This question is from a problem set on $L^p$ spaces in my undergraduate-level real analysis course. I said that $f_n$ converges if and only if it is Cauchy. Therefore, $\exists \, N\in\mathbb{N} \; \forall \, m,n>N \; ||f_m - f_n||_1 < \epsilon$. I then stated that we can choose a subsequence that will also be Cauchy; for example, let $f_{n_k}=f_{N+k}$. Since the $f_n$'s are Cauchy, we have that $\forall \, \ell,k \; ||f_{n_k}-f_{n_\ell}|| < \epsilon$. Therefore, $f_{n_k} \to f$ almost everywhere.
This proof is evidently erroneous, as I apparently cannot use the Cauchy condition in this way (I can't even read my grader's handwriting on this problem so I can't be sure of where exactly the inference fails, but I know it has to do with the Cauchy condition). I probably underestimated the difficulty of this problem. How can I fix this proof so it works?
For sake of simplicity, let's assume that we're working on the space $L^1([0,1])$, i.e. the space of all (complex) functions defined on $[0,1]$ with bounded $L^1$-norm. We are given that $f_n\rightarrow f$ in $L^1$-norm. Hence for all $\varepsilon>0$, there exists a $N\in \mathbb{N}$ such that for all $n\geq N$, $||f_n-f||<\varepsilon$, or equivalently $\int_{0}^1|f_n(x)-f(x)|\mathrm{d}x < \varepsilon$. Now this implies that $|f_n(x)-f(x)|<\varepsilon$ almost everywhere. So for each $n\geq N$, the measure of the set where $|f_n(x)-f(x)|\geq\varepsilon$ is zero.
You have to show that there exists a subsequence $f_{n_k}$ such that for almost all $x\in [0,1]$, $\lim_{k\rightarrow \infty}f_{n_k}(x)=f(x)$.