It is given a finite Borel measure $\mu$ on a polish space $E$. The claim is that $\mu$ is then a regular measure. In the proof, it is shown that for any closed set $A$ it holds$$(1) \quad \mu(A) = \sup \{ \mu(K):K\subseteq A,\, K \text{ compact}\}.$$ Then, it is shown that a set $B$ fulfills $$ \mu(B) = \sup \{ \mu(C): C\subseteq B,\, C \text{ closed}\}.$$ The author claims that it is clear that then $C$ also fulfills $(1)$. Could someone explain to me why this is the case?


To show that for a set $B\subseteq E$ we have

$$\mu(B) = \sup \left\{ \mu(K) : K \subseteq B,\, K\text{ compact}\right\},\tag{$\ast$}$$

we need to find, for an arbitrary $c < \mu(B)$, a compact $K\subseteq B$ with $\mu(K) > c$.

We know that $(\ast)$ holds for all closed $B$, and it was shown that the analogue of $(\ast)$ with "compact" replaced by "closed" holds [for all Borel sets $B$, presumably, or at least for the class of Borel sets under consideration at that point in the proof].

So, we fix a $c < \mu(B)$. Then it is known that there is a closed $F\subseteq B$ with $\mu(F) > c$. But since $(\ast)$ holds for all closed sets, and $c < \mu(F)$, there is a compact $K\subseteq F$ with $\mu(K) > c$. But of course $K\subseteq B$ by transitivity.

So, for all $c < \mu(B)$ we have

$$c < \sup \left\{ \mu(K) : K \subseteq B,\, K\text{ compact}\right\} \leqslant \mu(B),$$

which yields $(\ast)$ by taking the supremum over all $c < \mu(B)$.


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