I am trying to prove the following question:
Let $(E,\mathcal{A},\mu)$ be measure space, $g$ positive measurable function. Then we have that $\nu := \int_{A}gd\mu$ is a positive measure on $\mathcal{A}$. Suppose $E$ is now locally compact and Hausdorff and that open sets of $E$ are $\sigma$- compact. Suppose also that $\mathcal{A}$ is the Borel $\sigma$-algebra, $\mu$ is a regular measure, and $g\in L^p(\mu)$ for some $p\in [1,\infty]$. Then show that $\nu$ is regular.
As for this question, my attempt is as follows, and yet I am not sure if it is correct.
Let $V$ be an open set, $V = \cup_{n\ge 1} H_n$, where $H_n$ are compact set. Then, let $H_1\prec f_1\prec V$. Denote $K_i$ as the support for $f_i$ and define $H_1\cup \dots\cup H_n \cup K_1 \cup\dots \cup H_n \prec f_{n+1}\prec V$. Then $(f_n)$ is increasing sequence and its limit is $\mathbb{1}_{V}$. Now, by monotone convergence, if q is the conjugate component of $p$ $$\nu(V) = \lim_{n\to \infty} \int f_n d\nu = \lim \int f_ng\ d\mu = \int g \mathbb{1}_V \ d\mu \leq \|g\|_p \left(\int \mathbb{1}_V \ d\mu\right)^{1/q} = \|g\|_p (\nu(V))^{1/q}$$
Now let $A\in \mathcal{A}$ be a Borel set, $\epsilon >0$, then by regularity of $\mu$, $\exists K$ compact, $V$ open s.t. $K\subset A \subset V$ and $\nu(V\setminus K) < \epsilon$. Note that compact sets are closed, so $V\setminus K$ is open, then $\nu(V\setminus K) \leq \|g\|_p (\nu(V\setminus K))^{1/q} < \|g\|_p \epsilon ^{1/q}$, and this gives the regularity of $\nu$.
I do realize that there may be flaws in this proof and I would really appreciate it if someone could help. Thanks!