Integral defines a Radon measure Suppose that $\mu$ is a Radon measure on $X$. If $\phi\in L^{1}(\mu)$ and $\phi\geq 0$, show that $\nu(E)=\int_{E}\phi\;d\mu$ is a Radon measure.
$\nu$ is a finite measure so we just need to check the regularity conditions. I was showing the outer regularity on all Borel sets and I did it for sets with finite $\mu$-measure. I'm not sure about the argument for sets with infinite $\mu$-measure. Likewise for inner regularity on open sets.

This is just an elaboration of Martin's astute comment above:
Let $N = \phi^{-1} \{0 \}$. Then  $\nu A = \nu (A \setminus N)$. Furthermore, $N^c = \cup_k \Delta_k$, where $\Delta_k = \phi^{-1} (\frac{1}{k}, \infty)$.
We can bound $\mu \Delta_k$ as follows: $\|\phi\|_1 \ge \nu \Delta_k = \int_{\Delta_k} \phi d \mu \ge \frac{1}{k} \mu \Delta_k$, and so $\mu \Delta_k \le k \|\phi\|_1$.
We have $A \setminus N = \cup_k (A \cap \Delta_k)$, and since $\Delta_k$ is increasing, we have $\nu (A \cap \Delta_n) \to \nu(\cup_k (A \cap \Delta_k))$.
Let $\epsilon>0$, then choose $n$ such that $\nu(\cup_k (A \cap \Delta_k)) - \nu (A \cap \Delta_n)< \frac{\epsilon}{2}$.
Now we need to approximate $A \cap \Delta_n$  by a compact set.
Since $\mu$ is a Radon measure, there exists a sequence of compact sets $C_k \subset A \cap \Delta_n$ such that $\mu C_k \to \mu (A \cap \Delta_n)$. Without loss of generality (finite unions of compact sets are still compact) we may assume that the $C_k$ are increasing. We have $\mu((A \cap \Delta_n) \setminus \cup_k C_k) = 0$, hence $1_{C_k}(x) \to 1_{A \cap \Delta_n}(x)$ a.e. [$\mu$]. Since $1_{C_k}(x) \le 1$, the dominated convergence theorem gives $\int_{C_k} \phi d \mu \to \int_{A \cap \Delta_n} \phi d \mu $. Hence for some $k$, we have $\int_{A \cap \Delta_n} \phi d \mu - \int_{C_k} \phi d \mu < \frac{\epsilon}{2}$, and so 
\begin{eqnarray}
\nu A &=& \nu (A \setminus N) \\
&=& \nu ( \cup_k (A \cap \Delta_k)) \\
&<& \nu (A \cap \Delta_n) + \frac{\epsilon}{2} \\
&=& \int_{A \cap \Delta_n} \phi d \mu + \frac{\epsilon}{2} \\
&<& \int_{C_k} \phi d \mu + \frac{\epsilon}{2} + \frac{\epsilon}{2} \\
&=& \nu C_k + \epsilon
\end{eqnarray}
Hence $\nu A = \sup \{ \nu C | C \subset A, C \text{ compact} \}$.
$\nu$ is bounded by $\nu X = \|\phi\|_1$, hence is locally finite, and so $\nu$ is a Radon measure.
If you want to get outer regularity, let $A$ be a Borel set, and let $C_k \subset A^c$ be a sequence of compact sets such that $\nu C_k \to \nu A^c$. Now let $U_k = C_k^c$. Then $U_k$ is open and $A \subset U_k$. Since $\nu$ is finite, we have $\nu X  = \nu A + \nu A^c = \nu U_k + \nu C_k$. It follows that $\nu U_k \to \nu A$. It follows that  $\nu A = \inf \{ \nu U | A \subset U, U \text{ open} \}$, hence $\nu$ is outer regular.     
