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 theoremEdit
If <math>(S,\mathcal{B},m)</math> is a measure space with <math>m(S)<\infty,</math> and a sequence <math>\lambda_n</math> of complex measures. Assuming that each <math>\lambda_n</math> is absolutely continuous with respect to <math>m,</math> and that for all <math>B\in\mathcal{B}</math> the finite limits exist <math>\lim_{n\to\infty}\lambda_n(B)=\lambda(B)</math>. Then the absolute continuity of the <math>\lambda_n</math> with respect to <math>m</math> is uniform in <math>n</math>, that is, <math>\lim_B m(B)=0</math> implies that <math>\lim_{B}\lambda_n(B)=0</math> uniformly in <math>n</math>. Also <math>\lambda</math> is countably additive on <math>\mathcal{B}</math>.
PreliminariesEdit
Given a measure space <math>(S,\mathcal{B},m),</math> a distance can be constructed on <math>\mathcal{B}_0,</math> the set of measurable sets <math>B\in\mathcal{B}</math> with <math>m(B) < \infty.</math> This is done by defining
- <math>d(B_1,B_2) = m(B_1\Delta B_2),</math> where <math>B_1\Delta B_2 = (B_1\setminus B_2) \cup (B_2\setminus B_1)</math> is the symmetric difference of the sets <math>B_1,B_2\in\mathcal{B}_0.</math>
This gives rise to a metric space <math>\tilde{\mathcal{B}_0}</math> by identifying two sets <math>B_1,B_2\in \mathcal{B}_0</math> when <math>m(B_1\Delta B_2)=0.</math> Thus a point <math>\overline{B}\in\tilde{\mathcal{B}_0}</math> with representative <math>B\in\mathcal{B}_0</math> is the set of all <math>B_1\in\mathcal{B}_0</math> such that <math>m(B\Delta B_1) = 0.</math>
Proposition: <math>\tilde{\mathcal{B}_0}</math> with the metric defined above is a complete metric space.
Proof: Let <math display=block>\chi_B(x)=\begin{cases}1,&x\in B\\0,&x\notin B\end{cases}</math> Then <math display=block>d(B_1,B_2)=\int_S|\chi_{B_1}(s)-\chi_{B_2}(x)|dm</math> This means that the metric space <math>\tilde{\mathcal{B}_0}</math> can be identified with a subset of the Banach space <math>L^1(S,\mathcal{B},m)</math>.
Let <math>B_n\in\mathcal{B}_0</math>, with <math display=block>\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</math> Then we can choose a sub-sequence <math>\chi_{B_{n'}}</math> such that <math>\lim_{n'\to\infty}\chi_{B_{n'}}(x)=\chi(x)</math> exists almost everywhere and <math>\lim_{n'\to\infty}\int_S|\chi(x)-\chi_{B_{n'}(x)}|dm=0</math>. It follows that <math>\chi=\chi_{B_{\infty}}</math> for some <math>B_{\infty}\in\mathcal{B}_0</math> (furthermore <math>\chi (x) = 1</math> if and only if <math>\chi_{B_{n'}} (x) = 1</math> for <math>n'</math> large enough, then we have that <math>B_{\infty} = \liminf_{n'\to\infty}B_{n'} = {\bigcup_{n'=1}^\infty}\left({\bigcap_{m=n'}^\infty}B_m\right) </math> the limit inferior of the sequence) and hence <math>\lim_{n\to\infty}d(B_\infty,B_n)=0.</math> Therefore, <math>\tilde{\mathcal{B}_0}</math> is complete.
Proof of Vitali-Hahn-Saks theoremEdit
Each <math>\lambda_n</math> defines a function <math>\overline{\lambda}_n(\overline{B})</math> on <math>\tilde{\mathcal{B}}</math> by taking <math>\overline{\lambda}_n(\overline{B})=\lambda_n(B)</math>. This function is well defined, this is it is independent on the representative <math>B</math> of the class <math>\overline{B}</math> due to the absolute continuity of <math>\lambda_n</math> with respect to <math>m</math>. Moreover <math>\overline{\lambda}_n</math> is continuous.
For every <math>\epsilon>0</math> the set <math display=block>F_{k,\epsilon}=\{\overline{B}\in\tilde{\mathcal{B}}:\ \sup_{n\geq1}|\overline{\lambda}_k(\overline{B})-\overline{\lambda}_{k+n}(\overline{B})|\leq\epsilon\}</math> is closed in <math>\tilde{\mathcal{B}}</math>, and by the hypothesis <math>\lim_{n\to\infty}\lambda_n(B)=\lambda(B)</math> we have that <math display=block>\tilde{\mathcal{B}}=\bigcup_{k=1}^{\infty}F_{k,\epsilon}</math> By Baire category theorem at least one <math>F_{k_0,\epsilon}</math> must contain a non-empty open set of <math>\tilde{\mathcal{B}}</math>. This means that there is <math>B_0\in \mathcal{B}</math> and a <math>\delta>0</math> such that <math display=block>d(B,B_0)<\delta \Rightarrow\sup_{n\geq1}|\overline{\lambda}_{k_0}(\overline{B})-\overline{\lambda}_{k_0+n}(\overline{B})|\leq\epsilon.</math> On the other hand, any <math>B\in\mathcal{B}</math> with <math>m(B)\leq\delta</math> can be represented as <math>B=B_1\setminus B_2</math> with <math>d(B_1,B_0)\leq\delta</math> and <math>d(B_2,B_0)\leq \delta</math>. This can be done, for example by taking <math>B_1=B\cup B_0</math> and <math>B_2=B_0\setminus(B\cap B_0)</math>. Thus, if <math>m(B)\leq\delta</math> and <math>k\geq k_0</math> then <math display=block>\begin{align}|\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{align}</math> Therefore, by the absolute continuity of <math>\lambda_{k_0}</math> with respect to <math>m</math>, and since <math>\epsilon</math> is arbitrary, we get that <math>m(B)\to0</math> implies <math>\lambda_n(B) \to 0</math> uniformly in <math>n.</math> In particular, <math>m(B) \to 0</math> implies <math>\lambda(B) \to 0.</math>
By the additivity of the limit it follows that <math>\lambda</math> is finitely-additive. Then, since <math>\lim_{m(B) \to 0}\lambda(B) = 0</math> it follows that <math>\lambda</math> is actually countably additive.