Suppose $\sum_{n=1}^{\infty} \|f_n - f\|_{L^1} < \infty$, where
\begin{align} \|f_n - f\|_{L^1} = \int_{X}|f_n(x) - f(x)| dx. \end{align}
Let $\epsilon > 0$ and $\delta > 0$ be parameters at our disposal. By the absolute summability assumption, for a sufficiently large positive integer $N$, we have
\begin{align} \sum_{n=N}^{\infty} \|f_n - f\|_{L^1} < \epsilon \delta. \end{align}
Now, let
\begin{align} E_n = \{ x \in X : |f_n(x) - f(x)| > \delta \}. \end{align}
Then we have a simple bound
\begin{align} \sum_{n = N}^{\infty} m(E_n) \delta \leq \sum_{n=N}^{\infty} \|f_n - f\|_{L^1} < \epsilon \delta. \end{align}
So,
\begin{align} \sum_{n = N}^{\infty} m(E_n) < \epsilon. \end{align}
But,
\begin{align} m\left(\bigcup_{n = N}^{\infty} {E_n} \right) \leq \sum_{n = N}^{\infty} m(E_n) < \epsilon. \end{align}
Let $E = \bigcup_{n = N}^{\infty} {E_n}$. Then, we note
\begin{align} E^{C} &= \left( \bigcup_{n = N}^{\infty} {E_n} \right)^{C} \\ &= \bigcap_{n = N}^{\infty} {E_n^{C}} \\ &= \{ x \in X : |f_n(x) - f(x)| \leq \delta \hspace{5pt} \forall n \geq N \}. \end{align}
The parameters $\epsilon$ and $\delta$ are made arbitrarily small (simultaneously) as $N \to \infty$. As $\delta$ may be made arbitrarily small, we see that $f_n$ converges uniformly to $f$ on $E^{C}$. As $\epsilon$ may be made arbitrarily small, we have that $f_n$ (at worst) does not converge uniformly to $f$ on an exceptional set $E$ of measure zero.
Thus, $f_n$ converges almost uniformly to $f$.