If $f_n \to f$ in $L^1$ and $f_n \to g$ a.e. do we have $f = g$ a.e.?

Question

I was thinking about the following question regarding convergence almost everywhere (a.e.) and convergence in $L^1$:

Assume we know that there is an $f \in L^1$ and a measurable $g$ such that a sequence $f_n$ converges to $f$ in the $L^1$-norm and at the same time converges to $g$ a.e.. Can we conclude that $f=g$ a.e.?

Answer 1

Yes, $f_n \to f$ in $L^{1}$ implies that there is a subsequence which converges to $f$ almost everywhere and the subsequence also converges to $g$ almost everywhere. Hence $f=g$ almost everywhere.