Let $\Omega \subset \mathbb R^n$ be an arbitrary open set and consider the usual Lebesgue spaces of locally $p$-integrable functions, for $1 \leqslant p < \infty$. My goal is to prove the following claim:
Claim. Given $1 \leqslant p < \infty$, it follows that
$$ g \in L^p_{\operatorname{loc}}(\Omega) \Longleftrightarrow |g|^p \chi_\Omega \in L^1_{\operatorname{loc}}(\mathbb R^n). $$
My attempt. First suppose that $g \in L^p_{\operatorname{loc}}(\Omega).$ Writing it out, this means that
$$ \int_K |g(x)|^p \, dx < \infty, $$
for every compact set $K \subset \mathbb \Omega$. In order to prove that $|g|^p \, \chi_\Omega \in L^1_{\operatorname{loc}}(\mathbb R^n),$ one must show that
$$ \int_C |g(x)|^p \, \chi_\Omega(x) \, dx < \infty, $$
for every compact set $C \subset \mathbb R^n$. Clearly, if $C \cap \Omega$ is a compact subset of $\Omega$, we obtain the desired result. It's obvious that $C \cap \Omega$ is a subset of $\Omega$ and so everything we need to guarantee is that $C \cap \Omega$ is compact. Since $C$ is compact, we directly have that $C \cap \Omega$ is bounded but I don't think we can guarantee that $C \cap \Omega$ is closed since $\Omega$ is open.
Thanks for any help in advance.
You cannot prove this. Let $\Omega =(0,1),p=1$ and $g(x)=\frac 1 x$. Then $g \in L_{\text {loc}} ^{1}(\Omega)$ according to your definition. Take $C=[0,1]$. Then $\int_C|g(x)|\chi_{\Omega}(x)dx=\infty$.