Fall 2017 - Problem 3

Solution.

First, let $f = \sum_{k=1}^N c_k \chi_{\poc{a_k,b_k}}$, where $-\infty \leq a_k \leq b_k \leq \infty$ and the $\poc{a_k, b_k}$ are pairwise disjoint. Then

Since $\set{F_n}_n$ converges uniformly, it is uniformly Cauchy, so this quantity tends to $0$ as $n, m \to \infty$. This proves the result for simple functions.

Since $\set{F_n}_n$ is uniformly Cauchy, there exists $N \in \N$ such that if $n, m \geq N$, then $\abs{F_n\p{x} - F_m\p{x}} < \epsilon$ for any $x \in \R$. Since $\mu_N$ is a finite measure, we can apply downward continuity to get $R > 0$ such that $F_N\p{-R} = \mu_N\p{-\infty, -R} < \epsilon$. Thus, for $n \geq N$,

By applying the same argument to $1 - F_n\p{x}$ and making $R$ larger if necessary, we see that

for $n \geq N$.

Let $f$ be a bounded continuous function and $\epsilon > 0$. Since $f$ is bounded, we know that $f$ can be approximated uniformly by simple functions (regardless of sign of $f$) on compact sets, so let $g$ be a simple function such that $\norm{f - g}_{L^\infty\p{B\p{0,R}}} < \epsilon$.

Making $N$ larger if necessary, we may assume that if $n, m \geq N$, then $\abs{\int g \,\diff\mu_n - \int g \,\diff\mu_m} < \epsilon$. Hence, if $M$ is an upper bound for $f$,

Sending $\epsilon \to 0$, we get the claim.