Show that $\mu(A) = \sup \{ \mu(K) : K \subset A, K \text{ compact}\}$ is a measure Problem: Let $\mu : \mathcal{B}(\mathbb{R}^k) \rightarrow \mathbb{R}$ be a nonnegative and finitely additive set function with $\mu(\mathbb{R}^k) < \infty $.
Suppose that 
\begin{equation}
\mu(A) = \sup \{ \mu(K) : K \subset A, K \text{ compact}\} 
\end{equation}
for each $A \in \mathcal{B}(\mathbb{R}^k)$.
Then $\mu$ is a finite measure.
I'm stuck showing that $\mu$ is countably additive.
First, let $A_{i}$ disjoint, and $K_i \subset A_i$ compact such that $\mu(A_i) \leq \mu(K_i)+\epsilon/2^i$ for $\epsilon > 0$. 
Then
\begin{equation}
\mu\left(\bigcup_{i=1}^{\infty} A_i \right)
\geq \mu\left(\bigcup_{i=1}^{n} A_i \right)
\geq \mu\left(\bigcup_{i=1}^{n} K_i \right)
= \sum_{i=1}^n \mu(K_i)
\geq \sum_{i=1}^n \mu(A_i) + \frac{\epsilon}{2^i} ,
\end{equation}
which follows from monotonicity (obvious) and finite additivity (on the class of compact sets).
Hence, letting $n \rightarrow \infty$, and since $\epsilon$ was arbitrary,
\begin{equation}
\mu\left(\bigcup_{i=1}^{\infty} A_i \right)
\geq \sum_{i=1}^\infty \mu(A_i) .
\end{equation}
Next, I want to establish the reverse inequality.
This is where I'm stuck. Any ideas?
Thanks in advance!
Chris
 A: First, $\mu$ is $\sigma$-additive iff it is continuous at $\emptyset$. Suppose that $\mu$ is not $\sigma$-additive, i.e. there exists a decreasing sequence of sets $\{A_n\}$ s.t. $\cap_{n\ge 1}A_n=\emptyset$ but $\inf_n\mu(A_n)=\epsilon>0$. Let $K_n\subset A_n$ compact s.t. $\mu(K_n)\le \mu(A_n)+\epsilon2^{-(n+1)}$. Then, letting $B_n=\cap_{k=1}^nK_n$, we get
$$
\mu(A_n\setminus B_n)\le \sum_{k=1}^n\mu(A_k\setminus K_k)<\frac{\epsilon}{2}.
$$
Thus, $\mu(B_n)>0$ so that $\{B_n\}$ is a decreasing sequence of nonempty compact sets. Hence, $\cap_{n\ge 1}K_n\ne \emptyset$, a contradiction.
A: This is enough:
Claim: Let $(K_{n})_{k\in\mathbb{N}}$ a decreasing sequence of compacts subsets such that $\bigcap_{n\in\mathbb{N}}K_{n}=\emptyset$. Then, $\lim_{m\to+\infty}\mu(K_{m})=0$.
If not, there is some $\delta>0$ such that $\mu(K_{n})>\delta$, for all $n$. In particular, every $K_{n}$ is non-empty and so, by the finite intersection property, $\bigcap_{n\in\mathbb{N}}K_{n}\not=\emptyset$.
A: $\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\irun}[1]{_{#1 \in \N}}$
Unfortunately math.stackexchange hast problems with the \intertext command.
This is a very detailed proof for your problem
$\Omega = \R^n$,
$\mathfrak{A} = \mathfrak{B}(\R^n)$
At first I want to show some regularity with open sets
\begin{align*}
\mu(A) &= \mu(\Omega\backslash A^{c}) = \mu(\Omega) - \mu(A^{c}) \\
  &= \mu(\Omega) - \sup\{ \mu(K) : K \in \mathfrak{A},K \subseteq A^{c}, K \text{ compact}\}
\end{align*}
when you take in the minus sign then you receive a $\inf$ instead of the $\sup$
\begin{align*}
     = \mu(\Omega) + \inf\{ -\mu(K) : K \in \mathfrak{A}, K \subseteq A^{c}, K \text{ compact}\} \\
     = \inf\{ \mu(\Omega) -\mu(K) : K \in \mathfrak{A}, K \subseteq A^{c}, K \text{ compact}\}
\end{align*}
Have in mind, that $\mu(\Omega) -\mu(K)$ is nothing else than $\mu(K^{c})$ and that $K\in \mathfrak{A}$ is equivalent to $K^{c}\in\mathfrak{A}$
\begin{align*}
= \inf\{ \mu(K^{c}) : K^{c} \in \mathfrak{A}, (K^{c})^{c} \subseteq A^{c}, (K^{c})^{c} \text{ compact}\}
\end{align*}
now replace $K^{c}$ with $O$
\begin{align*}
= \inf\{ \mu(O) : O \in \mathfrak{A}, O^{c} \subseteq A^{c}, O^{c} \text{ compact}\} \\
= \inf\{ \mu(O) : O \in \mathfrak{A}, A \subseteq O, O^{c} \text{ compact}\}
\end{align*}
    That means that for every $A \in \mathfrak{A}$ and every $\epsilon>0$ there is a $O\in\mathfrak{A}$ such that $A \subseteq O^{}$ and $\mu(O) - \epsilon \leq \mu(A)$. Clearly $O$ is open since $O^{c}$ is compact.
Let $(A_{n})\irun{n}$ be a sequence of disjoint Sets of $\mathfrak{A}$, such that the union $\bigcup\irun{n} A_{n}$ is in $\mathfrak{A}$.
Then there is regarding to the already proven for every $A_{n}$ a set $O_{n}$ such that $A_n \subseteq O_n$ and $\mu(O_{n}) - \frac{\epsilon}{2^{n}} \leq \mu(A_{n})$. 
Now choose an arbitrary
\begin{align*}
K \in \bigg\{  K \in \mathfrak{A} \;\,\bigg|\;\, K \subseteq \bigcup\irun{n} A_{n}, K \text{ compact}\bigg\}
\end{align*}
It is easy to see $K \subseteq \bigcup\irun{n} A_{n}\subseteq \bigcup\irun{n} O_{n}$. Since $K$ is compact and all $O_{n}$ are open, there must be a $n_{0}\in \N$, such that ${K} \subseteq \bigcup_{n=1}^{n_{0}} O_{n}^{}$ is true. Now we receive
\begin{align*}
   K  \subseteq \bigcup_{n=1}^{n_{0}} O_{n}
  \end{align*}
    because of the monotony of $\mu$
        \begin{align*}
   \mu(K)  \leq \mu\bigg(\bigcup_{n=1}^{n_{0}} O_{n}\bigg)
     = \sum_{n=1}^{n_{0}} \mu(O_{n})
     \leq \sum_{n=1}^{n_{0}} \Big(\mu(A_{n}) + \frac{\epsilon}{2^{n}}\Big)
     \leq \Big(\sum\irun{n} \mu(A_{n}) \Big)+ \epsilon
  \end{align*}
for all $\epsilon > 0$. Since the right-hand-side is independ of $K$ this inequality is also true for the supremum of all such $K$. This means nothing else than
        \begin{align*}
   \mu\Big(\bigcup\irun{n} A_{n}\Big) \leq \Big(\sum\irun{n} \mu(A_{n}) \Big)+ \epsilon \quad \text{for}\quad \epsilon > 0.
  \end{align*}
Now this inequality must also hold for $\epsilon = 0$ and this shows that $\mu$ is $\sigma$-subadditiv.
