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Fall 2018 - Problem 3

weak convergence

Let $\p{X, \rho}$ be a compact metric space and let $P\p{X}$ be the set of all probability measures on the Borel sigma-algebra of $X$ (i.e., $\mu \in P\p{X}$ if $\mu$ is a positive Borel measure and $\mu\p{X} = 1$ ). Assume $\set{\mu_n}_n$ is a sequence in $P\p{X}$ and $\mu$ is another element of $P\p{X}$ such that for all continuous $\func{f}{X}{\R}$ ,

\int_X f\p{x} \,\diff\mu_n \xrightarrow{n\to\infty} \int_X f\p{x} \,\diff\mu.

Prove that

\mu_n\p{E} \xrightarrow{n\to\infty} \mu\p{E}

whenever $E$ is a Borel subset of $X$ such that $\mu\p{\cl{E}} = \mu\p{E^\itr}$ , where $\cl{E}$ is the closure of $E$ and $E^\itr$ is the interior of $E$ .

Solution.

Observe that $\chi_{E^\itr} \leq \chi_E \leq \chi_{\cl{E}}$ . Since $\cl{E}$ is closed, $\chi_{\cl{E}}$ is upper-semicontinuous and because $E^\itr$ is open, $\chi_{E^\itr}$ is lower-semicontinuous. Hence, because $X$ is a metric space, we get sequences of continuous functions $\set{f_n}_n$ and $\set{g_n}_n$ such that $f_n$ decreases to $\chi_{\cl{E}}$ and $g_n$ increases to $\chi_{E^\itr}$ . Thus,

\begin{gathered} \int_X g_k \,\diff\mu_n \leq \mu_n\p{E^\itr} \leq \mu_n\p{E} \leq \mu_n\p{\cl{E}} \leq \int_X f_k \,\diff\mu_n \\ \implies \int_X g_k \,\diff\mu \leq \liminf_{n\to\infty} \mu_n\p{E} \leq \limsup_{n\to\infty} \mu_n\p{E} \leq \int_X f_k \,\diff\mu \end{gathered}

by weak-* convergence.

By continuity, $g_1$ is bounded from below by $M > 0$ , so $g_k + M \geq 0$ . By monotone convergence,

\lim_{k\to\infty} \int_X g_k \,\diff\mu = \lim_{k\to\infty} \int_X \p{g_k + M} - M \,\diff\mu = \int_X \chi_{E^\itr} + M \,\diff\mu - \int_X M \,\diff\mu = \int_X \chi_{E^\itr} \,\diff\mu,

since $\mu$ is a probability is a probability measure, i.e., constants are integrable. Similarly, $f_1$ is bounded, hence $\mu$ -integrable, so by dominated convergence,

\lim_{k\to\infty} \int_X f_k \,\diff\mu = \int_X \chi_{\cl{E}} \,\diff\mu.

Altogether, we get

\mu\p{E^\itr} \leq \liminf_{n\to\infty} \mu_n\p{E} \leq \limsup_{n\to\infty} \mu_n\p{E} \leq \mu\p{\cl{E}} = \mu\p{E^\itr},

so these are equalities, and so

\lim_{n\to\infty} \mu_n \p{E} = \mu\p{E},

as desired.