Supppose $\mu$ is nonnegative and finite additive on $\Bbb{R}^k$ and that $\mu(\mathbb{R}^k)<\infty$. Suppose further that $\mu(A)= \sup(\mu(K))$, where $K$ ranges over the compact subsets of $A$. Show that $\mu$ is countably additive.
To show, $A_i$ disjoint, $\mu(\cup_i A_i)=\sum_i \mu(A_i)$
Let, $K_i\subset A_i$ are compact such that $\mu(A_i-K_i)< \frac{\epsilon}{2^i}$. For $\epsilon$ there exists a compact set $K$ such that $\mu(\cup_i A_i-K)<\epsilon$.
We also have $\sum_i^n \mu(A_i) – \sum_i^n \mu(K_i)= \mu(\cup_i^n A_i – \cup_i^n K_i)<\epsilon$, for each $n$.(finite additivity)
What should I next?
Best Answer
HINTS:
Intermediate step: If $A_n $ decreases to empty set then $\mu (A_n) \to 0$. To prove this choose compact sets $K_n \subseteq A_n$ such that $\mu(A_n\setminus K_n) <\frac {\epsilon} {2^{n}}$. Since $\cap_n K_n \subseteq \cap_n A_n$ is empty there exists $N$ such that $\cap_{i=1}^{N} K_i$ is empty. [This is a basic result for compact sets].
Hence $\mu (A_N) =\mu (\cap_{i=1}^{N} A_i)\leq \mu (\cup_{i=1}^{N} (A_i \setminus K_i))\leq \epsilon$. It follows that $\mu (A_n) \to 0$ as claimed.
Rest is easy. If $A$ is the disjoint union of $A_n$'s apply above result to $B_i=A\setminus \cup _{i=1}^{n} A_i$. I will leave the rest to you.