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$\sum_{n \in \mathbb{N}} ||f_{n}-f||_{1} < \infty$ 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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    @Mark Your hint seems preferable to the brute force proof, but I can't quite figure out what you mean. You're integrating with respect to a new measure $\mu_F(A) = \int_X F \mathbb{I}_A d\mu$ right? What is the 'appropriate function'? I noticed that $f_n \rightarrow f$ pointwise with respect to $\mu_F$, but applying Egorov doesn't seem to help.2017-12-10

3 Answers 3

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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.