A question on Tight probability measures (regular measure) 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.
 A: 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. 
A: Thanks to Davide Giraudo for this:
The remaining steps are as follows. Let $K_n$, be a collection of compact sets such that $\forall n \geq 1$
$$P(K_n) > 1-\frac{1}{2n}$$
Now as per the provided theorem, given $A \in \mathscr{B}$, and $n \geq 1$ we have $F\subset A$ closed such that
$$P(A) - P(F) = P(A \setminus F) < \frac{1}{2n}$$
Now as Davide Giraudo showed, $P(A \cap K_n) \geq P(A) - \frac{1}{2n}$
Add the previous two statements, and use the fact that $F \subset A$ to get
$$P(A) - \frac{1}{n} \leq P(A \cap K_n) - P(A) + P(F)$$
$$ = P(F) - P(A \cap K_n^c) = P(F \cap K_n) + P(F \cap K_n^c)- P(A \cap K_n)$$
$$\leq P(F \cap K_n)$$
Now $F \cap K_n$ is a closed subset of a compact set in a metric space and hence is compact. Additionally $F \cap K_n \subset A$ !!. 
Therefore we have 
$$P(A) \leq P(F \cap K_n) + \frac{1}{n} \leq \sup\{P(K): K \subset A, \mbox{K compact}\} + \frac{1}{n}$$
$$\Rightarrow P(A) \leq \sup\{P(K): K\subset A, \mbox{K compact}\} + \frac{1}{n}$$
Take $n \to \infty$ to get the desired inequality.
QED
A: An alternative approach to prove 
$P(A) \leq \sup\{P(K): K \subset A, K \hspace{1mm}  compact\}$.
For any given $n$, we can find $F_n$ closed such that 
$P(A) \leq P(F_n) + \frac{1}{n}$. (By regularity of the measure P)
Also, we can find $K_n$ compact such that $P(K_n) > 1 - \frac{1}{n}$ (since we assume P is tight). Now $F_n \cap K_n$ is compact with $P(F_n \cap K_n) = P(F_n) + P(K_n) - P(F_n \cup K_n)$. 
Hence we have, $P(F_n) = P(F_n \cap K_n) - P(K_n) + P(F_n \cup K_n) \leq P(F_n \cap K_n) - 1 + \frac{1}{n} + 1$.
We thus have $P(A) \leq P(F_n \cap K_n) + \frac{1}{2n}$.
Therefore, for any $n$, we have  $J_n := F_n \cap K_n \subset A$ and $J_n$ compact such that
$P(A) \leq P(J_n) + \frac{1}{2n}$.
Since the above is true for all $n$, we have 
$P(A) \leq \sup\{P(J): J \subset A, J \hspace{1mm}  compact\}$.
A: Let $\epsilon>0$ be given. There are closed F (By Theorem. 1.1 of billingsly) and compact $K$ such that
        $$P(K)>1-\frac{\epsilon}{2},$$ i.e. $$P(K')<\epsilon/2,$$
and     $$P(A-F)<\epsilon/2,$$
Then we have $P(A-(F\cap K ))<\epsilon$, which proposes $F\cap K$ as desired compact set.
