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Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is bounded almost everywhere. Then is it bounded?

If so, what is the main idea or method in tis proof, and can I generalize this for upto what?

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It's not. $\operatorname{id}_{\Bbb R}\cdot \chi_{\Bbb Q}$ is bounded a.e. (it's a.e. $0$) but certainly not bounded. –  Lord_Farin Oct 23 '12 at 10:31
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This is not true. Controlling almost everywhere does not give anything pointwisely. –  Hui Yu Oct 23 '12 at 10:31
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A more interesting question might be whether a continuous function which is bounded a.e. is bounded. –  fgp Oct 23 '12 at 10:57
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1 Answer

up vote 4 down vote accepted

Let $f$ map the irrationals to $0$ and the rationals to themselves. Since the rational numbers form a countable set, it has measure $0$. Then f is bounded almost everywhere but is not bounded.

More generally, on any infinite set, one can define a function that is bounded almost everywhere but is not bounded. Simply take a countable subset $x_1,x_2,x_3,\ldots$ and send $x_i$ to $i$ and other elements to $0$.

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