Vitali–Hahn–Saks theorem

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In mathematics, the Vitali–Hahn–Saks theorem, introduced by Vitali (1907), Hahn (1922), and Saks (1933), proves that under some conditions a sequence of measures converging point-wise does so uniformly and the limit is also a measure.

Statement of the theorem

If [math]\displaystyle{ (S,\mathcal{B},m) }[/math] is a measure space with [math]\displaystyle{ m(S)\lt \infty, }[/math] and a sequence [math]\displaystyle{ \lambda_n }[/math] of complex measures. Assuming that each [math]\displaystyle{ \lambda_n }[/math] is absolutely continuous with respect to [math]\displaystyle{ m, }[/math] and that a for all [math]\displaystyle{ B\in\mathcal{B} }[/math] the finite limits exist [math]\displaystyle{ \lim_{n\to\infty}\lambda_n(B)=\lambda(B). }[/math] Then the absolute continuity of the [math]\displaystyle{ \lambda_n }[/math] with respect to [math]\displaystyle{ m }[/math] is uniform in [math]\displaystyle{ n, }[/math] that is, [math]\displaystyle{ \lim_B m(B)=0 }[/math] implies that [math]\displaystyle{ \lim_{B}\lambda_n(B)=0 }[/math] uniformly in [math]\displaystyle{ n. }[/math] Also [math]\displaystyle{ \lambda }[/math] is countably additive on [math]\displaystyle{ \mathcal{B}. }[/math]

Preliminaries

Given a measure space [math]\displaystyle{ (S,\mathcal{B},m), }[/math] a distance can be constructed on [math]\displaystyle{ \mathcal{B}_0, }[/math] the set of measurable sets [math]\displaystyle{ B\in\mathcal{B} }[/math] with [math]\displaystyle{ m(B) \lt \infty. }[/math] This is done by defining

[math]\displaystyle{ d(B_1,B_2) = m(B_1\Delta B_2), }[/math] where [math]\displaystyle{ B_1\Delta B_2 = (B_1\setminus B_2) \cup (B_2\setminus B_1) }[/math] is the symmetric difference of the sets [math]\displaystyle{ B_1,B_2\in\mathcal{B}_0. }[/math]

This gives rise to a metric space [math]\displaystyle{ \tilde{\mathcal{B}_0} }[/math] by identifying two sets [math]\displaystyle{ B_1,B_2\in \mathcal{B}_0 }[/math] when [math]\displaystyle{ m(B_1\Delta B_2)=0. }[/math] Thus a point [math]\displaystyle{ \overline{B}\in\tilde{\mathcal{B}_0} }[/math] with representative [math]\displaystyle{ B\in\mathcal{B}_0 }[/math] is the set of all [math]\displaystyle{ B_1\in\mathcal{B}_0 }[/math] such that [math]\displaystyle{ m(B\Delta B_1) = 0. }[/math]

Proposition: [math]\displaystyle{ \tilde{\mathcal{B}_0} }[/math] with the metric defined above is a complete metric space.

Proof: Let [math]\displaystyle{ \chi_B(x)=\begin{cases}1,&x\in B\\0,&x\notin B\end{cases} }[/math] Then [math]\displaystyle{ d(B_1,B_2)=\int_S|\chi_{B_1}(s)-\chi_{B_2}(x)|dm }[/math] This means that the metric space [math]\displaystyle{ \tilde{\mathcal{B}_0} }[/math] can be identified with a subset of the Banach space [math]\displaystyle{ L^1(S,\mathcal{B},m) }[/math].

Let [math]\displaystyle{ B_n\in\mathcal{B}_0 }[/math], with [math]\displaystyle{ \lim_{n,k\to\infty}d(B_n,B_k)=\lim_{n,k\to\infty}\int_S|\chi_{B_n}(x)-\chi_{B_k}(x)|dm=0 }[/math] Then we can choose a sub-sequence [math]\displaystyle{ \chi_{B_{n'}} }[/math] such that [math]\displaystyle{ \lim_{n'\to\infty}\chi_{B_{n'}}(x)=\chi(x) }[/math] exists almost everywhere and [math]\displaystyle{ \lim_{n'\to\infty}\int_S|\chi(x)-\chi_{B_{n'}(x)}|dm=0 }[/math]. It follows that [math]\displaystyle{ \chi=\chi_{B_{\infty}} }[/math] for some [math]\displaystyle{ B_{\infty}\in\mathcal{B}_0 }[/math] (furthermore [math]\displaystyle{ \chi (x) = 1 }[/math] if and only if [math]\displaystyle{ \chi_{B_{n'}} (x) = 1 }[/math] for [math]\displaystyle{ n' }[/math] large enough, then we have that [math]\displaystyle{ B_{\infty} = \liminf_{n'\to\infty}B_{n'} = {\bigcup_{n'=1}^\infty}\left({\bigcap_{m=n'}^\infty}B_m\right) }[/math] the limit inferior of the sequence) and hence [math]\displaystyle{ \lim_{n\to\infty}d(B_\infty,B_n)=0. }[/math] Therefore, [math]\displaystyle{ \tilde{\mathcal{B}_0} }[/math] is complete.

Proof of Vitali-Hahn-Saks theorem

Each [math]\displaystyle{ \lambda_n }[/math] defines a function [math]\displaystyle{ \overline{\lambda}_n(\overline{B}) }[/math] on [math]\displaystyle{ \tilde{\mathcal{B}} }[/math] by taking [math]\displaystyle{ \overline{\lambda}_n(\overline{B})=\lambda_n(B) }[/math]. This function is well defined, this is it is independent on the representative [math]\displaystyle{ B }[/math] of the class [math]\displaystyle{ \overline{B} }[/math] due to the absolute continuity of [math]\displaystyle{ \lambda_n }[/math] with respect to [math]\displaystyle{ m }[/math]. Moreover [math]\displaystyle{ \overline{\lambda}_n }[/math] is continuous.

For every [math]\displaystyle{ \epsilon\gt 0 }[/math] the set [math]\displaystyle{ F_{k,\epsilon}=\{\overline{B}\in\tilde{\mathcal{B}}:\ \sup_{n\geq1}|\overline{\lambda}_k(\overline{B})-\overline{\lambda}_{k+n}(\overline{B})|\leq\epsilon\} }[/math] is closed in [math]\displaystyle{ \tilde{\mathcal{B}} }[/math], and by the hypothesis [math]\displaystyle{ \lim_{n\to\infty}\lambda_n(B)=\lambda(B) }[/math] we have that [math]\displaystyle{ \tilde{\mathcal{B}}=\bigcup_{k=1}^{\infty}F_{k,\epsilon} }[/math] By Baire category theorem at least one [math]\displaystyle{ F_{k_0,\epsilon} }[/math] must contain a non-empty open set of [math]\displaystyle{ \tilde{\mathcal{B}} }[/math]. This means that there is [math]\displaystyle{ \overline{B_0}\in\tilde{\mathcal{B}} }[/math] and a [math]\displaystyle{ \delta\gt 0 }[/math] such that [math]\displaystyle{ d(B,B_0)\lt \delta }[/math] implies [math]\displaystyle{ \sup_{n\geq1}|\overline{\lambda}_{k_0}(\overline{B})-\overline{\lambda}_{k_0+n}(\overline{B})|\leq\epsilon }[/math] On the other hand, any [math]\displaystyle{ B\in\mathcal{B} }[/math] with [math]\displaystyle{ m(B)\leq\delta }[/math] can be represented as [math]\displaystyle{ B=B_1\setminus B_2 }[/math] with [math]\displaystyle{ d(B_1,B_0)\leq\delta }[/math] and [math]\displaystyle{ d(B_2,B_0)\leq \delta }[/math]. This can be done, for example by taking [math]\displaystyle{ B_1=B\cup B_0 }[/math] and [math]\displaystyle{ B_2=B_0\setminus(B\cap B_0) }[/math]. Thus, if [math]\displaystyle{ m(B)\leq\delta }[/math] and [math]\displaystyle{ k\geq k_0 }[/math] then [math]\displaystyle{ \begin{align}|\lambda_k(B)|&\leq|\lambda_{k_0}(B)|+|\lambda_{k_0}(B)-\lambda_k(B)|\\&\leq|\lambda_{k_0}(B)|+|\lambda_{k_0}(B_1)-\lambda_k(B_1)|+|\lambda_{k_0}(B_2)-\lambda_k(B_2)|\\&\leq|\lambda_{k_0}(B)|+2\epsilon\end{align} }[/math] Therefore, by the absolute continuity of [math]\displaystyle{ \lambda_{k_0} }[/math] with respect to [math]\displaystyle{ m }[/math], and since [math]\displaystyle{ \epsilon }[/math] is arbitrary, we get that [math]\displaystyle{ m(B)\to0 }[/math] implies [math]\displaystyle{ \lambda_n(B) \to 0 }[/math] uniformly in [math]\displaystyle{ n. }[/math] In particular, [math]\displaystyle{ m(B) \to 0 }[/math] implies [math]\displaystyle{ \lambda(B) \to 0. }[/math]

By the additivity of the limit it follows that [math]\displaystyle{ \lambda }[/math] is finitely-additive. Then, since [math]\displaystyle{ \lim_{m(B) \to 0}\lambda(B) = 0 }[/math] it follows that [math]\displaystyle{ \lambda }[/math] is actually countably additive.

References



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