Let $\Omega\subseteq\mathbb{R}^n$ be open such that $\mathcal{L}^n(\Omega)<\infty$ and $f_1,f_2,\ldots \in L^2(\Omega)$ a Cauchy sequence. Does there always exist an $f\in L^1(\Omega)$ such that
$\hspace{1cm}\lim_{j\to\infty}\int_{\Omega}f_jd\mathcal{L^n}=\int_{\Omega}fd\mathcal{L}^n$
holds?
This seems like it is correct by using the fact that $L^2$ is complete, hence $\lim_{j\to\infty}f_j$ exists and one should probably use that as the function $f$. Furthermore, since $\mathcal{L}^n(\Omega)<\infty,\, L^2(\Omega)\subseteq L^1(\Omega)$ holds and so the integrals are actually well-defined. However, why should $L^2$ convergence imply convergence of integrals? I also haven't been able to come up with a counterexample since canonical things like $\Omega=(0,2],f=0,\,f_j=\frac{1}{x^{i_j}}$ with $i_j$ such that $f_j$ is square integrable but not integrable don't work since such $f_j$ don't exist.
$L^{2}(\Omega)$ is complete. So there exists $f \in L^{2}(\Omega)$ such that $\int |f_n-f|^{2} \to 0$. By Holder's inequality $\int|f_n-f| \leq \sqrt {\mathcal L^{n} (\Omega)} \sqrt {\int |f_n-f|^{2} }\to 0$.