TITLE: Fast $L^{1}$ Convergence implies almost uniform convergence
QUESTION [8 upvotes]: $\sum_{n \in \mathbb{N}} \lVert f_{n}-f \rVert_{1} < \infty$ implies $f_{n}$ converges almost uniformly to $f$, how to show this?
EDIT: Egorov's theorem is available. I have been able to show pointwise a.e. convergence using Chebyshev and Borel-Cantelli, I am having trouble trying to pass to almost uniform convergence using the absolute summability condition...

REPLY [3 votes]: 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 $k \geq 1$. By the absolute summability assumption, for a sufficiently large positive integer $N_k$,
\begin{align} \sum_{n=N_k}^{\infty} \|f_n - f\|_{L^1} < 4^{-k}. \end{align}
Now, let
\begin{align} E_{n,k} = \{ x \in X : |f_n(x) - f(x)| > 2^{-k} \}. \end{align}
Then we have a simple bound
\begin{align} \sum_{n = N_k}^{\infty} m(E_{n,k})2^{-k} \leq \sum_{n=N}^{\infty} \|f_n - f\|_{L^1} < 4^{-k}. \end{align}
So,
\begin{align} m\left(\bigcup_{n = N_k}^{\infty} {E_{n,k}} \right) \leq \sum_{n = N_k}^{\infty} m(E_{n,k}) < 2^{-k}. \end{align}
Let $E = \bigcap_{k=1}^{\infty}\bigcup_{n = N_k}^{\infty} {E_{n,k}}$. This is a set of measure zero. Moreover,
\begin{align*}
E^{C} = \bigcup_{k=1}^{\infty}\bigcap_{n = N_k}^{\infty} {E_{n,k}^{C}} = \bigcup_{k=1}^{\infty}\{ x \in X : |f_n(x) - f(x)| \leq 2^{-k} \hspace{5pt} \forall n \geq N_k \}
\end{align*}
That is, $f_n$ converges uniformly to $f$ on the set $E^C$. Thus, $f_n$ converges almost uniformly to $f$.