Theorem 3.3.7 (Compact-metric measure selection)2C91786CB405

Proof. If $X$ is empty or $M=0$, every measure is zero and the assertion is immediate. First construct a countable dense subset of $C(X)$. For each positive integer $n$, choose a finite $1/n$-net $x_{n,1},\ldots ,x_{n,N_ n}$, and put

The denominator is everywhere positive. These continuous functions form a partition of unity, and $p_{n,i}(x)>0$ implies $\operatorname {dist}(x,x_{n,i})<2/n$. For $f\in C(X)$, choose rational $q_{n,i}$ within $1/n$ of $f(x_{n,i})$. Then $\sum _ iq_{n,i}p_{n,i}$ converges uniformly to $f$ by uniform continuity. All rational combinations for all $n$ form a countable dense family.

Enumerate that family as $(g_ l)$. For every $l$, the sequence $\int g_ l\, d\mu _ j$ is bounded by $M\| g_ l\| _\infty $. Successive subsequences and diagonal extraction yield a subsequence along which all these integrals converge. If $f\in C(X)$ and $g_ l$ is uniformly close to $f$, then

Hence the integrals of every $f$ are Cauchy. Their limit $L(f)$ is a real linear functional with $|L(f)|\leq M\| f\| _\infty $. Riesz representation gives the required $\mu $ with $\| \mu \| \leq M$. If the original measures are positive, $L(f)\geq 0$ for $f\geq 0$; the positivity part of Riesz representation makes $\mu $ positive. $\square$