4
$\begingroup$

Lets say you have a locally comapct metric space. And you have a complex Borel measure on that space. And the space is given by $\Omega=\bigcup\limits_{n=1}^\infty A_n$, where each $A_n$ is compact. Then I am supposed to show that this is a regular measure. That is if $\nu$ is the measure and $|\nu|$ is the total variation. Then for any Borel set E, and any $\epsilon> 0$, there is a compact set K and an open set O, such that $K \subset E\subset O$, and $|\nu|(O\backslash K)<\epsilon.$

I dont really know how to show this.

We know that $|\nu|$ must be a finite measure. First I thought that by using the continuity of measures we can choose a set $\cup_{n=1}^NA_n=A$ , such that $|\nu|(E\backslash A)<\epsilon/4$. But one problem is that even if A is compact $A\cap E$ may not be.

I guess I must use the local compactness in some way.

Any hints?

$\endgroup$
0
1
$\begingroup$

Since each $A_n$ is compact and metrizable, it is separable. Hence, so is $\Omega = \bigcup_n A_n$. Since $\Omega$ is metric, we see that $\Omega$ is second countable.

Now, Theorem 7.8 in Folland's book "Real Analysis" implies that every Borel measure on $\Omega$ which is finite on compact sets (in particular every finite Borel measure) is regular.

The proof idea of the theorem is to consider the functional $f \mapsto \int f \, d\mu$ (with $\mu = |\nu|$) in your case. By the Riesz representation theorem, we then have $\int f \, d\mu = \int f \, d\gamma$ for all $f \in C_c(\Omega)$ and some regular measure $\gamma$.

Now, for an arbitrary open set $U \subset \Omega$, we have $U = \bigcup_n K_n$ for compact sets $K_n$ by local compactness and second countability. Now, choose $f_n \in C_c$ with $\rm {supp} f_n \subset U$ and $0 \leq f_n \leq 1$ and $f_n |_{K_n} \equiv 1$. We get (by the monotone convergence theorem, since $f_n \uparrow 1_U$) $$ \mu(U) = \lim_n \int f_n \, d\mu = \lim_n \int f_n \, d\gamma = \gamma(U). $$

Now, if two finite measures agree on the $\pi$ system of all open sets, they agree on the whole Borel sigma algebra, as a Corollary of Dynkins $\pi$-$\lambda$ theorem.

Final comment: Your claim would be false if $\Omega$ was not metrizable.

$\endgroup$
2
  • $\begingroup$ Do you have an example for a non-metric space for which this is not true? $\endgroup$ – Cronus Feb 9 '18 at 2:26
  • $\begingroup$ @Cronus: Good question. See point (3) in the following question: math.stackexchange.com/questions/1234859/…. If I find the time I might post the details tomorrow. $\endgroup$ – PhoemueX Feb 10 '18 at 21:39

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.