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 $(S,\mathcal,m)$ 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 $B\in\mathcal$ the finite limits exist $\lim_\lambda_n(B)=\lambda(B).$ 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 $\lim_\lambda_n(B)=0$ uniformly in $n.$ Also $\lambda$ is countably additive on $\mathcal.$

Preliminaries

Given a measure space $(S,\mathcal,m),$ a distance can be constructed on $\mathcal_0,$ the set of measurable sets $B\in\mathcal$ 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 $B_1,B_2\in\mathcal_0.$
This gives rise to a metric space $\tilde$ by identifying two sets $B_1,B_2\in \mathcal_0$ when $m(B_1\Delta B_2)=0.$ Thus a point $\overline\in\tilde$ with representative $B\in\mathcal_0$ is the set of all $B_1\in\mathcal_0$ such that $m(B\Delta B_1) = 0.$

Proposition: $\tilde$ with the metric defined above is a complete metric space.

''Proof:'' Let
$\chi_B(x)=\begin1,&x\in B\\0,&x\notin B\end$
Then
$d(B_1,B_2)=\int_S, \chi_(s)-\chi_(x), dm$
This means that the metric space $\tilde$ can be identified with a subset of the Banach space $L^1(S,\mathcal,m)$.

Let $B_n\in\mathcal_0$, with
$\lim_d(B_n,B_k)=\lim_\int_S, \chi_(x)-\chi_(x), dm=0$
Then we can choose a sub-sequence $\chi_$ such that $\lim_\chi_(x)=\chi(x)$ exists almost everywhere and $\lim_\int_S, \chi(x)-\chi_, dm=0$. It follows that $\chi=\chi_$ for some $B_\in\mathcal_0$ (furthermore $\chi (x) = 1$ if and only if $\chi_ (x) = 1$ for $n'$ large enough, then we have that $B_ = \liminf_B_ = \left(B_m\right)$ the limit inferior of the sequence) and hence $\lim_d(B_\infty,B_n)=0.$ Therefore, $\tilde$ is complete.

Proof of Vitali-Hahn-Saks theorem

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

For every $\epsilon>0$ the set
$F_=\$
is closed in $\tilde$, and by the hypothesis $\lim_\lambda_n(B)=\lambda(B)$ we have that
$\tilde=\bigcup_^F_$
By Baire category theorem at least one $F_$ must contain a non-empty open set of $\tilde$. This means that there is $\overline\in\tilde$ and a $\delta>0$ such that
$d(B,B_0)<\delta$ implies $\sup_, \overline_(\overline)-\overline_(\overline), \leq\epsilon$
On the other hand, any $B\in\mathcal$ 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
$\begin, \lambda_k(B), &\leq, \lambda_(B), +, \lambda_(B)-\lambda_k(B), \\&\leq, \lambda_(B), +, \lambda_(B_1)-\lambda_k(B_1), +, \lambda_(B_2)-\lambda_k(B_2), \\&\leq, \lambda_(B), +2\epsilon\end$
Therefore, by the absolute continuity of $\lambda_$ 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 $\lim_\lambda(B) = 0$ it follows that $\lambda$ is actually countably additive.

References

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{{DEFAULTSORT:Vitali-Hahn-Saks Theorem

Theorems in measure theory