Vitali–Hahn–Saks theorem

In mathematics, the Vitali–Hahn–Saks theorem, introduced by Vitali (1907), Hahn (1922), and Saks (1933), proves that a pointwise convergent sequence of finite measures is uniformly absolutely continuous with respect to any common dominating finite measure, and its limit is also a finite measure. This holds in the more general setting of finite signed measures or complex measures.

Preliminaries

Let $(S,{ℬ})$ be a measurable space and let $\lambda$ be a complex measure. The variation of $\lambda$ is the finite measure $|\lambda |:{ℬ}\to \mathbb{ℝ}$ given by $$|\lambda |(B)=\underset{\pi }{\sup }\sum _{A\in \pi }|\lambda (A)|,$$ where the supremum is taken over all partitions $\pi$ of $B$ into a countable number of disjoint measurable subsets. The total variation is then $\Vert \lambda {\Vert }_{TV}=|\lambda |(S)$.

A complex measure $\lambda$ is called absolutely continuous with respect to some measure $m$ (denoted $\lambda \ll m$), if $|\lambda |(B)=0$ for all measurable sets $B\in {ℬ}$ with $m(B)=0$. This is equivalent to the following: For every $\epsilon >0$ there is a $\delta >0$ so that $m(B)<\delta$ implies $|\lambda |(B)<\epsilon$. Since $$|\lambda (B)|\le |\lambda |(B)\le 4\underset{A\subseteq B}{\sup }|\lambda (A)|,$$ absolute continuity is also equivalent to the following: For every $\epsilon >0$ there is a $\delta >0$ so that $m(B)<\delta$ implies $|\lambda (B)|<\epsilon$.

If $\lambda \ll m$, then, by the Radon–Nikodym theorem there is an integrable function $f:S\to \mathbb{ℂ}$, such that $\lambda (B)=\underset{B}{\int }fdm$ for every measurable $B\in {ℬ}$. This function is uniquely defined up to an $m$-null set and called a Radon–Nikodym derivative of $\lambda$ with respect to $m$. It is usually denoted by $f=\frac{d\lambda }{dm}$.

Statement of the theorem

Let $(S,{ℬ})$ be a measurable space and let $({\lambda }_n{)}_{n\ge 1}$ be a sequence of finite complex measures such that the pointwise limit $\underset{n\to \mathrm{\infty }}{\lim }{\lambda }_n(B)$ exists in $\mathbb{ℂ}$ for every measurable set $B\in {ℬ}$. Define the function $\lambda :{ℬ}\to \mathbb{ℂ}$ by $$\lambda (B)=\underset{n\to \mathrm{\infty }}{\lim }{\lambda }_n(B).$$ Then, the following holds:

• Countable Additivity and Boundedness: $\lambda$ is countably additive and thus a finite complex measure. Furthermore, $\underset{n\ge 1}{\sup }\Vert {\lambda }_n{\Vert }_{TV}<\mathrm{\infty }$ and $\Vert \lambda {\Vert }_{TV}<\mathrm{\infty }$.
• Absolute Continuity: Assume that $m$ is a finite measure on $(S,{ℬ})$ satisfying ${\lambda }_n\ll m$ for all $n\ge 1$, such as $$m=\sum _{n\ge 1}\frac{|{\lambda }_n|}{2^n(1+\Vert {\lambda }_n{\Vert }_{TV})}.$$Then, $\lambda$ is also absolutely continuous with respect to $m$, that is $\lambda \ll m$, and the sequence $({\lambda }_n{)}_{n\ge 1}$ is uniformly absolutely continuous with respect to $m$, that is, for every $\epsilon >0$ there is a $\delta >0$ so that $m(B)<\delta$ implies $|{\lambda }_n(B)|<\epsilon$ uniformly for all $n\ge 1$.
• Densities: Again, if $m$ is a finite measure on $(S,{ℬ})$ satisfying ${\lambda }_n\ll m$ for all $n\ge 1$, set $f_n=\frac{d{\lambda }_n}{dm}$ for all $n\ge 1$ and $f=\frac{d\lambda }{dm}$. Then, $(f_n{)}_{n\ge 1}$ is uniformly integrable and $f_n$ converges in $L^1(S,{ℬ},m)$ with respect to the weak topology $\sigma (L^1,L^{\mathrm{\infty }})$, that is, $$\int g{f}_{n}dm\to \int gfdm$$ for all bounded measurable $g:S\to \mathbb{ℂ}$.

Preparations for the Proof

Let $(S,{ℬ},m)$ be a measure space and write ${{ℬ}}_0$ to denote the set of measurable sets $B\in {ℬ}$ with $m(B)<\mathrm{\infty }$: $${{ℬ}}_0=\{B\in {ℬ}:m(B)<\mathrm{\infty }\}.$$ Write $B_1△B_2=(B_1\setminus B_2)\cup (B_2\setminus B_1)$ to denote the symmetric difference of the sets $B_1,B_2$. Define the equivalence relation $\approx$ on ${{ℬ}}_0$ by $B_1\approx B_2$ if $m(B_1△B_2)=0$. Write ${\overline{{ℬ}}}_0={{ℬ}}_0/\approx$ to denote the quotient set of ${{ℬ}}_0$ with respect to the equivalence relation $\approx$: $${\overline{{ℬ}}}_0=\{\overline{B}:B\in {{ℬ}}_0\},$$ where $\overline{B}=\{A\in {{ℬ}}_0:A\approx B\}$ is the equivalence class of $B$. Then, $m$ induces a well-defined quotient map on ${\overline{{ℬ}}}_0$ which is denoted by $\overline{m}$: $\overline{m}(\overline{B})=m(B)$.

Proposition: Define $d_m:{\overline{{ℬ}}}_0\times {\overline{{ℬ}}}_0\to \mathbb{ℝ}$ by $$d_m({\overline{B}}_1,{\overline{B}}_2)=m(B_1△B_2).$$ Then, $d_m$ is well-defined and $({\overline{{ℬ}}}_0,d_m)$ is a complete metric space.

Proof of the Proposition

Using indicator functions as before, we have $$d_m({\overline{B}}_1,{\overline{B}}_2)=m(B_1△B_2)=\int |{\mathbf{𝟏}}_{B_1}-{\mathbf{𝟏}}_{B_2}|dm=\Vert {\mathbf{𝟏}}_{B_1}-{\mathbf{𝟏}}_{B_2}{\Vert }_1.$$ This means that the metric space $({\overline{{ℬ}}}_0,d_m)$ can be identified with a subset of the space $L^1(S,{ℬ},m)$ which is a Banach space. Hence, $d_m$ is a well-defined metric. Consider a Cauchy sequence $({\overline{B}}_n{)}_{n\ge 1}$. Then, $$\underset{k,n\to \mathrm{\infty }}{\lim }{d}_m({\overline{B}}_k,{\overline{B}}_n)=\underset{k,n\to \mathrm{\infty }}{\lim }\Vert {\mathbf{𝟏}}_{B_k}-{\mathbf{𝟏}}_{B_n}{\Vert }_1=0.$$ Hence, we can choose a subsequence $({\overline{B}}_{n_k}{)}_{k\ge 1}$ such that $({\mathbf{𝟏}}_{B_{n_k}}{)}_{k\ge 1}$ converges almost everywhere. Define $g:S\to \mathbb{ℝ}$ by $$g(x)=\underset{k\to \mathrm{\infty }}{\lim }{\mathbf{𝟏}}_{B_{n_k}}.$$ Then, $g$ is measurable, has values in $\{0,1\}$ almost everywhere and $$\underset{k\to \mathrm{\infty }}{\lim }\Vert {\mathbf{𝟏}}_{B_{n_k}}-g{\Vert }_1=0.$$ Therefore, $g={\mathbf{𝟏}}_{B_{\mathrm{\infty }}}$ almost everywhere for some $B_{\mathrm{\infty }}\in {{ℬ}}_0$ and $$\underset{k\to \mathrm{\infty }}{\lim }{d}_m({\overline{B}}_{n_k},{\overline{B}}_{\mathrm{\infty }})=\underset{k\to \mathrm{\infty }}{\lim }\Vert {\mathbf{𝟏}}_{B_{n_k}}-{\mathbf{𝟏}}_{B_{\mathrm{\infty }}}{\Vert }_1=0,$$ that is, the Cauchy sequence $({\overline{B}}_n{)}_{n\ge 1}$ contains a convergent subsequence and is convergent itself. Therefore, ${\overline{{ℬ}}}_0$ is complete. We remark that $B_{\mathrm{\infty }}$ is explicitly given by the limit inferior of the sequence $({\overline{B}}_{n_k}{)}_{k\ge 1}$: $$B_{\mathrm{\infty }}=\underset{k\to \mathrm{\infty }}{\mathrm{lim\; inf}}{B}_{n_k}=\bigcup _{k=1}^{\mathrm{\infty }}\bigcap _{j=k}^{\mathrm{\infty }}{B}_{n_j}.$$

Proposition: Let $({\lambda }_n{)}_{n\ge 1}$ be a sequence of complex measures such that $(\Vert {\lambda }_n{\Vert }_{TV}{)}_{n\ge 1}$ is bounded. Then, $({\lambda }_n{)}_{n\ge 1}$ converges pointwise, that is, the sequence $({\lambda }_n(B){)}_{n\ge 1}$ converges for every $B\in {ℬ}$, if and only if, for every bounded measurable function $g:S\to \mathbb{ℂ}$, the sequence $(\int gd{\lambda }_n{)}_{n\ge 1}$ converges.

Proof of the Proposition

The implication $\Leftarrow$ is obvious, since, for every $B\in {ℬ}$, the indicator function ${\mathbf{𝟏}}_B$ with $${\mathbf{𝟏}}_B(x)=\{\begin{array}{ll}1,& \text{if }x\in B,\\ 0,& \text{if }x\notin B,\end{array}$$ is bounded and measurable and ${\lambda }_n(B)=\int {\mathbf{𝟏}}_{B}d{\lambda }_n$. For the other implication $\Rightarrow$, we first note that, by linearity of limits, the sequence $(\int sd{\lambda }_n{)}_{n\ge 1}$ converges for all measurable simple functions $s:S\to \mathbb{ℂ}$. Write $K$ for the bound of total variations: $$K=\underset{n\ge 1}{\sup }\Vert {\lambda }_n{\Vert }_{TV}.$$ Given a bounded measurable function $g:S\to \mathbb{ℂ}$ and an $\epsilon >0$ there is a measurable simple function $s:S\to \mathbb{ℂ}$ with $\Vert g-s{\Vert }_{\mathrm{\infty }}<\epsilon$. Furthermore, $(\int sd{\lambda }_n{)}_{n\ge 1}$ converges, thus there exists $N$ such that $$|\int sd{\lambda }_k-\int sd{\lambda }_n|<\epsilon$$ for all $k,n>N$. Then, $$\begin{array}{rl}|\int gd{\lambda }_k-\int gd{\lambda }_n|& \le |\int gd{\lambda }_k-\int sd{\lambda }_k|+|\int sd{\lambda }_k-\int sd{\lambda }_n|+|\int sd{\lambda }_n-\int gd{\lambda }_n|\\ & \le \int |g-s|d|{\lambda }_k|+|\int sd{\lambda }_k-\int sd{\lambda }_n|+\int |g-s|d|{\lambda }_n|\\ & <\epsilon (2K+1)\end{array}$$ for all $k,n>N$. Consequently, the sequence $(\int gd{\lambda }_n{)}_{n\ge 1}$ converges.

Let $m$ be a measure on $(S,{ℬ})$. Fix some $\delta >0$. A set $B\in {ℬ}$ is called $\delta$-indivisible, if it cannot be partitioned into two disjoint measurable subsets $B_1,B_2\in {ℬ}$ satisfying $m(B_1)\ge \delta$ and $m(B_2)\ge \delta$.

Proposition: Let $m$ be a finite measure on $(S,{ℬ})$ and fix some $\delta >0$. Then, there exists a partition $\pi$ of $S$ into at most $\lfloor m(S)/\delta \rfloor +1$ disjoint $\delta$-indivisible, measurable sets.

Proof of the Proposition

Start with the partition ${\pi }_1=\{S\}$. Assume that ${\pi }_n$ is already defined. If all sets in ${\pi }_n$ are already $\delta$-indivisible, then set ${\pi }_{n+1}={\pi }_n$. If not, pick one set $B\in {\pi }_n$ that is not $\delta$-indivisible. Then, $B$ can be partitioned into two disjoint measurable subsets $B_1,B_2\in {ℬ}$ satisfying $m(B_1)\ge \delta$ and $m(B_2)\ge \delta$. In this case, set ${\pi }_{n+1}=({\pi }_n\setminus \{B\})\cup \{B_1,B_2\}$. This process must stabilize after at most $\lfloor m(S)/\delta \rfloor$, since each splitting creates a new set of measure at least $\delta$.

Proof of Vitali-Hahn-Saks theorem

Let $m$ be a finite measure on $(S,{ℬ})$ satisfying ${\lambda }_n\ll m$ for all $n\ge 1$. If none is given, take for instance $$m=\sum _{n\ge 1}\frac{|{\lambda }_n|}{2^n(1+\Vert {\lambda }_n{\Vert }_{TV})},$$ as mentioned above. Since $m$ is finite, we have ${{ℬ}}_0={ℬ}$. Furthermore, we write $\overline{{ℬ}}$ for the quotient set ${\overline{{ℬ}}}_0$ above. Each ${\lambda }_n$ defines a quotient map ${\overline{\lambda }}_n$ on $\overline{{ℬ}}$ by ${\overline{\lambda }}_n(\overline{B})={\lambda }_n(B)$. This quotient map ${\overline{\lambda }}_n$ is well-defined: If $A\in \overline{B}$, then $m(A△B)=0$ and thus $|{\lambda }_n|(A△B)=0$, since ${\lambda }_n\ll m$. Therefore, ${\lambda }_n(A)={\lambda }_n(B)$. Moreover, ${\overline{\lambda }}_n$ is continuous as a function from the metric space $(\overline{{ℬ}},d_m)$ to $\mathbb{ℂ}$.

Fix $\epsilon >0$. For every $n,k\ge 1$ the function $|{\overline{\lambda }}_n-{\overline{\lambda }}_{n+k}|$ is again continuous, so that $$F_{n,k,\epsilon }=\{\overline{B}\in \overline{{ℬ}}:|{\overline{\lambda }}_n(\overline{B})-{\overline{\lambda }}_{n+k}(\overline{B})|\le \epsilon \}$$ is a closed subset of $\overline{{ℬ}}$. Consequently, $$F_{n,\epsilon }=\bigcap _{k\ge 1}{F}_{n,k,\epsilon }=\{\overline{B}\in \overline{{ℬ}}:\underset{k\ge 1}{\sup }|{\overline{\lambda }}_n(\overline{B})-{\overline{\lambda }}_{n+k}(\overline{B})|\le \epsilon \}$$ is also a closed subset of $\overline{{ℬ}}$. By the hypothesis the sequence $({\lambda }_n(B){)}_{n\ge 1}$ converges for every $B\in {ℬ}$, so that $$\underset{k\ge 1}{\sup }|{\overline{\lambda }}_n(\overline{B})-{\overline{\lambda }}_{n+k}(\overline{B})|=\underset{k\ge 1}{\sup }|{\lambda }_n(B)-{\lambda }_{n+k}(B)|\to 0.$$ Therefore, we obtain $$\overline{{ℬ}}=\bigcup _{n=1}^{\mathrm{\infty }}{F}_{n,\epsilon }.$$ By Baire category theorem at least one of the sets $(F_{n,\epsilon }{)}_{n\ge 1}$ must contain a non-empty open set of $\overline{{ℬ}}$. This means that there is $n_0\ge 1$ and $B_0\in {ℬ}$ and an $r_0>0$ such that $F_{n_0,\epsilon }$ contains the open ball in $\overline{{ℬ}}$ with center ${\overline{B}}_0$ and radius $r_0$: $$\{\overline{B}\in \overline{{ℬ}}:d_m(\overline{B},{\overline{B}}_0)<r_0\}\subseteq F_{n_0,\epsilon }.$$ In other words: $$d_m(\overline{B},{\overline{B}}_0)<r_0⟹\underset{k\ge 1}{\sup }|{\overline{\lambda }}_{n_0}(\overline{B})-{\overline{\lambda }}_{n_0+k}(\overline{B})|\le \epsilon .$$ Since ${\lambda }_n\ll m$ for all $n<n_0$, there is a $\delta >0$, so that $m(B)<\delta$ implies $|{\lambda }_n(B)|<\epsilon$ for all $n<n_0$. We may choose $\delta >0$ sufficiently small, so that $\delta \le r_0$.

Now, consider a measurable set $B\in {ℬ}$ with $m(B)<\delta$. Define $B_1=B\cup B_0$ and $B_2=B_0\setminus B$. Then, we have $$B=(B\cup B_0)\setminus (B_0\setminus B)=B_1\setminus B_2$$ and $$\begin{array}{rl}{d}_m({\overline{B}}_1,{\overline{B}}_0)& =m(B_1△B_0)=m((B\cup B_0)△B_0)\le m(B)<\delta ,\\ d_m({\overline{B}}_2,{\overline{B}}_0)& =m(B_2△B_0)=m((B_0\setminus B)△B_0)\le m(B)<\delta .\end{array}$$ Thus, if $n<n_0$, then $|{\lambda }_n(B)|<\epsilon$ and, if $n\ge n_0$, then $$\begin{array}{rl}|{\lambda }_n(B)|& \le |{\lambda }_{n_0}(B)|+|{\lambda }_{n_0}(B)-{\lambda }_n(B)|\\ & \le |{\lambda }_{n_0}(B)|+|{\lambda }_{n_0}(B_1)-{\lambda }_n(B_1)|+|{\lambda }_{n_0}(B_2)-{\lambda }_n(B_2)|\\ & <3ϵ\end{array}$$ Therefore, $({\lambda }_n{)}_{n\ge 1}$ is uniformly absolutely continuous with respect to $m$.

By the linearity of the limit $\lambda$ is finitely additive. And by uniformly absolute continuity of $({\lambda }_n{)}_{n\ge 1}$ the limit $\lambda$ is actually countably additive.

Next, we prove boundedness of the total variations: Fix $\epsilon =1$. By uniformly absolute continuity of $({\lambda }_n{)}_{n\ge 1}$ there is $\delta >0$ such that $m(B)<\delta$ implies $|{\lambda }_n(B)|<1$. By the third proposition above, there is a finite partition $\pi$ of $S$ into $\delta$-indivisible sets with respect to the finite measure $m$. For any $B\in \pi$ and any measurable $A\subseteq B$, we cannot have both, $m(A)\ge \delta$ and $m(B\setminus A)\ge \delta$. If $m(A)<\delta$, then, by uniformly absolute continuity, $$|{\lambda }_n(A)|<1\le |{\lambda }_n(B)|+1$$ for all $n\ge 1$. If, on the other hand $m(B\setminus A)<\delta$, then, again by uniform absolute continuity, $|{\lambda }_n(B\setminus A)|<1$ for all $n\ge 1$, which implies $$|{\lambda }_n(A)|=|{\lambda }_n(B)-{\lambda }_n(B\setminus A)|\le |{\lambda }_n(B)|+|{\lambda }_n(B\setminus A)|<|{\lambda }_n(B)|+1$$ for all $n\ge 1$. In both cases, for every measurable subset $A\subseteq B$ and every $n\ge 1$: $$|{\lambda }_n(A)|<|{\lambda }_n(B)|+1$$ Thus, $$|{\lambda }_n|(B)\le 4\underset{A\subseteq B}{\sup }|{\lambda }_n(A)|\le 4|{\lambda }_n(B)|+4$$ Since $({\lambda }_n(B){)}_{n\ge 1}$ converges for every $B\in \pi$ and $\pi$ is finite, we have $$C=\underset{B\in \pi }{\max }\underset{n\ge 1}{\sup }|{\lambda }_n(B)|<\mathrm{\infty }$$ and therefore $$\Vert {\lambda }_n{\Vert }_{TV}=\sum _{B\in \pi }|{\lambda }_n|(B)\le \sum _{B\in \pi }(4|{\lambda }_n(B)|+4)\le (4C+4)|\pi |<\mathrm{\infty }.$$

The remaining assertions of the theorem follow directly using the second proposition above and boundedness of all total variations.

References

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