The collection of all regular sets is a $\sigma$-algebra iff $X$ is regular Let $X$ be a Hausdorff topological space with finite Borel measure $\mu$. Let $\mathcal{T}$ be the collection of all Borel sets $A$ with
$$\mu(A) = \sup\{\mu(C): C \subset A, C \text{ compact}\},$$
$$\mu(X \backslash A) = \sup\{\mu(C): C \subset X \backslash A, C \text{ compact}\}.$$
Prove that $\mathcal{T}$ is a $\sigma$-algebra if and only if $X \in \mathcal{T}$.
The "$\Rightarrow$"-part is kind of trivial.
Now for "$\Leftarrow$" let $X \in \mathcal{T}$. For $\mathcal{T}$ to be a $\sigma$-algebra, we need:
(i) $X \in \mathcal{T}$.
(ii) $A \in \mathcal{T} \Rightarrow X \backslash A \in \mathcal{T}$.
(iii) $A_1, A_2, ... \in \mathcal{T} \Rightarrow \bigcup_{n=1}^{\infty} A_n \in \mathcal{T}$.
I only have trouble proving the last property. (i) follows by assumption and (ii) follows because we have $A = X \backslash (X \backslash A)$.
For (iii):
Let $A_1, A_2, ... \in \mathcal{T}$. Then for each $n \in \mathbb{N}$: $A_n \in \mathcal{B}(X)$ and
$$\mu(A_n) = \sup\{\mu(C): C \subset A_n, C \text{ compact}\},$$
$$\mu(X \backslash A_n) = \sup\{\mu(C): C \subset X \backslash A_n, C \text{ compact}\}.$$
Since $\mathcal{B}(X)$ is a $\sigma$-algebra, we have $\bigcup_{n=1}^{\infty} A_n \in \mathcal{B}(X)$.
Now there is only left to prove
$$\mu(\bigcup_{n=1}^{\infty} A_n) = \sup\{\mu(C): C \subset \bigcup_{n=1}^{\infty} A_n, C \text{ compact}\},$$
$$\mu(X \backslash (\bigcup_{n=1}^{\infty} A_n)) = \sup\{\mu(C): C \subset X \backslash (\bigcup_{n=1}^{\infty} A_n), C \text{ compact}\}.$$
But I don't know how to show this. Could you help me?
Thanks in advance!
 A: Set $A := \bigcup_{n \in \mathbb{N}} A_n$ and fix $\epsilon>0$. The continuity of the measure $\mu$ implies
$$\mu(A) = \lim_{N \to \infty} \mu \left( \bigcup_{n=1}^N A_n \right).$$
Choose $N$ sufficiently large such that
$$0 \leq \mu(A)- \mu \left( \bigcup_{n=1}^N A_n \right) \leq \epsilon. \tag{1}$$
By assumption, there exists $C_n$ compact, $C_n \subset A_n$, such that
$$\mu(A_n) \leq \mu(C_n) + \frac{\epsilon}{2^n} \tag{2}$$
for each $n \in \{1,\ldots,N\}$. If we set $C := \bigcup_{n=1}^N C_n$, then $C$ is compact, $C \subset \bigcup_{n=1}^N A_n \subset A$ and
$$\begin{align*} \mu \left( \bigcup_{n=1}^N A_n \backslash C \right)=\mu \left( \bigcup_{n=1}^N A_n \backslash \bigcup_{n=1}^N C_n \right) &\leq \mu \left( \bigcup_{n=1}^N (A_n \backslash C_n) \right) \\ &\leq \sum_{n=1}^N \mu(A_n \backslash C_n) \\ &\stackrel{(2)}{\leq} \epsilon. \end{align*}$$
Combining this with $(1)$, we get
$$\mu(A) \leq \mu \left( \bigcup_{n=1}^N A_n \right) + \epsilon = \mu(C)+ \mu \left( \bigcup_{n=1}^N A_n \backslash C \right)+\epsilon \leq \mu(C) + 2 \epsilon.$$
Since $\epsilon>0$ is arbitrary, this proves
$$\mu(A) = \sup\{\mu(C); C \subset A, C \, \text{compact}\}.$$
A similar argument works for the complement; I leave it to you.
