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I was reading this post Understanding proof of completeness of $L^{\infty}$ and it is mentioned that the sequence $(f_n)_{n\in\mathbb N}$ converges uniformly on a conegligible set $N^C$. Could someone point out any useful consequences of this observation?

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    It shows that a) a pointwise a.e. limit exists, b) that limit is measurable, and c) that limit is essentially bounded. – Daniel Fischer Jun 19 '14 at 17:19
  • @DanielFischer I see the first two points you listed but in the following way: there is a measurable conegligible set $E$ such that the sequence$(f_n(x)){n\in\mathbb{N}}$ is Cauchy for all $x\in E$, so simply define $f=\left(\lim{n\rightarrow\infty}f_n(x)\right)\chi_E$. Not sure if this is correct. What about point c)? – bibo_extreme Jun 19 '14 at 17:26
  • You want the limit to be in $L^\infty$, so it ought to be essentially bounded. The uniform convergence on a set whose complement has measure $0$ is a quick way to get that. (And you want the sequence to converge to the limit function in the norm, so it's necessary that it converges uniformly on a conegligible set.) – Daniel Fischer Jun 19 '14 at 17:32
  • Are you perhaps using some standard result in the last part of your post? – bibo_extreme Jun 19 '14 at 17:34
  • Suppose $\lVert f_n - f_m\rVert_\infty \leqslant 1$ for all $n,m \geqslant K$. Then $$\lvert f(x)\rvert = \lim_{n\to\infty} \lvert f_n(x) - f_K(x) + f_K(x)\rvert \leqslant \lvert f_K(x)\rvert + \lim_{n\to\infty} \lvert f_n(x) - f_K(x)\rvert \leqslant \lVert f_K\rVert_\infty + 1.$$ Since that holds for all $x\in N^c$, it follows that $\lVert f\rVert_\infty \leqslant \lVert f_K\rVert_\infty + 1 < \infty$. – Daniel Fischer Jun 19 '14 at 17:39
  • OK it seems clear to me now. Had we not had uniform convergence then you wouldn't have been able to write "for all $x\in N^C$"? – bibo_extreme Jun 19 '14 at 17:45
  • Well, to conclude the essential boundedness, it would suffice to have a pointwise a.e. convergent sequence that is uniformly essentially bounded. But the norm of $L^\infty$ is that of uniform convergence on some conegligible set, so it's necessary to conclude the convergence of the sequence in $L^\infty$. For everything else, it's just a convenient thing to avoid looking at individual points. – Daniel Fischer Jun 19 '14 at 17:50

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