Vitali–Hahn–Saks theorem

In mathematics, the Vitali–Hahn–Saks theorem, introduced by, , and , 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 $$(S,\mathcal{B},m)$$ is a measure space with $$m(S)<\infty,$$ and a sequence $$\lambda_n$$ of complex measures. Assuming that each $$\lambda_n$$ is absolutely continuous with respect to $$m,$$ and that a for all $$B\in\mathcal{B}$$ the finite limits exist $$\lim_{n\to\infty}\lambda_n(B)=\lambda(B).$$ Then the absolute continuity of the $$\lambda_n$$ with respect to $$m$$ is uniform in $$n,$$ that is, $$\lim_B m(B)=0$$ implies that $$\lim_{B}\lambda_n(B)=0$$ uniformly in $$n.$$ Also $$\lambda$$ is countably additive on $$\mathcal{B}.$$

Preliminaries
Given a measure space $$(S,\mathcal{B},m),$$ a distance can be constructed on $$\mathcal{B}_0,$$ the set of measurable sets $$B\in\mathcal{B}$$ with $$m(B) < \infty.$$ This is done by defining
 * $$d(B_1,B_2) = m(B_1\Delta B_2),$$ where $$B_1\Delta B_2 = (B_1\setminus B_2) \cup (B_2\setminus B_1)$$ is the symmetric difference of the sets $$B_1,B_2\in\mathcal{B}_0.$$

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

Proposition: $$\tilde{\mathcal{B}_0}$$ with the metric defined above is a complete metric space.

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

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

Proof of Vitali-Hahn-Saks theorem
Each $$\lambda_n$$ defines a function $$\overline{\lambda}_n(\overline{B})$$ on $$\tilde{\mathcal{B}}$$ by taking $$\overline{\lambda}_n(\overline{B})=\lambda_n(B)$$. This function is well defined, this is it is independent on the representative $$B$$ of the class $$\overline{B}$$ due to the absolute continuity of $$\lambda_n$$ with respect to $$m$$. Moreover $$\overline{\lambda}_n$$ is continuous.

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

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