a} w_{a}e^{−in_{a}x})(\overline{∑_{b} w_{b}e^{−in_{b}x})}dμ \\ & = \frac{μ̂(0)}{k} + \frac{2}{k^{2}} ∑_{a<b} Re w_{b}w̄_{a}μ̂(n_{b} − n_{a}). \end{aligned}

⊣

Lemma 2. If ν, μ are as in Lemma 1, then R(ν) ≤ R(μ)^{1/2}.

Proof. Suppose that for some 0 < n_{1} < n_{2} < ⋯ we have |ν̂(n_{j})| ≥ t. (Similarly we can deal with the case 0 > n_{1} > n_{2} · · ·.) Using the preceding lemma for n_{1}, ⋯, n_{k} we get

\frac{2}{k^{2}} ∑_{1≤a<b≤k} |μ̂(n_{b} − n_{a})| ≥ t^{2} − \frac{μ̂(0)}{k},

so

R(μ) ≥ \overline{lim_{k}}(\frac{2}{k^{2}} ∑_{1≤a<b≤k} |μ̂(n_{b} − n_{a})|) ≥ t^{2}.

⊣

So let us assume that E, F ∈ U′_{0}, in order to show that E ∪ F ∈ U′_{0}. Pick ϵ such that 1 > ϵ > 0 and R(μ) ≥ ϵ for any μ ∈ P (E) ∪ P (F). Now consider any μ ∈ P (E ∪ F).

If μ(E) = 0 it follows that μ ∈ P (F), thus R(μ) ≥ ϵ. So we can assume that μ(E) > 0, μ(F) > 0. Let μ_{1} = μ|E, μ_{2} = μ|F, so that μ_{i} ≤ μ and μ_{i}(𝕋) ≤ 1. By Lemma 2,

\begin{matrix}R(μ)^{1/2} ≥ R(μ_{1}) ≥ ϵ · μ(E) \\ R(μ)^{1/2} ≥ R(μ_{2}) ≥ ϵ · μ(F). \end{matrix}

By adding, we get that

R(μ)^{1/2} ≥ ϵ/2,

so R(μ) ≥ ϵ^{2}/4. Thus, in any case, R(μ) ≥ ϵ^{2}/4 for any μ ∈ P (E ∪ F) and we are done.⊣

It is clear that U′_{0} consists of nowhere dense sets, so the final (and main) claim, required to prove the theorem, is to verify that for every open non-∅ V ⊆ 𝕋, there is K ⊆ V, K ∈ I \ U′_{0}.

The main lemma is the following, where for any probability Borel measure μ, supp(μ) is the smallest closed set supporting μ, i.e., supp(μ) = 𝕋 \ ⋃{V : V open and μ(V) = 0}.

Lemma 3. Let E = [a, b] be a closed interval, a < b and let μ = (λ|E)/λ(E) (so that μ ∈ R and supp(μ) = E). Let J ⊆ K(E) be G_{δ}, hereditary and assume it contains all finite subsets of E. Then for any N > 0, ϵ > 0 there is ν ∈ P (E) with supp(ν) = E_{0} ∪⋯∪E_{N−1}, where E_{n} ∈ J, and

sup_{j∈ℤ} |μ̂(j) − ν̂(j)| ≤ \frac{(1 + ϵ)}{N}.

Granting this, the proof of the above claim can be completed as follows:

Fix an open non-∅ interval V ⊆ 𝕋 and let E = \overline{V}. Let μ = (λ|E)/λ(E). Let, in Lemma 3, J = K(E) ∩ I. This is G_{δ}, hereditary in K(E), and contains all finite subsets of K(E). So, by Lemma 3, since μ̂(n) → 0, there is, for any given ϵ, N, ν ∈ P (E) with supp(ν) ∈ I ∩ K(E) and R(ν) ≤ \frac{(1+ϵ)}{N}.

Thus for any open non-empty V ⊆ 𝕋 if we choose x ∈ V we can find for each n ∈ ℕ a closed set K_{n} ∈ I such that dist(x, E_{n}) < 1/n and a probability measure ν_{n} ∈ P (K_{n}) with R(ν_{n}) ≤ \frac{1}{n}. Let K = \{x\} ∪ ⋃_{n} K_{n}. Then K ∈ I and clearly inf\{R(μ) : μ ∈ P (K)\} = 0, so K ∈ I \ U′_{0}.

So it only remains to give the

Proof of Lemma 3. The proof uses methods of Körner and Kaufman.

First, let J = ⋂_{n} G_{n}, with G_{n} decreasing and open in K(E). Let
G_{n}^{∗} = {K ∈ K(E) : ∀L ⊆ K(L ∈ G_{n})} ⊆ G_{n}.

Clearly G_{n}^{∗} is hereditary and J = ⋂_{n} G_{n}^{∗}. It is also easy to see that G_{n}^{∗} is open too. (We prove that K(E) \ G_{n}^{∗} is closed. Let K_{p} ∈ K(E) \ G_{n}^{∗} and K_{p} → K. Then there exists L_{p} ⊆ K_{p} with L_{p} ∉ G_{n}. By compactness, there is a converging subsequence L_{p_{i}} → L. As L_{p_{i}} → L, K_{p_{i}} → K and L_{p_{i}} ⊆ K_{p_{i}}, we have L ⊆ K. But L_{p_{i}} ∉ G_{n}, so as G_{n} is open, L ∉ G_{n}, i.e., K ∉ G_{n}^{∗}.) So, to avoid complicated notation, we assume that each G_{n} is open hereditary to start with.

Now notice that if G is open hereditary in K(E) and K ∈ G, then there is open V in E with K ⊆ V, so that K(V) ⊆ G. (Otherwise for any such V, there is L_{V} ∈ K(V) with L_{V} ∉ G. Letting V_{n} be such that \overline{V_{n+1}} ⊆ V_{n} and K = ⋂_{n} V_{n} and L_{n} = L_{V_{n}}, we can find a convergent subsequence L_{n_{i}} → L. Then for any n, L ⊆ \overline{V_{n}}, so L ⊆ K and L ∉ G, as G is open, a contradiction.)

Before we proceed to the construction of ν we will need a nice observation of Körner.

Lemma 4. Let Δ = [a, b + ℓ] be a closed interval and let ρ, σ be probability measures with ρ(Δ) = σ(Δ). Then

| ∫_{Δ} e^{−inx}dρ − ∫_{Δ} e^{−inx}dσ| ≤ 2ρ(Δ) sup_{x∈Δ}|e^{−inx} − e^{−ina}|.

Proof. Notice that if ρ_{Δ} = ρ|Δ, σ_{Δ} = σ|Δ, then

∫ d|ρ_{Δ} − σ_{Δ}| = ||ρ_{Δ} − σ_{Δ}||_{M} ≤ ||ρ_{Δ}||_{M} + ||σ_{Δ}||_{M} = ρ_{Δ}(𝕋) + σ_{Δ}(𝕋) = 2ρ(Δ).

Now, since ∫ d(ρ_{Δ} − σ_{Δ}) = ρ_{Δ}(𝕋) − σ_{Δ}(𝕋) = ρ(Δ) − σ(Δ) = 0, we have

| ∫_{Δ} e^{−inx}dρ − ∫_{Δ} e^{−inx}dσ|
=| ∫ e^{−inx}dρ_{Δ} − ∫ e^{−inx}dσ_{Δ}|
=| ∫ (e^{−inx} − e^{ina})d(ρ_{Δ} − σ_{Δ})|
≤ sup_{x∈Δ} |e^{inx} − e^{−ina}| · ∫ d|ρ_{Δ} − σ_{Δ}|
≤2ρ(Δ) · sup_{x∈Δ} |e^{−inx} − e^{ina}|.
⊣

We are now ready to start the construction: We will define probability measures μ_{k} ∈ R and numbers p_{k} ∈ ℕ such that

(i) 0 < p_{0} = p_{1} < ⋯ < p_{N−1} < p_{N} < p_{N+1} < ⋯; (ii) μ_{0} = μ_{1} = ⋯ = μ_{N−1} = μ; (iii) (|j| ≤ p_{k+N−1} or |j| ≥ p_{k+N}) ⇒ |μ̂_{k+N}(j) − μ̂_{k}(j)| ≤ \frac{1}{2}ϵ · 2^{−k−1}; (iv) p_{k} ≤ |j| ⇒ |μ̂_{k}(j)| < ϵ/2;

(v) supp(μ_{k+N}) ⊆ supp(μ_{k}); (vi) supp(μ_{n+ℓ·N}) ∈ G_{ℓ}, ℓ = 1, 2, ⋯; n = 0, ⋯, N − 1; (vii) supp(μ_{k}) is a finite union of disjoint closed intervals contained in E, and on each

one of these intervals μ_{k} is a multiple of Lebesgue measure on that interval.

Assume this can be done. For n = 0, 1, ⋯, N − 1 let

μ^{n} = lim_{ℓ}^{w^{∗}}μ_{n+ℓ·N} ∈ P (E).

(To see that this limit exists, it is enough to show that lim_{ℓ} μ̂_{n+ℓ·N}(j) exists for each j ∈ ℤ. But, by (iii), given j, if ℓ is so large that |j| ≤ p_{n+ℓ·N−1}, then we have |μ̂_{n+ℓ·N}(j) − μ̂_{n+(ℓ−1)N}(j)| ≤ \frac{1}{2}ϵ · 2^{−n−(ℓ−1)N−1}, so that \{μ̂_{n+ℓ·N}(j)\}_{ℓ} is a Cauchy sequence.)

By (v), supp(μ^{n}) ⊆ supp(μ_{n+ℓN}) for any ℓ, so supp(μ^{n}) ∈ ⋂_{ℓ} G_{ℓ} = J. Let ν = \frac{1}{N} (μ^{0} + ⋯ μ^{N−1}). Then supp(ν) ⊆ ⋃_{n=0}^{N−1} supp(μ^{n}), so, as J is hereditary, supp(ν) = E_{0} ∪⋯∪E_{N−1} with E_{n} ∈ J. Finally, fix j ∈ ℤ, in order to show that |μ̂(j)−ν̂(j)| ≤ \frac{(1+ϵ)}{N}. We have

|μ̂(j) − ν̂(j)|

=|\frac{1}{N} ( ∑_{n=0}^{N−1}(μ̂_{n}(j) − lim_{ℓ} μ̂_{n+ℓN}(j)))|

= lim_{ℓ}|\frac{1}{N} ∑_{n=0}^{N−1}(μ̂_{n}(j) − μ̂_{n+ℓN}(j))|

≤\frac{1}{N} ∑_{k=0}^{∞} |μ̂_{k+N}(j) − μ̂_{k}(j)|.

Now if |j| ≤ p_{k+N−1} or |j| ≥ p_{k+N}, we have

|μ̂_{k+N}(j) − μ̂_{k}(j)| ≤ \frac{ϵ}{2} · 2^{−k−1},

and if p_{k+N−1} < |j| < p_{k+N}, then |μ̂_{k+N}(j) − μ̂_{k}(j)| ≤ 1 + ϵ/2, since |μ̂_{k}(j)| < ϵ/2 for |j| ≥ p_{k} and n_{k+N−1} > n_{k}. So

\begin{aligned}|μ̂(j) − ν̂(j)| & ≤ \frac{1}{N} (1 + ϵ/2 + ∑_{k=0}^{∞} ϵ/2 · 2^{−k−1}) \\ & = \frac{(1 + ϵ)}{N}. \end{aligned}

To construct μ_{n}, p_{n} satisfying (i) – (vii) we start with μ_{0} = μ_{1} = ⋯ μ_{N−1} = μ and 0 < p_{0} = p_{1} = ⋯ = p_{N−1}, so that |j| ≥ p_{0} ⇒ |μ̂(j)| < ϵ/2 (which can be found as μ̂(j) → 0.)

Now we assume the construction has been done up to k + N − 1 (k = 0, 1, 2, ⋯), and we will construct μ_{k+N}, p_{k+N}. Let supp(μ_{k}) = Δ_{1} ∪ ⋯ ∪ Δ_{r_{k}}, Δ_{m} pairwise disjoint closed intervals contained in E. Let also k = n + ℓN (0 ≤ n ≤ N − 1, ℓ ≥ 0), so that k + N = n + (ℓ + 1)N.

Using Lemma 4, split each Δ_{m} into finitely many small enough closed subintervals, Δ_{m,q}, with only endpoints in common, so that the oscillation of e^{−ijx} for |j| ≤ n_{k+N−1} in each one of them is ≤ \frac{1}{4}ϵ · 2^{−k−1}. Then by Lemma 4, if ρ, σ are continuous (i.e., ρ({x}) = σ({x}) = 0) probability Borel measures supported by ⋃_{m} Δ_{m} = ⋃_{m,q} Δ_{m,q} and ρ(Δ_{m,q}) = σ(Δ_{m,q}) for every m, q, then we have

|j| ≤ p_{k+N−1} ⇒ |ρ̂(j) − σ̂(j)| ≤ \frac{1}{2}ϵ · 2^{−k−1}.

Choose one point x_{m,q} in the interior of each Δ_{m,q} and denote the resulting finite set by K = {x_{m,q}}. Then K ∈ G_{ℓ+1}, so, as G_{ℓ+1} is open hereditary, we can find a small closed interval Γ_{m,q} contained in the interior of Δ_{m,q} with x_{m,q} ∈ Γ_{m,q}, so that all closed subsets of ⋃_{m,q} Γ_{m,q} are contained in G_{ℓ+1}. We can of course assume that the Γ_{m,q} are pairwise disjoint.

Define then the probability Borel measure μ_{k+N} as follows: μ_{k+N} has support ⋃_{m,q} Γ_{m,q} and

μ_{k+N}|Γ_{m,q} = μ_{k}|Γ_{m,q} · \frac{μ_{k}(Δ_{m,q})}{μ_{k}(Γ_{m,q})}.

It is clear that (v)–(vii) are true. Now

\begin{aligned}μ_{k+N}(Δ_{m,q}) & = μ_{k}(Γ_{m,q}) · \frac{μ_{k}(Δ_{m,q})}{μ_{k}(Γ_{m,q})} \\ & = μ_{k}(Δ_{m,q}), \end{aligned}

so (iii) holds for |j| ≤ p_{k+N−1}. Finally, choose p_{k+N} > p_{k+N−1} large enough so that |j| ≥ p_{k+N} ⇒ |μ̂_{k+N}(j)| < ϵ/2( which can be done since μ̂_{k+N}(j) → 0, as |j| → ∞), and also |j| ≥ p_{k+N} ⇒ |μ̂_{k+N}(j) − μ̂_{k}(j)| ≤ \frac{1}{2}ϵ · 2^{−k−1} (which again can be done since μ̂_{k}(j) → 0 as |j| → ∞).

This completes the construction, the proof and these lectures. ⊣

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