# A locally finite Borel measure on a locally and $\sigma$ compact metric space is a Radon measure

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 $X$.

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 $$\mu( U \backslash L ) \le \sum \mu(U'_n \backslash L_n) < \epsilon$$
• @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, – Orest Bucicovschi Dec 20 '15 at 13:59