6
$\begingroup$

Let $X$ be a locally compact metric space which is $\sigma$-compact, and let $\mu$ is an unsigned Borel measure which is finite on every compact set. Show that $\mu$ is a Radon measure.

I know that every unsigned Borel measure on a compact metric space which is $\sigma$-compact is a Radon measure. From the assumption, we only need to verify outer regularity and inner regularity. Inner regularity is easier, since we can write $X$ as a countable union and compacts sets $K_n$, and a closed set in each $K_n$ is also a closed set in $X$. I have difficulty verifying outer regularity, open set in $K_n$ may not be open in $X$.

$\endgroup$
2
  • $\begingroup$ My answer here math.stackexchange.com/questions/1574021/… also answers your question. $\endgroup$
    – PhoemueX
    Commented Dec 19, 2015 at 14:07
  • $\begingroup$ @PhoemueX I have seen it. I don't want to invoke the Reisz representation theorem, can we prove it if we have known the compact case? $\endgroup$
    – Xiang Yu
    Commented Dec 19, 2015 at 14:12

1 Answer 1

3
$\begingroup$

HINT:

The regularity in the compact case ( see this post) implies the regularity in the $\sigma$-compact case. Indeed:

Take an exhaustion with compacts $K_n \subset \overset{\circ} K_{n+1}$, $\cup_n K_n = X$. Take a Borel set $A$ in $X$. Then $A \cap K_n$ is Borel in $K_{n+1}$. Take $L_n\subset A \cap K_n \subset U_n$, with $U_n$ open in $K_{n+1}$, $L_n$ compact and $\mu(U_n \backslash L_n) < \epsilon/2^n$. Modify $U_n$ to $U'_n = U_n \cap \overset{\circ} K_{n+1}$, open in $X$. We still have $$L_n\subset A \cap K_n \subset U'_n$$ and $\mu(U'_n \backslash L_n) < \epsilon/2^n$. Now take $U = \cup_n U'_n$ open and $L = \cup_n L_n$ countable union of compacts. We have $L \subset A \subset U$ and $$\mu( U \backslash L ) \le \sum \mu(U'_n \backslash L_n) < \epsilon$$

$\endgroup$
3
  • 1
    $\begingroup$ Nice answer. The key point is that every compact set in a LCH space has an open neighborhood whose closure is compact. $\endgroup$
    – Xiang Yu
    Commented Dec 20, 2015 at 1:54
  • $\begingroup$ @XiangYu: Thanks. Yes, we managed to approximate pieces of $A$ of the form $A \cap K$ by using what you noticed. and then put together the approximations, $\endgroup$
    – orangeskid
    Commented Dec 20, 2015 at 13:59
  • $\begingroup$ Is this true in the non-metric case? $\endgroup$
    – Cronus
    Commented Feb 9, 2018 at 2:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .