<Home>Qualifying Exams><Analysis>Fall 2011 - Problem 6

Fall 2011 - Problem 6

Urysohn's lemma, weak convergence

Let $\p{X, d}$ be a compact metric space. Let $\mu_n$ be a sequence of positive Borel measures on $X$ that converge in the weak-* topology to a finite positive Borel measure $\mu$ , that is,

\int_X f \,\diff\mu_n \to \int_X f \,\diff\mu \quad\text{for all}\quad f \in C\p{X}.

Here, $C\p{X}$ denotes the space of bounded continuous functions on $X$ . Show that

\mu\p{K} \geq \limsup_{n\to\infty}\, \mu_n\p{K} \quad\text{for all compact sets}\quad K \subseteq X.

Solution.

If $K = X$ , then $\chi_K$ is continuous, and so

\mu\p{K} = \int_X \chi_K \,\diff\mu = \lim_{n\to\infty} \chi_K \,\diff\mu_n = \lim_{n\to\infty} \mu_n\p{K}

by weak convergence, so we have equality. Now suppose $X \setminus K \neq \emptyset$ and let

F_n = \set{x \in X \st d\p{x, K} \geq \frac{1}{n}}.

For $n$ large enough, $F_n \neq \emptyset$ since $X \setminus K \neq \emptyset$ and because the distance between two disjoint closed sets is positive. Since $x \mapsto d\p{x, K}$ is continuous, $F_n$ is closed. Hence, $K$ and $F_n$ are disjoint closed sets, so by Urysohn's lemma, there exists a continuous function $\func{f_n}{X}{\br{0,1}}$ such that $f\p{x} = 1$ for $x \in K$ and $f\p{x} = 0$ for $x \in F_n$ .

Set $g_n = \min\,\set{f_1, \ldots, f_n}$ , which is continuous since it is the minimum of finitely many continuous functions. Thus, $\set{g_n}_n$ is a decreasing sequence of functions. Moreover, $x \in K$ , then $f_1\p{x} = \cdots = f_n\p{x} = 1 \implies g_n\p{x} = 1$ and if $x \in F_n$ , then $g_n\p{x} \leq f_n\p{x} = 0$ , so $g_n$ still has the same separation property as the $f_n$ . Finally, $g_n \to \chi_K$ pointwise: if $x \in K$ , then $g_n\p{x} = 1$ for all $n \geq 1$ . Otherwise, $d\p{x, K} > 0$ , so there exists $n \geq 1$ such that $x \in F_n$ , and so $g_k\p{x} = 0$ for all $k \geq n$ . Hence,

\tag{1} \int_X g_k \,\diff\mu_n \geq \int_X \chi_K \,\diff\mu_n \implies \int_X g_k \,\diff\mu \geq \limsup_{n\to\infty} \int_X \chi_K \,\diff\mu_n,

since the limit on the left-hand side exists as $n \to \infty$ . Since $g_k$ decreases to $\chi_K$ , $1 - g_k \geq 0$ increases to $\chi_K$ , so by the monotone convergence theorem,

\int_X g_k \,\diff\mu = \int_X 1 - \p{1 - g_k} \,\diff\mu \xrightarrow{k\to\infty} \mu\p{X} - \int_X 1 - \chi_K \,\diff\mu = \int_X \chi_K \,\diff\mu.

Thus, sending $k \to \infty$ in (1) and noting that the right-hand side is independent of $k$ ,

\mu\p{K} = \int_X \chi_K \,\diff\mu \geq \limsup_{n\to\infty}\int_X \chi_K \,\diff\mu_n = \limsup_{n\to\infty} \mu_n\p{K}

as required.