Vitali–Hahn–Saks theorem

In mathematics, the Vitali–Hahn–Saks theorem, introduced by Template:Harvs, Template:Harvs, and Template:Harvs, 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 $(S,ℬ,m)$ is a measure space with $m(S)<\mathrm{\infty },$ 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 $B\in ℬ$ the finite limits exist $\underset{n\to \mathrm{\infty }}{\lim }{\lambda }_n(B)=\lambda (B).$ Then the absolute continuity of the ${\lambda }_n$ with respect to $m$ is uniform in $n,$ that is, $\underset{B}{\lim }m(B)=0$ implies that $\underset{B}{\lim }{\lambda }_n(B)=0$ uniformly in $n.$ Also $\lambda$ is countably additive on $ℬ.$

Preliminaries

Given a measure space $(S,ℬ,m),$ a distance can be constructed on ${ℬ}_0,$ the set of measurable sets $B\in ℬ$ with $m(B)<\mathrm{\infty }.$ This is done by defining

$d(B_1,B_2)=m(B_1\mathrm{\Delta }{B}_2),$ where $B_1\mathrm{\Delta }{B}_2=(B_1\setminus B_2)\cup (B_2\setminus B_1)$ is the symmetric difference of the sets $B_1,B_2\in {ℬ}_0.$

This gives rise to a metric space $\widetilde{{ℬ}_0}$ by identifying two sets $B_1,B_2\in {ℬ}_0$ when $m(B_1\mathrm{\Delta }{B}_2)=0.$ Thus a point $\bar{B}\in \widetilde{{ℬ}_0}$ with representative $B\in {ℬ}_0$ is the set of all $B_1\in {ℬ}_0$ such that $m(B\mathrm{\Delta }{B}_1)=0.$

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

Proof: Let $${\chi }_B(x)=\{\begin{array}{cc}1,& x\in B\\ 0,& x\notin B\end{array}$$ Then $$d(B_1,B_2)=\underset{S}{\int }|{\chi }_{B_1}(s)-{\chi }_{B_2}(x)|dm$$ This means that the metric space $\widetilde{{ℬ}_0}$ can be identified with a subset of the Banach space $L^1(S,ℬ,m)$.

Let $B_n\in {ℬ}_0$, with $$\underset{n,k\to \mathrm{\infty }}{\lim }d(B_n,B_k)=\underset{n,k\to \mathrm{\infty }}{\lim }\underset{S}{\int }|{\chi }_{B_n}(x)-{\chi }_{B_k}(x)|dm=0$$ Then we can choose a sub-sequence ${\chi }_{B_{n^{\prime }}}$ such that $\underset{n^{\prime }\to \mathrm{\infty }}{\lim }{\chi }_{B_{n^{\prime }}}(x)=\chi (x)$ exists almost everywhere and $\underset{n^{\prime }\to \mathrm{\infty }}{\lim }\underset{S}{\int }|\chi (x)-{\chi }_{B_{n^{\prime }}(x)}|dm=0$. It follows that $\chi ={\chi }_{B_{\mathrm{\infty }}}$ for some $B_{\mathrm{\infty }}\in {ℬ}_0$ (furthermore $\chi (x)=1$ if and only if ${\chi }_{B_{n^{\prime }}}(x)=1$ for $n^{\prime }$ large enough, then we have that $B_{\mathrm{\infty }}=\underset{n^{\prime }\to \mathrm{\infty }}{\mathrm{lim\; inf}}{B}_{n^{\prime }}=\bigcup _{n^{\prime }=1}^{\mathrm{\infty }}\left(\bigcap _{m=n^{\prime }}^{\mathrm{\infty }}{B}_m\right)$ the limit inferior of the sequence) and hence $\underset{n\to \mathrm{\infty }}{\lim }d(B_{\mathrm{\infty }},B_n)=0.$ Therefore, $\widetilde{{ℬ}_0}$ is complete.

Proof of Vitali-Hahn-Saks theorem

Each ${\lambda }_n$ defines a function ${\bar{\lambda }}_n(\bar{B})$ on $\widetilde{ℬ}$ by taking ${\bar{\lambda }}_n(\bar{B})={\lambda }_n(B)$. This function is well defined, this is it is independent on the representative $B$ of the class $\bar{B}$ due to the absolute continuity of ${\lambda }_n$ with respect to $m$. Moreover ${\bar{\lambda }}_n$ is continuous.

For every $ϵ>0$ the set $$F_{k,ϵ}=\{\bar{B}\in \widetilde{ℬ}:\underset{n\ge 1}{\sup }|{\bar{\lambda }}_k(\bar{B})-{\bar{\lambda }}_{k+n}(\bar{B})|\le ϵ\}$$ is closed in $\widetilde{ℬ}$, and by the hypothesis $\underset{n\to \mathrm{\infty }}{\lim }{\lambda }_n(B)=\lambda (B)$ we have that $$\widetilde{ℬ}=\bigcup _{k=1}^{\mathrm{\infty }}{F}_{k,ϵ}$$ By Baire category theorem at least one $F_{k_0,ϵ}$ must contain a non-empty open set of $\widetilde{ℬ}$. This means that there is $\overline{B_0}\in \widetilde{ℬ}$ and a $\delta >0$ such that $$d(B,B_0)<\delta$$ implies $\underset{n\ge 1}{\sup }|{\bar{\lambda }}_{k_0}(\bar{B})-{\bar{\lambda }}_{k_0+n}(\bar{B})|\le ϵ$ On the other hand, any $B\in ℬ$ with $m(B)\le \delta$ can be represented as $B=B_1\setminus B_2$ with $d(B_1,B_0)\le \delta$ and $d(B_2,B_0)\le \delta$. 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 $m(B)\le \delta$ and $k\ge k_0$ then $$\begin{array}{cc}|{\lambda }_k(B)|& \le |{\lambda }_{k_0}(B)|+|{\lambda }_{k_0}(B)-{\lambda }_k(B)|\\ & \le |{\lambda }_{k_0}(B)|+|{\lambda }_{k_0}(B_1)-{\lambda }_k(B_1)|+|{\lambda }_{k_0}(B_2)-{\lambda }_k(B_2)|\\ & \le |{\lambda }_{k_0}(B)|+2ϵ\end{array}$$ Therefore, by the absolute continuity of ${\lambda }_{k_0}$ with respect to $m$, and since $ϵ$ is arbitrary, we get that $m(B)\to 0$ 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 $\underset{m(B)\to 0}{\lim }\lambda (B)=0$ it follows that $\lambda$ is actually countably additive.

References

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