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Let $(f_n)$ be a sequence of measurable functions that converges almost everywhere to a measurable function $f$. Assume that there is an integrable function $g$ such that $|f_n|\leq g$ for all $n$ almost everywhere. Show that $(f_n)$ converges almost uniformly to $f$.

Now I don't know of any other sufficient condition for almost uniform convergence other that Ergorov's theorem. However, to apply it I need my space to be of finite measure, and somehow the existence of a dominating integrable function $g$ has to play a role. But I wouldn't know where to start, so I'd like some hint on how to start working on the problem. Thank you in advance!

John Griffin
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user313212
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2 Answers2

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Hint: Work through the proof of Egoroff's theorem. In order to apply continuity from above, one needs that one of the sets has finite measure. In the usual setting, this is given by the fact that the space has finite measure. Here, we instead use the fact that the sequence is dominated.

After you finish proving this, you can redo the proof a couple more times to prove the following results.

1) If $\sum_{n=1}^\infty \|f_n-f\|_1 < \infty$, then $f_n\to f$ almost uniformly.

2) If $\sum_{n=1}^\infty \mu[|f_n-f|>1/n] < \infty$, then $f_n\to f$ almost uniformly.

They show that, in particular, if $f_n\to f$ in $L^1$ or in measure, then there is a subsequence such that $f_{n_k}\to f$ almost uniformly.

John Griffin
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  • I'm trying to follow your hint. The sets involved are $$E_{n,k}= \bigcup_{m\geq n}{ x : |f_m-f|\geq 1/k }$$ But when I try to calculate the measure of one of those sets, I still can't get rid of the finite measure space hypothesis of the original version. I get $$\mu(E_{n,k}) \leq \sum_{m\geq n} \mu({x : |f_m-f|\geq 1/k})$$. Now I believe triangle inequality to the $|f_m - f|$ part is needed to apply use the fact that the $f_n$'s are dominated by $g$, but I don't see how – user313212 Aug 16 '17 at 19:41
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    @user313212 Note that $$ E_{1,k}=\bigcup_{m\geq 1}[|f_m-f|\geq 1/k] \subseteq [2g\geq 1/k] $$ and $$ \mu[2g\geq 1/k] \leq 2k|g|_1 < \infty. $$ – John Griffin Aug 16 '17 at 20:47
  • Perfect! I was comsidering each set independently. Thank you very much for your help! :) – user313212 Aug 16 '17 at 20:56
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My classmate has told me the following interesting proof:

We may assume $f_n\to f$ and $|f_n|\le g$ everywhere by discarding a set of measure zero. Let $E_n$ be the set $\{x: \frac 1n\le g(x)< \frac{1}{n-1}\}$ if $n\ge 2$ and $E_1= \{x: g(x)\ge 1\}$, $E_0= \{x: g(x)=0\}$. On $E_0$ we must have $f= f_n=0$, and $f_n\to f$ uniformly on $E_0$. If $f_m\to f$ uniformly on $A$ and $B$, then so does $f_n$ tend to $f$ uniformly on $A\cup B$. Therefore we may only consider $\cup_{n=1}^{+\infty} E_n$ in what follows. Then $(E_n)$ are disjoint and each $E_n$ has finite measure as $g\in L^1 (X)$. Given $\epsilon>0$, for each $n$, we can use Egorov's Theorem to select $A_n\subseteq E_n$,, $f_m\to f$ uniformly on $A_n$, and $m(E_n- A_n)< \frac{\epsilon}{2^n}$. Then $m(X- \cup_{n=1}^{+\infty} A_n)\le\epsilon$. We claim $f_m\to f$ uniformly on $\cup_{n=1}^{+\infty} A_n$.

By construction of $E_n$, if $n\ge 2$ and $x\in E_n$, then $$ |f_n(x)- f(x)|\le 2g(x)< \frac{2}{n-1} . $$ Given $\eta>0$ sufficiently small, we select $N_1$ subject to $\frac{2}{N_1-1} <\eta$. Then for $m\ge N_1$, $|f_n(x)- f(x)|< \eta$ for all $n\in\mathbb Z_+$ and $x\in E_m$ (and a fortiori when $x\in A_m$). As $f_n\to f$ uniformly on each $A_1,A_2,\cdots, A_{N_1-1}$ and these are finitely many sets, we can uniformly select $N\in\mathbb Z_+$ so that $$ \sup\nolimits_{x\in A_1\cup A_2\cup\cdots\cup A_{N_1-1}} |f_n -f| <\eta $$ for all $n> N$. Consequently, $\sup_{x\in \cup_{n=1}^{+\infty} A_n} |f_m- f|< \eta$ for all $m >N$. This indicates $f_m\to f$ uniformly on $\cup_{n=1}^{+\infty} A_n$.

Asigan
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