I know that $g\in L^1[a,b]$ implies that $g<\infty$ almost everywhere. It seems like it should follow that $g$ is bounded almost everywhere, but I can't think of a proof or a counterexample.
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3What about $g(x) = x^{-1/2}$ on $(0, 1)$? – Martin R May 05 '20 at 19:40
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1Or this: https://math.stackexchange.com/q/3313919/42969. – Martin R May 05 '20 at 19:43
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What does $g$ bounded a.e. mean? – zhw. May 05 '20 at 20:36
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@zhw I was thinking in the sense that there is some real number $M$ such that $|g|\leq M$ everywhere on $[a,b]$ except possibly on a subset of measure zero. – smileemote May 05 '20 at 20:42
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1The way to say that is $g\in L^\infty.$ – zhw. May 05 '20 at 20:44