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This is Rudin's RCA book. He discusses completeness of $L^\infty(\mu)$, but there is a part of proof that I don't understand.

In $L^\infty(\mu)$, suppose $\{f_n\}$ is a Cauchy sequence, and let $A_k, B_{m,n}$ be the sets where $|f_k(x)| > \|f_k\|_\infty$ and $|f_n(x)-f_m(x)| > \|f_n-f_m\|_\infty$, and let $E$ be the union of these sets, for $k, m ,n = 1,2,3, \cdots$. Then, $\mu(E)=0$, and on the completement of $E$, the sequence $\{f_n\}$ converges uniformly to a bounded function $f$. Define $f(x) = 0$ on $E$. Then, $f \in L^\infty(\mu)$ and $\|f_n - f\|_\infty \rightarrow 0$ as $n \rightarrow \infty$.

Since $f$ is a complex function, I understand that $f_k$ converges to some measurable complex function $f$. The part that I do not understand is 1. (almost everywhere) boundedness, and 2. (a.e.) uniform convergence. Any help would be appreciated.

Jose Avilez
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James C
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  • THe set $E$ being the countable union of sets of measure zero has also measure zero. If $x\notin E$ then $|f_k(x)|\leq |f_k|\infty$ for all $k$, and also $|f_n(x)-f_m(x)|\leq |f_n-f_m|$ for all $n,m$, Since $f_m$ is Cauchy in $L\infty$, then the numeric sequence $f_m(x)$ is bounded and Cauchy and so it converges to a finite number, say $f(x)$. This happens for each $x\notin E$ (hence the almost everywhere) – Mittens Jun 29 '21 at 03:24
  • Thanks. It showed 1. What about 2.? For a.e. $x$ (for simplicity, I'll denote it is $\forall x$), $|f_n(x)-f_m(x)| \leq |f_n-f_m|$. Hence, I have a uniform Cauchy $\forall x \in X$. How does this imply that ${f_n}$ uniformly converges to $f$? – James C Jun 29 '21 at 03:31
  • @Resolved. $|f(x) - f_n(x)| = \lim_{m \rightarrow \infty} |f_m(x) - f_n(X)| < \epsilon$. – James C Jun 29 '21 at 03:34
  • Given $\varepsilon>0$, there is $N$ such that $n,m\geq N$ implies $|f_n-f_m|\infty$. Now, for all $x\notin E$, for all $n\geq N$, you have $$|f(x)-f_n(x)|=\lim{m\rightarrow\infty}|f_n(x)-f_m(x)|\leq\liminf_m|f_m-f_n|_\infty\leq\varepsilon$$ This implies that $f_n$ converges to $f$ uniformly on $E^c$. – Mittens Jun 29 '21 at 03:36
  • You may rewrite your comment as an answer, and I'll accept it. – James C Jun 29 '21 at 03:37
  • Glad to know all is resolved now. – Mittens Jun 29 '21 at 03:37
  • This has been asked before and I rather keep it as comment. Here is a good answer to your question. You may upvote this one if you want. – Mittens Jun 29 '21 at 03:38

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