I am trying to understand one implication of the Prokhorov Theorem. I do not understand justify the claims in the highlighted part. Any comments would be much appreciated.
If a family $\mathcal{A}\subset \mathcal{P}(X)$ of probability measures is tight then for any sequence $(\mu_k)_{k\in\mathbb{N}}\subset \mathcal{A}$ there exists a subsequence $\mu_{k_j}$ and a probability measure $\mu$ such that $\mu_{k_j}\to \mu$ narrowly.
For the proof, it is being assumed that $X$ is complete, separable, and locally compact. Then first we construct a sequence of compact sets $K_n$ such that $\mu(X\setminus K_n)\leq n^{-1}$ for all $\mu\in \mathcal{A}.$ Since $X$ is locally compact then by an inductive construction we can assume that $K_n\subset \text{int}(K_{n+1})$ where $\text{int}(A)$ denotes the interior of set $A$.
Given a sequence $\mu_k$ by Banach-Alaoglu's theorem we get that restriction of the measure to $K_n$ satisfies, ${\mu_{k}}_{|K_n}\to \mu^{(n)}$ weak-* converges upto a subsequence as $k\to \infty$ where $\mu^{(n)}$ is a finite signed measure. Then by a diagonal argument we can extract subsequence such that,
$${\mu_{k_j}}_{|K_n}\to \mu^{(n)},\quad \text{weak-* as }j\to \infty $$ for all $n\in \mathbb{N}.$ Now observe that $\mu^{(n)}$ vanishes outside $K_n$. I think that this holds true because if we take any set $E$ outside $K_n$ then ${\mu_{k}}_{|K_n}(E)=\mu_{k_j}(E\cap K_n) =0\to 0$ as $j\to\infty$ and thus $\mu^{(n)}(E)=0$ for all $E$ outside $K_n.$ Also we have that, $\mu^{(n)}(X\setminus K_n)\leq n^{-1}.$
Here is the part that I don't understand:
Testing against functions with compact support in $\text{int}(K_n)$ we get $\mu^{(n+1)}_{\text{int}(K_n)}=\mu^{(n)}_{\text{int}(K_n)}.$ Also since $\mu^{(n)}$ vanishes outside $K_n$ it follows $\mu^{(n+1)}\geq \mu^{(n)}.$
$ \mu^{(n+1)} (E)=\lim_j \mu_{k_j}(E \cap K_{n+1}) \geq \lim_j \mu_{k_j}(E \cap K_{n})=\mu^{(n)} (E)$ for continuity sets $E$. But that is enough to say that $\mu^{(n+1)} \geq \mu^{n}$.