Given that $\{f_n\}$ be a sequence of Lebesgue-measurable real-valued functions on $[0, 1]$ such that \begin{equation} \lim_{n \rightarrow \infty}\int_0^1 |f_n(x)| dx = 0 \end{equation}
I want to show there is a subsequence of $\{f_n\}$ such that $\{f_{n_i}(x)\}$ converges to $0$ for a.e. $x$.
Im thinking of using the inequality $\int_0^1 |f_n(x)| dx > \mu(\{x: |f_n(x)| > \epsilon\})\epsilon$ but I didnt' get very far past that.
Note that with $A_n = \{x : |f_n(x)| > \epsilon\}$,
$$\mu(A_n) \leqslant \frac{1}{\epsilon}\int_{A_n} |f_n| \leqslant \frac{1}{\epsilon}\int_{[0,1]} |f_n|\underset{n \to \infty} \longrightarrow 0 ,$$
and $f_n \to 0$ in measure. Now you can use the well-known theorem that convergence in measure implies a.e. convergence of a subsequence proved here.
There exist $n_1,n_2<\cdots$ such that $\int |f_{n_k}|d\mu <\frac 1 {2^{k}}$. This gives $\int \sum_k |f_{n_k}|d\mu=\sum_k \int |f_{n_k}|d\mu<\infty$. Hence, $\sum_k |f_{n_k}|<\infty$ almost everywhere. This implies that $|f_{n_k}| \to 0$ almost everyhwere.
The interchange of sum and integral above can be justified using either Tonelli's Theorem or Montone Convergence Theorem.
