This is somewhat a basic question, but I'm having difficulty proceeding with a certain part of the proof. I was reading Billingsley "Convergence of Probability Measures", and I encountered the following question:
Question:
Given a metric space $X$ and $\mathscr{B}$ the Borel sigma algebra, prove that a probability measure $P$ on $X$ is tight iff
$$\forall A \in \mathscr{B},\quad \sup \{P(K): K\subset A, K \mbox{ compact}\} = P(A)$$
My Proof:
($\Rightarrow$)
Put $A=X$. Then we have
$$1 = P(X) = \sup \{P(K): K\subset X, K \mbox{ compact}\}$$
By definition of sup, we get $\forall \epsilon > 0$, $\exists K$ compact such that
$$P(K) \geq 1-\epsilon$$ Thus P is tight.
($\Leftarrow$)
Billingsley suggests using the following theorem:
Probability measures on metric spaces are regular. That is
$$\forall A \in \mathscr{B}, \forall \epsilon > 0, \exists F \mbox{ closed}, G \mbox{ open such that}$$
$$F \subset A \subset G, \quad P(G \setminus F) < \epsilon$$
I do not know how to use compactness here. I reasoned that if I could prove that $\forall \epsilon > 0$, $\exists K \subset A$ compact such that $P(A\setminus K) < \epsilon$, I would be done. Since the compact sets in "tightness", needn't be subsets of A, I don't know how to use it. I was able to show the easy inequality which is:
$$\sup \{P(K): K\subset A, K \mbox{ compact}\} \leq P(A)$$
(since $K\subset A$)
I would appreciate any ideas, hints and tips (if not answers) to this. References are also welcome. I searched for "Tight Regular Measures" and "Tight Probability" but to no useful results.
Best Answer
First, we will see how Billingsley's claim helps to conclude. For each $n$, consider $K_n$ compact such that $\mu(K_n)>1-n^{-1}$. We can assume the sequence $(K_n,n\geqslant 1)$ to be increasing. Take $B\in\mathcal B(X)$ and $\varepsilon>0$. By the claim, there is $F\subset B$ such that $\mu(B\setminus F)<\varepsilon$. Fix $n$ such that $\varepsilon>n^{-1}$. Then $$\mu(B\cap K_n)=\mu(B)-\mu(B\setminus (B\cap K_n))=\mu(B)-\mu(B\setminus K_n)\\\geqslant \mu(B)-\mu(K_n^c)\geqslant \mu(B)-n^{-1}\geqslant \mu(B)-\varepsilon.$$ We thus get $\mu(F\cap K_n)>\mu(B)-2\varepsilon$, and $F\cap K_n$ is a compact subset of $B$.
Now we shall see the claim is true. First, note that in a metric space, an closed set is a countable intersection of open sets, and using the fact that the measure is finite, we deduce that the assertion is true for $B$ open.
Then the remaining task is to show that the collection of the $B$ for which the property holds is a $\sigma$-algebra.