Fast $L^{1}$ Convergence implies almost uniform convergence $\sum_{n \in \mathbb{N}} \lVert f_{n}-f \rVert_{1} < \infty$ 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...
 A: 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 $f$.
A: Put $g_n:=|f_n-f|$, and fix $\delta>0$. We have $\sum_{n\in\mathbb N}\lVert g_n\rVert_{L^1}<\infty$ so we can find a strictly increasing sequence $N_k$ of integers such that $\sum_{n\geq N_k}\lVert g_n\rVert_1\leq \delta 4^{-k}$. Put $A_k:=\left\{x\in X:\sup_{n\geq N_k}g_n(x)>2^{1-k}\right\}$. Then $A_k\subset\bigcup_{n\geq N_k}\left\{x\in X: g_n(x)\geq 2^{-k}\right\}$ so 
$$2^{-k}\mu(A_k)\leq \sum_{n\geq N_k}2^{-k}\mu\left\{x\in X: g_n(x)\geq 2^{-k}\right\}\leq \sum_{n\geq N_k}\lVert g_n\rVert_1\leq \delta 4^{-k},$$
so $\mu(A_k)\leq \delta 2^{-k}$. Put $A:=\bigcup_{k\geq 1}A_k$. Then $\mu(A)\leq \sum_{k\geq 1}\mu(A_k)\leq \delta\sum_{k\geq 1}2^{-k}=\delta$, and if $x\notin A$ we have for all $k$: $\sup_{n\geq N_k}g_n(x)\leq 2^{1-k}$ so $\sup_{n\geq N_k}\sup_{x\notin A}g_n(x)\leq 2^{1-k}$. It proves that $g_n\to 0$ uniformly on $A^c$, since for a fixed $\varepsilon>0$, we take $k$ such that $2^{1-k}$, so for $n\geq N_k$ we have $\sup_{x\notin A}g_n(x)\leq\varepsilon$.
A: Put $g_n:=|f_n-f|$ and let $\delta >0$ be given. We need to show that there exists $A\subset X$ with $\mu (A)<\delta$, such that for any $k>0$ there exists $N>0$ so that for all $n>N$ and all $x\notin A$, $g_n(x)<\frac{1}{k}$.
Define $B_{n,k} := \left\{x\in X:g_n(x)\ge \frac{1}{k} \right\}$ and $A_{N,k}:=\bigcup_{n>N}B_{n,k}$. By Markov, we have:
$$\mu (A_{N,k})\le \sum_{n>N}\mu (B_{n,k})\le k\sum_{n>N} ||g_n||_1\mathrel{\mathop{\longrightarrow}_{\mathrm{N\to\infty}}} 0. $$
Thus for each $k$, let $N_k$ be an integer such that $\mu (A_{N_k,k})\le \frac{\delta}{2^k}$. Let $$A:=\bigcup_{k>0}A_{N_k,k}.$$
Then $\mu(A)\le \delta$. Moreover, for a given $k>0$, take $N=N_k$, so for any $n>N$ and $x\notin A$, $g_n(x)<\frac{1}{k}$ as required.
