Reference Request: Finite Borel measure are Radon Original:
Let $\mu$ be a finite measure on $\mathbb{R}^k$ dominated by the Lebesgue measure.  Does $\mu$ need to be a Radon measure?
Updated Question:
Does anyone have a reference to the fact that if $\mu$ is a finite Borel measure on $\mathbb{R}^k$ then it is a Radon measure?
 A: So to verify that some measure is a Radon measure we need to verify, that it is inner regular, outer regular and is finite on all compact sets. In your case, since $\mu$ is assumed to be finite, we only need to verify the other two properties. Since we can do that in an even more general setting than $\mathbb{R}^k$ I will do so and prove the following:

If $(X, d)$ is a metric space and $\mu$ is a finite measure on the Borel sigma algeba $B_X$, then $$\mu(A) = \inf\{\mu(O) \mid A \subset O, O \text{ open}\} = \sup\{\mu(F) \mid F \subset A, F \text{ closed}\}$$
  In particular, if $(X,d)$ is complete and separable, then $$\mu(A) = \sup\{\mu(K) \mid K \subset A, K \text{ compact}\}$$

Proof: Define the set $$\mathcal{G} = \{A \in B_X \mid \forall \epsilon>0 \colon\exists F \subset A \subset O \colon \mu(O\backslash F) < \epsilon\} \tag{$\star$}$$ where the sets $F$ are closed, and the sets $O$ are assumed open in the definition of $(\star)$. First note, that $\mathcal{G}$ contains all closed sets. Indeed, if $A$ is a closed set, just choose $F = A$ in $(\star)$ and consider the open sets $O_k = \{x \in X \mid \text{dist}(x, A) < 1/k\}$, where $\text{dist}(x, A) = \inf\limits_{a \in A} d(x,a)$. Since $O_k \searrow A$ (since $A$ is closed) we may find an index $K$ such that $\mu(O_K) < \mu(A) + \epsilon$ by continuity of finite measures from above. So set $O = O_K$. Thus all closed sets are indeed in $\mathcal{G}$. Furthermore, $\mathcal{G}$ is a sigma algebra: It is clear, that $X \in \mathcal{G}$. If $A \in \mathcal{G}$, then let $\epsilon > 0$ and let $F \subset A \subset O$ be given such that $\mu(O \backslash F) < \epsilon$. Now consider the closed set $C = O^c$ and the open set $U = F^c$, then $C \subset A^c \subset U$ and $\mu(U \backslash C) = \mu(O \backslash F) < \epsilon$, so $A^c \in \mathcal{G}$. Last but not least, if $\{A_n\}_n \subset \mathcal{G}$ is a countable collection of sets, then set $A = \bigcup A_n$. Let $\epsilon > 0$, then for each $n \geq 1$ we may find $F_n \subset A_n \subset O_n$ such that $\mu(O_n \backslash F_n) < 2^{-(n+1)} \epsilon$. Set $O = \bigcup O_n$ and $F = \bigcup\limits_{n = 1}^N F_n$ where $N$ is so big, that $\mu(\bigcup F_n) < \epsilon/2 + \mu(F)$ (this is possible by continuity from below). Now note, that \begin{eqnarray}
\mu(O \backslash F) = \mu(O) - \mu(F) < \mu(O) - \mu(\bigcup F_n) + \epsilon/2 = \mu(\bigcup O_n \backslash \bigcup F_n) + \epsilon/2 \\ \leq \mu(\bigcup O_n\backslash F_n) + \epsilon/2 \leq \sum \mu(O_n \backslash F_n) + \epsilon/2 < \epsilon
\end{eqnarray} 
Hence $\mathcal{G}$ is a sigma algebra containing all closed sets, which form a generator for $B_X$. Thus $B_X = \mathcal{G}$.
Now assume $(X, d)$ to be complete and separable. Consider a countable dense subset $\{x_n\}_n \subset X$ and the closed delta-balls $K(x_n, \delta) = \{x \in X \mid d(x_n,x) \leq \delta\}$. For all $m \geq 1$ fixed we have $X = \bigcup\limits_{n \geq 1} K(x_n, 1/m)$. Thus if $\epsilon > 0$ is given, then for all $m \geq 1$ there exists $N_m$ such that $$\mu \big(X \backslash \bigcup\limits_{n = 1}^{N_m} K(x_n, 1/m) \big)< 2^{-m}\epsilon$$
Let $K = \bigcap\limits_{m \geq 1} \bigcup\limits_{n=1}^{N_m} K(x_n, 1/m)$, then $\mu(K^c) \leq \epsilon$ by construction (check this). Moreover, it is clear, that $K$ is closed and it is in particular totally bounded. Thus $K$ is compact. Now if $A \in B_X$ we can find a closed set $F \subset A$ such that $\mu(F) > \mu(A) - \epsilon$. But then $F \cap K$ is compact with $F \cap K \subset A$ and satisfies $\mu(F \cap K) > \mu(A) -2 \epsilon$.
This answers your question, since $\mathbb{R}^k$ is of course a separable, complete metric space.
