Let $(X, \rho)$ be a metric spac and let $\mu$ be a finite Borel measure on it. Assume thay $(X, \rho)$ is complete and separable. Prove that for every $\epsilon > 0$ there is a complact subset $K \subset X$ such that $\mu(K^c) < \epsilon$.
I am not sure what to do here. We are to use that a subset of a complete metric space is compact iff it is closed and totally bounded. I managed to prove that for any $\delta > 0$ there is a finite subset $\{x_1,…,x_n\}$ such that $\mu((B(x_1,\delta) \cup… \cup B(x_n,\delta))^c) < \epsilon$. It seems like I have to take something like closure, but I am not sure how.
Best Answer
For each $n$ take $\delta =\frac 1 n$ and replace $\epsilon$ by $\epsilon /2^{n}$ in your argument to get a finite set $\{x_{n1},x_{n2},...,x_{nk_n}\}$. Consider $H=\cap_n [B(x_{n1},\frac 1 n)\cup B(x_{n2},\frac 1 n),...,\cup B(x_{nk_n},\frac 1 n)]$ This set is totally bounded and hence its closure $K$ is compact. The measure of $H^{c}$ is less than $\epsilon$. Hence $\mu(K^{c}) <\epsilon$.