1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $c < \mu(B)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.