# 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?

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.

• Thanks for your answer! The only thing I don't understand is why the continuity of $\mu$ implies that first equation. Don't we need that the $A_n$ are increasing? Commented Nov 21, 2015 at 17:51
• @anova No... set $B_N:=\bigcup_{n=1}^N A_n$, then $B_N \uparrow A$, i.e. we can use the continuity of the measure for the sequence $(B_N)_N$.
– saz
Commented Nov 21, 2015 at 18:03
• Yes, of course. Thanks. When I try to do it for the complement I have some trouble because $X \backslash \bigcup_{n=1}^N A_n \downarrow X \backslash \bigcup_{n=1}^{\infty} A_n$. The objective is to obtain the inequality $\mu(X \backslash \bigcup_{n=1}^{\infty} A_n) \leq \mu(\bigcup_{n=1}^N C_n) + k\cdot \epsilon$ (for some $k$), right? (where $C_n \subset X \backslash A_n$) Commented Nov 21, 2015 at 18:15
• @anova Yeah. So what kind of trouble do you have? Since $\mu$ is a finite measure, it is also continuous from above, i.e. for any sequence $B_N \downarrow B$, we have $$\mu(B) = \lim_{N \to \infty} \mu(B_N).$$
– saz
Commented Nov 21, 2015 at 18:56
• Yes, now I have it. It is true that we don't need the analog of (1) in the proof for the complement, right? Commented Nov 21, 2015 at 21:23