Let $(f_n)_n$ be a sequence converging to $f$ in $L^p(\Omega,\mathfrak{F},\mu)$ for some $1\leq p\leq\infty$. Assume all the $f_n$ are measurable with respect to $\mathfrak{G}\subset\mathfrak{F}$. Does this imply that $f$ is measurable with respect to $\mathfrak{G}$?
I think if I am working with a probability space (or a finite measure space) $(\Omega,\mathfrak{F},\mu)$ I can prove that the limit is again measurable with respect to $\mathfrak{G}$ using conditional expectations. Is there any way to conclude something similar for non-finite measure spaces?
In general, the a.e. pointwise limit of a sequence of measurable function is measurable. The proof is quite straightforward. Since you are working with a complete measure space, every null set is measurable. If you call $\Omega_0$ the null set where you don't have pointwise convergence and $f$ the a.e. pointwise limit, the measurability of $f$ is equivalent to the measurability of $f|_{\Omega \setminus \Omega_0}$, and now it's easy because the pointwise limit of a sequence of measurable functions is measurable. You can find the proof with all the details, for example, in Royden, Fitzpatrick "Real Analysis"