Consider a sequence of functions $\{f_n \}\in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ , convergent to $f$ in $L^1(\mathbb{R})$ and to $g$ in $L^2(\mathbb{R})$. Prove that $f=g$ a.e.
What I understood is that there are 3 different kinds of convergence used in this problem: convergence in $L^1(\mathbb{R})$, convergence in $L^2(\mathbb{R})$, and pointwise convergence in $\mathbb{R}$.
Since an $L^p$ space is complete, every convergent sequence is Cauchy therefore $\{f_n\}$ is Cauchy in $L_p$.
According to The Riesz-Fischer Theorem, I can pick one subsequence that is convergent to $f$ in $L^1(\mathbb{R})$ as well as a.e on $\mathbb{R}$ to $f$. Similarly, there is a subsequence does the same to $g$. But I do not know how to show that $f$ and $g$ are the same a.e because I am not sure these two subsequences that I pick are the same or going to the same limit.
Thank you in advance.
After you take a subsequence converging to $f$ a.e., you can take a subsequence of that subsequence converging to $g$ a.e.
However, I don't understand how you are supposed to be applying the Riesz-Fischer theorem, and I'd be interested to know what you meant by that. The fact that convergence in $L^p$ implies the existence of a subsequence converging a.e. to the same limit can be proved (among other ways) using the notion of convergence in measure.
(Incidentally, every convergent sequence in a metric space is Cauchy, regardless of completeness.)