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?


1 Answer 1


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.

  • $\begingroup$ Do you have an example for a non-metric space for which this is not true? $\endgroup$
    – Cronus
    Commented Feb 9, 2018 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
    Commented Feb 10, 2018 at 21:39

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