# Fast $L^{1}$ Convergence implies almost uniform convergence

$\sum_{n \in \mathbb{N}} ||f_{n}-f||_{1} < \infty$ 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...

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Do you have Egorov's theorem available, or are you being asked to show this directly? It might help if you provided some more background context about where this question comes from, what you already know, and so forth. –  user16299 Jan 16 '12 at 9:13
Let $F:=\sum_{n \in \mathbb{N}} (f_n -f)$. The assumption implies that $F$ is defined almost everywhere and is integrable. If $\mu$ is your original measure, consider now the measure $dF=Fd\mu$ and apply Egorov's theorem to an appropriate function in this new (finite!) measure space. –  Mark Jan 16 '12 at 12:37

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 $\sup_{x\notin A}g_n(x)\leq\varepsilon$.

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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 $f$.

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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.

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