CLT

3 Separating subalgebras

Definition 3.1
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A set \(\mathcal F\) of functions \(E \to F\) separates points in \(E\) if for all \(x, y \in E\) with \(x \ne y\), there exists \(f \in \mathcal F\) with \(f(x) \ne f(y)\).

Definition 3.2
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A set \(\mathcal F\) of functions \(E \to F\) is separating in \(\mathcal P(E)\) if for all probability measures \(\mu , \nu \) on \(E\) with \(\mu \ne \nu \), there exists \(f \in \mathcal F\) with \(\mu [f] \ne \nu [f]\).

Lemma 3.3

In a Borel space \(E\), the set \(C_b(E, \mathbb {C})\) of bounded continuous functions from \(E\) to \(\mathbb {C}\) is separating in \(\mathcal P(E)\).

Proof

The Mathlib lemma MeasureTheory.FiniteMeasure.ext_of_forall_lintegral_eq shows that \(C_b(E, \mathbb {R}_+)\) is separating. Since \(C_b(E, \mathbb {R}_{+}) \subseteq C_b(E, \mathbb {C})\), the larger set is also separating.

Lemma 3.4
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\(x \mapsto (t \mapsto \exp (i \langle t, x \rangle ))\) is a monoid homomorphism from \((E,+)\) to \(C_b(E, \mathbb {C})\).

Proof
Lemma 3.5
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The functions of the form \(x \mapsto \sum _{k=1}^n a_k e^{i\langle t_k, x\rangle }\) for \(n \in \mathbb {N}\), \(a_1, \ldots , a_n \in \mathbb {R}\) and \(t_1, \ldots , t_n \in E\) are a star-subalgebra of \(C_b(E, \mathbb {C})\).

Proof

The monoid homomorphism can be lifted to a homomorphism of algebras from AddMonoidAlgebra ℝ E to \(C_b(E, \mathbb {C})\) using AddMonoidAlgebra.lift. \(\mathcal M\) is the range of this homomorphism. The “star” part can be checked easily.

Let \(\mathcal M\) be the set of these functions, which we call exponential polynomials.

Lemma 3.6
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The star-subalgebra \(\mathcal M\) separates points.

Proof

TODO

Lemma 3.7

Let \(\mu \) be a finite measure on a Borel space \(E\) and let \(K\) be a measurable compact set such that \(\mu (K^c) \le \varepsilon \). Let \(f \in C(E, \mathbb {R})\). Then

\begin{align*} \left\vert \mu [fe^{-\varepsilon \Vert f \Vert ^2}] - \mu [f e^{-\varepsilon \Vert f \Vert ^2} \mathbb {I}_K] \right\vert \le \sqrt{\varepsilon } \: . \end{align*}
Proof
Lemma 3.8
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For any \(f \in C_b(E, \mathbb {R})\) and a probability measure \(\mu \),

\begin{align*} \mu [f] = \lim _{\varepsilon \to 0} \mu \left[f e^{-\varepsilon \Vert f \Vert ^2} \right] \: . \end{align*}
Proof
Lemma 3.9
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Let \(\mu , \nu \) be two finite measures in a standard Borel space \(E\). Let \(\mathcal{A}\) be a subalgebra of \(C_b(E, \mathbb {R})\) that separates points. If for all \(g \in \mathcal A\), \(\mu [g] = \nu [g]\), then for all \(f \in C_b(E, \mathbb {R})\) and all \(\varepsilon {\gt} 0\),

\begin{align*} \left\vert \mu [fe^{-\varepsilon \Vert f \Vert ^2}] - \nu [f e^{-\varepsilon \Vert f \Vert ^2}] \right\vert \le 6\sqrt{\varepsilon } \: . \end{align*}
Proof

Let \(\mu , \nu \) be two finite measures in a standard Borel space \(E\). Let \(\mathcal{A}\) be a star-subalgebra of \(C_b(E, \mathbb {k})\) that separates points, where \(\mathbb {k}\) is \(\mathbb {R}\) or \(\mathbb {C}\). If for all \(g \in \mathcal A\), \(\mu [g] = \nu [g]\), then for all \(f \in C_b(E, \mathbb {k})\), \(\mu [f] = \nu [f]\).

Proof

Proved in Mathlib PR #19782

Theorem 3.11 Subalgebras separating points

Let \(E\) be a complete separable pseudo-metric space. Let \(\mathcal A \subseteq C_b(E, \mathbb {C})\) be a star-subalgebra that separates points. Then \(\mathcal A\) is separating in \(\mathcal P(E)\).

Proof

Let \(\mu \) and \(\nu \) be two probability measures such that \(\mu [f] = \nu [f]\) for all \(f \in \mathcal A\). We want to prove that \(\mu = \nu \). Since \(C_b(E, \mathbb {C})\) is separating, it suffices to prove that for \(g \in C_b(E, \mathbb {C})\), \(\mu [g] = \nu [g]\). This is proved in Lemma 3.10

Lemma 3.12

The exponential polynomials \(\mathcal M\) are separating.

Proof

Apply Theorem 3.11, using Lemma 3.6.

Lemma 3.13

Let \(\mathcal A\) be a separating subalgebra of \(C_b(E, \mathbb {C})\). Let \(\mu , \mu _1, \mu _2, \ldots \) be probability measures, with \((\mu _n)\) tight. If for all \(f \in \mathcal A\), \(\mu _n[f] \to \mu [f]\), then \(\mu _n \xrightarrow {w} \mu \).

Proof

To show convergence to \(\mu \), it suffices to show that every subsequence has a subsequence that converges to \(\mu \) (Filter.tendsto_of_subseq_tendsto). Since the family is tight, it is relatively compact by Prokhorov’s theorem. We get that each subsequence has a converging subsequence. Let then \(\mu '\) be a weak limit of \(\mu _n\). We have \(\mu _n[f] \to \mu '[f]\) and \(\mu _n[f] \to \mu [f]\) for all \(f \in \mathcal A\), hence \(\mu '[f] = \mu [f]\) for all \(f \in \mathcal A\). Since \(\mathcal A\) is separating, \(\mathcal\mu ' = \mu \).