Probability laws on a compact metric space have weakly convergent subsequences

K cannot be empty because the given laws have mass one. List a countable dense test family $f_1,f_2,\dots$ by F1. Each numerical sequence $\int f_{j}d{\mu }_n$ is bounded by $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}$. F2 supplies nested infinite subsequences along which the first j integrals converge. AC supplies these successive selections; taking the jth index of the jth subsequence gives one increasing diagonal subsequence $n_j$ with convergence for every listed test.

For any continuous f and $\eta$>0 choose a listed test h with $\mathrm{\parallel }f-h{\mathrm{\parallel }}_{\mathrm{\infty }}<\eta$. The inequality $\mathrm{\mid }\int fd{\mu }_{n_j}-\int fd{\mu }_{n_k}\mathrm{\mid }\le 2\eta +\mathrm{\mid }\int hd{\mu }_{n_j}-\int hd{\mu }_{n_k}\mathrm{\mid }$ shows the f integrals are Cauchy. Define L(f) as their finite limit. Taking limits in finite linear combinations gives linearity; nonnegative f has nonnegative integrals and hence L(f)>=0; also L(1)=1.

The compact metric space K is Hausdorff and locally compact (K itself is a compact neighborhood of each point), and $C_c(K)=C(K)$. The positive functional in step 1.2 therefore satisfies F3. Its representing Borel measure has total mass L(1)=1. The defining identity L(f)=integral f against that measure, combined with step 1.2, is weak convergence of the extracted subsequence.