Tightness and Inner Regularity Let $P$ be a probability measure on a Borel $\sigma$-algebra (on some metric space, $\Omega$).
It is called tight if for every $\epsilon >0$, there exists a compact $K$ such that $P(X \in K) \geq 1 - \epsilon$. 
It is called inner regular if $P(A) = \sup \{P(K) | K \subset A, K \text{ compact}\}$ for every $A$ in the Borel $\sigma$-algebra. 
It is clear that inner regularity implies tightness if we choose $A=\Omega$. For an arbitrary set $A$, using tightness we can obtain a
compact $K(\epsilon)$ such that $P(A\cap K(\epsilon)) \geq P(A) - \epsilon$. If $A$ is closed then $K \cap A$ is compact and hence we get
inner regularity for closed sets $A$. How can I prove this for general $A$?
Thanks.
 A: Hint: Denote by $\mathcal{C}$ the closed sets and by $\mathcal{O}$ the open sets.


*

*Show that
$$\Sigma := \{A; \forall \epsilon>0 \, \exists F \in \mathcal{C}, U \in \mathcal{O}: F \subseteq A \subseteq U, \mathbb{P}(U \backslash F) \leq \epsilon\}$$
is a $\sigma$-algebra and $\mathcal{C} \subseteq \Sigma$.

*Conclude that $\Sigma$ equals the Borel $\sigma$-algebra.

*Deduce that $\mathbb{P}$ is inner regular. (Use the compact set $K(\epsilon)$  mentioned in your question.)



Some more hints for the first step:


*

*$\Sigma$ is closed under countable unions: Fix $\epsilon>0$ and some sequence $(A_j)_{j \in \mathbb{N}}$. By assumption, there exist $F_j \in \mathcal{C}$ and $U_j \in \mathcal{O}$ such that $F_j \subseteq A_j \subseteq U_j$ and $$\mathbb{P}(U_j \backslash F_j) \leq \frac{\epsilon}{2} 2^{-j}. \tag{1}$$ If we define $$A := \bigcup_{j \in \mathbb{N}} A_j \qquad F := \bigcup_{j=1}^N F_j \qquad U := \bigcup_{j \in \mathbb{N}} U_j,$$ for some fixed $N \in \mathbb{N}$, then $F \subseteq A \subseteq U$, $F$ is closed and $U$ open. Using $(1)$ show that we can choose $N$ sufficiently large such that $\mathbb{P}(U \backslash F) \leq \epsilon$. This proves $A \in \Sigma$.

*$\mathcal{C} \subseteq \Sigma$: Fix some closed set $A \in \mathcal{C}$ and $\epsilon>0$. Set $F := A$. For any $\delta>0$ the set $$U_{\delta} := A+B_{\delta}(0)= \{x+y; x \in A, y \in B_{\delta}(0)\}$$ is open and $$A = \bigcap_{n \in \mathbb{N}} U_{\frac{1}{n}}.$$ Show that we can choose $n$ sufficiently large such that $\mathbb{P}(U_{\frac{1}{n}} \backslash F)<\epsilon$.


Reference: René Schilling: Maß und Intergal, Satz A.10. (in German)
A: I am trying to fill in the minor details from saz's answer. 
Countable Union: From $\sigma$-subadditivity of $P$,we have $P(\cup_j U_j \backslash \cup_j F_j) \leq \epsilon$, i.e., $\lim_{N \to \infty} 
P(\cup_j U_j \backslash \cup_{j=1}^N F_j) \leq \epsilon$. Hence, by continuity of $P$ there exists an $N$ such that $P(\cup_j U_j \backslash \cup_{j=1}^N F_j) < \epsilon + \delta$.
For closed set the argument follows again from continuity/monotonicity of $P$.
Finally, let $K(\epsilon)$ be as in the question. Let $F \subset A \subset U$ be such that $P(U\backslash F) \leq \epsilon$. Then,
$P(A\cap K(\epsilon)) = P(A\cap F \cap K(\epsilon)) +  P(A \backslash F \cap K(\epsilon))$ which implies that $P(F \cap K(\epsilon)) = P(A\cap K(\epsilon)) - P(A \backslash F \cap K(\epsilon)) \geq P(A)-\epsilon-\epsilon$ 
