Radon measure is inner regular if and only if it is semi-finite Let $X$ be a locally compact Hausdorff space and $\lambda: \mathcal{B}(X) \to [0, \infty]$ an outer regular Borel measure (finite on compact sets) that is inner regular on open sets. It is well known that $\lambda$ is inner regular on every Borel set $B \in \mathcal{B}(X)$ with $\lambda(B) < \infty$.
Show the following: $\lambda$ is inner regular on every Borel set if and only if $\lambda$ is semi-finite, i.e. for any $B \in \mathcal{B}(X)$ with $\lambda(B) = \infty$ there is a $C \in \mathcal{B}(X)$ with $C \subseteq B$ and $0 < \lambda(C) < \infty$.
 A: We need a combination of the two usual versions of the Riesz-Markov-Kakutani representation theorem, see for example Salamon's Measure and Integration, Theorem 3.15:

Theorem Let $X$ be a locally compact Hausdorff space and $I: C_c(X) \to \mathbb{R}$ a positive functional.
Then there is a unique inner regular Borel measure $\mu_1: \mathcal{B}(X) \to [0, \infty]$ and a unique outer regular Borel measure $\mu_2: \mathcal{B}(X) \to [0, \infty]$ that is inner regular on open sets such that $$I(f) = \int_X f \, d\mu_1 = \int_X f \, d\mu_2$$ for all $f \in C_c(X)$. Furthermore $\mu_1$ and $\mu_2$ agree on open and compact sets and we have $\mu_1 \leq \mu_2$.

Now let $\mu$ denote the inner regular Borel measure representing the same positive functional as $\lambda$ according to the above theorem. For $B \in \mathcal{B}(X)$ with $\mu(B) < \infty$ we have $$\mu(B) = \sup \{\mu(K) \mid K \subseteq B \text{ compact}\},$$
so we find nested compact sets $K_n \subseteq B$ such that $\mu(B) - \mu(K_n) < \frac{1}{n}$. Hence $A := \bigcup_{n \in \mathbb{N}} K_n$ satisfies $$\mu(B) = \lim_{n \to \infty} \mu(K_n) = \mu(A) \text{ and } \mu(B) = \lim_{n \to \infty} \mu(K_n) = \lim_{n \to \infty} \lambda(K_n) = \lambda(A).$$
Moreover we have $\lambda(B) = \mu(B)$ for every $B \in \mathcal{B}(X)$ with $\lambda(B) < \infty$ since $\lambda$ and $\mu$ are inner regular on such $B$.
Now we can show the claimed equivalence:
First let $\lambda$ be semi-finite and let $B \in \mathcal{B}(X)$. We want to show that $\lambda(B) = \mu(B)$. Since this is the case if $\mu(B) = \infty$ or $\lambda(B) < \infty$, assume now that $\mu(B) < \infty = \lambda(B)$.
By the above we find an $A \in \mathcal{B}(X)$ with $$\infty > \mu(B) = \mu(A) = \lambda(A), \text{ so } \mu(B \setminus A) = 0 \text{ and } \lambda(B \setminus A) = \infty.$$
Semi-finiteness of $\lambda$ now gives a $C \in \mathcal{B}(X)$ with $C \subseteq B \setminus A$ such that $\lambda(C) \in (0, \infty)$ and we obtain the contradiction $$0 < \lambda(C) = \mu(C) \leq \mu(B \setminus A) = 0.$$
Hence $\lambda(B) = \mu(B)$ and we conclude that $\lambda = \mu$ is inner regular.
Conversely assume that $\lambda$ is inner regular. Then for every $A \in \mathcal{B}(X)$ with $\lambda(A) = \infty$ we have $$\infty = \lambda(A) = \sup\{\lambda(K) \mid K \subseteq A \text{ compact}\},$$
so we find a compact $K \subseteq A$ with $1 < \lambda(K) < \infty$. Therefore $\lambda$ is semi-finite.
