# Vitali–Hahn–Saks theorem

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 $(S,{\mathcal {B}},m)$ 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 $F_{k,\epsilon }=\{{\overline {B}}\in {\tilde {\mathcal {B}}}:\ \sup _{n\geq 1}|{\overline {\lambda }}_{k}({\overline {B}})-{\overline {\lambda }}_{k+n}({\overline {B}})|\leq \epsilon \}$ 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 $\sup _{n\geq 1}|{\overline {\lambda }}_{k_{0}}({\overline {B}})-{\overline {\lambda }}_{k_{0}+n}({\overline {B}})|\leq \epsilon$ 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 ${\begin{aligned}|\lambda _{k}(B)|&\leq |\lambda _{k_{0}}(B)|+|\lambda _{k_{0}}(B)-\lambda _{k}(B)|\\&\leq |\lambda _{k_{0}}(B)|+|\lambda _{k_{0}}(B_{1})-\lambda _{k}(B_{1})|+|\lambda _{k_{0}}(B_{2})-\lambda _{k}(B_{2})|\\&\leq |\lambda _{k_{0}}(B)|+2\epsilon \end{aligned}}$ Therefore, by the absolute continuity of $\lambda _{k_{0}}$ with respect to $m$, and since $\epsilon$ 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 $\lim _{m(B)\to 0}\lambda (B)=0$ it follows that $\lambda$ is actually countably additive.