Vitali–Hahn–Saks theorem

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

Given a measure space ${\displaystyle (S,{\mathcal {B}},m),}$ a distance can be constructed on ${\displaystyle {\mathcal {B}}_{0},}$ the set of measurable sets ${\displaystyle B\in {\mathcal {B}}}$ with ${\displaystyle m(B)<\infty .}$ This is done by defining

${\displaystyle d(B_{1},B_{2})=m(B_{1}\Delta B_{2}),}$ where ${\displaystyle B_{1}\Delta B_{2}=(B_{1}\setminus B_{2})\cup (B_{2}\setminus B_{1})}$ is the symmetric difference of the sets ${\displaystyle B_{1},B_{2}\in {\mathcal {B}}_{0}.}$

This gives rise to a metric space ${\displaystyle {\tilde {{\mathcal {B}}_{0}}}}$ by identifying two sets ${\displaystyle B_{1},B_{2}\in {\mathcal {B}}_{0}}$ when ${\displaystyle m(B_{1}\Delta B_{2})=0.}$ Thus a point ${\displaystyle {\overline {B}}\in {\tilde {{\mathcal {B}}_{0}}}}$ with representative ${\displaystyle B\in {\mathcal {B}}_{0}}$ is the set of all ${\displaystyle B_{1}\in {\mathcal {B}}_{0}}$ such that ${\displaystyle m(B\Delta B_{1})=0.}$

Proposition: ${\displaystyle {\tilde {{\mathcal {B}}_{0}}}}$ with the metric defined above is a complete metric space.

Proof: Let

$${\displaystyle \chi _{B}(x)={\begin{cases}1,&x\in B\\0,&x\notin B\end{cases}}}$$

Then

$${\displaystyle d(B_{1},B_{2})=\int _{S}|\chi _{B_{1}}(s)-\chi _{B_{2}}(x)|dm}$$

This means that the metric space ${\displaystyle {\tilde {{\mathcal {B}}_{0}}}}$ can be identified with a subset of the Banach space ${\displaystyle L^{1}(S,{\mathcal {B}},m)}$.

Let ${\displaystyle B_{n}\in {\mathcal {B}}_{0}}$, with

$${\displaystyle \lim _{n,k\to \infty }d(B_{n},B_{k})=\lim _{n,k\to \infty }\int _{S}|\chi _{B_{n}}(x)-\chi _{B_{k}}(x)|dm=0}$$

Then we can choose a sub-sequence ${\displaystyle \chi _{B_{n'}}}$ such that ${\displaystyle \lim _{n'\to \infty }\chi _{B_{n'}}(x)=\chi (x)}$ exists almost everywhere and ${\displaystyle \lim _{n'\to \infty }\int _{S}|\chi (x)-\chi _{B_{n'}(x)}|dm=0}$. It follows that ${\displaystyle \chi =\chi _{B_{\infty }}}$ for some ${\displaystyle B_{\infty }\in {\mathcal {B}}_{0}}$ (furthermore ${\displaystyle \chi (x)=1}$ if and only if ${\displaystyle \chi _{B_{n'}}(x)=1}$ for ${\displaystyle n'}$ large enough, then we have that ${\displaystyle B_{\infty }=\liminf _{n'\to \infty }B_{n'}={\bigcup _{n'=1}^{\infty }}\left({\bigcap _{m=n'}^{\infty }}B_{m}\right)}$ the limit inferior of the sequence) and hence ${\displaystyle \lim _{n\to \infty }d(B_{\infty },B_{n})=0.}$ Therefore, ${\displaystyle {\tilde {{\mathcal {B}}_{0}}}}$ is complete.

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

For every ${\displaystyle \epsilon >0}$ the set

$${\displaystyle F_{k,\epsilon }=\{{\overline {B}}\in {\tilde {\mathcal {B}}}:\ \sup _{n\geq 1}|{\overline {\lambda }}_{k}({\overline {B}})-{\overline {\lambda }}_{k+n}({\overline {B}})|\leq \epsilon \}}$$

is closed in ${\displaystyle {\tilde {\mathcal {B}}}}$, and by the hypothesis ${\displaystyle \lim _{n\to \infty }\lambda _{n}(B)=\lambda (B)}$ we have that

$${\displaystyle {\tilde {\mathcal {B}}}=\bigcup _{k=1}^{\infty }F_{k,\epsilon }}$$

By Baire category theorem at least one ${\displaystyle F_{k_{0},\epsilon }}$ must contain a non-empty open set of ${\displaystyle {\tilde {\mathcal {B}}}}$. This means that there is ${\displaystyle {\overline {B_{0}}}\in {\tilde {\mathcal {B}}}}$ and a ${\displaystyle \delta >0}$ such that

$${\displaystyle d(B,B_{0})<\delta }$$

implies ${\displaystyle \sup _{n\geq 1}|{\overline {\lambda }}_{k_{0}}({\overline {B}})-{\overline {\lambda }}_{k_{0}+n}({\overline {B}})|\leq \epsilon }$

On the other hand, any ${\displaystyle B\in {\mathcal {B}}}$ with ${\displaystyle m(B)\leq \delta }$ can be represented as ${\displaystyle B=B_{1}\setminus B_{2}}$ with ${\displaystyle d(B_{1},B_{0})\leq \delta }$ and ${\displaystyle d(B_{2},B_{0})\leq \delta }$. This can be done, for example by taking ${\displaystyle B_{1}=B\cup B_{0}}$ and ${\displaystyle B_{2}=B_{0}\setminus (B\cap B_{0})}$. Thus, if ${\displaystyle m(B)\leq \delta }$ and ${\displaystyle k\geq k_{0}}$ then

$${\displaystyle {\begin{aligned}|\lambda _{k}(B)|&\leq |\lambda _{k_{0}}(B)|+|\lambda _{k_{0}}(B)-\lambda _{k}(B)|\\&\leq |\lambda _{k_{0}}(B)|+|\lambda _{k_{0}}(B_{1})-\lambda _{k}(B_{1})|+|\lambda _{k_{0}}(B_{2})-\lambda _{k}(B_{2})|\\&\leq |\lambda _{k_{0}}(B)|+2\epsilon \end{aligned}}}$$

Therefore, by the absolute continuity of ${\displaystyle \lambda _{k_{0}}}$ with respect to ${\displaystyle m}$, and since ${\displaystyle \epsilon }$ is arbitrary, we get that ${\displaystyle m(B)\to 0}$ implies ${\displaystyle \lambda _{n}(B)\to 0}$ uniformly in ${\displaystyle n.}$ In particular, ${\displaystyle m(B)\to 0}$ implies ${\displaystyle \lambda (B)\to 0.}$

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

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