Let $\phi \geq 0$ be a function in $L^1(\mu)$ where $\mu$ is a Radon measure (= a Borel measure on $X$ that is finite on compact sets, inner regular on open sets and outer regular on all compact sets) on the locally compact Hausdorff space $X$. Show that $$\nu(E) = \int_E \phi \,d \mu$$ is again a Radon measure.
Attempt:
Trivially, since $\nu(X) <\infty$, $\nu$ is finite on compact sets. It remains to show that $$\nu(E) = \inf\{\nu(U): U \supseteq E, U \text{ open}\}$$ $$\nu(U) = \sup\{\nu(K): K \subseteq U, K \text{ compact}\}$$
where $E$ is a Borel set of $X$ and $U$ is an open set of $X$. I managed to show that this is true when $\mu(E), \mu(U) < \infty$ by invoking the following:
$$\forall \varepsilon > 0: \exists \delta > 0: \mu(F) < \delta\implies \nu(F) < \varepsilon$$
However, I'm stuck if $\mu(E) = \infty = \nu(U)$. How can I proceed?
Here is a complete solution (with the help of @Sebastian Spindler):
(STEP 1) We show that if $E$ is a Borel set with $\mu(E) < \infty$, then $\nu$ is inner regular at $E$.
Since $\phi \in L^1(\mu)$, we have the following property:
$$\forall \varepsilon > 0: \exists \delta > 0: \mu(F) < \delta \implies \nu(F) = \int_F \phi \,d \mu < \varepsilon$$
Fix $\varepsilon > 0$ and choose the corresponding $\delta > 0$, as above. Then, since $\mu$ is inner regular at $\sigma$-finite Borelsets, we can choose a compact $K \subseteq E$ with $\mu(E\setminus K) < \delta$. Consequently,
$$\nu(E) \leq \nu(K) + \nu(E\setminus K) < \varepsilon + \nu(K) $$
and we conclude that $$\nu(E) = \sup\{\nu(K): K \subseteq E, K \text{ compact}\}$$
Thus $\nu$ is inner regular at $E$.
(STEP 2) We show $\nu$ is inner regular on all Borel subsets.
Let $E$ be a Borel subset of $X$. Note that we have the pointwise limit $$\lim_{n \to \infty} \phi \chi_{E \cap \{\phi > 1/n\}} = \phi\chi_{E \cap \{\phi > 0\}}$$
Hence, invoking the dominated convergence theorem, $$\nu(E) = \int_E \phi \, d \mu = \int_{E \cap \{\phi > 0\}}\phi \, d \mu = \lim_n \int_{E \cap \{\phi > 1/n\}} \phi \, d \mu = \lim_n \nu(E \cap \{\phi > 1/n\})$$
However, $$\infty > \Vert \phi \Vert_1 \geq \nu(\{\phi > 1/n\}) = \int_{\{\phi>1/n\}} \phi \, d \mu\geq n^{-1} \mu(\{\phi> 1/n\})$$
Hence, $\mu(E \cap \{\phi> 1/n\}) < \infty$. By (STEP 1), we may select compact sets $K_n \subseteq E \cap \{\phi > 1/n\}$ with $ \nu(E\cap \{\phi > 1/n\})-\nu(K_n) < 1/n$.
Thus, we see that $\lim_n \nu(K_n)= \nu(E)$ and thus $$\nu(E) = \sup \{ \mu(K): K \subseteq E, K \text{ compact}\}$$
(STEP 3) We show $\nu$ is outer regular on all Borel sets.
Let $E$ be a Borel set. Let $\varepsilon > 0$. By (STEP 2), there is a compact subset $K \subseteq E^c$ with $\nu(E^c)- \nu(K) < \varepsilon$. The set $U:= K^c$ is open and contains $E$ and $$\nu(U) - \nu(E) = \nu(X)- \nu(K) - \nu(E) = \nu(E^c) - \nu(K) < \varepsilon$$ Hence, $$\nu(E) = \inf\{\nu(U): U \supseteq E, U \text{ open}\}$$ as desired.