Let $(X,\mathscr{M},\mu)$ be a measure space and $f$ be an essentially bounded function, i.e. $f\in L^{\infty}$. How do I show that
$\lim_{p\rightarrow 0}\int_X|f|^p\,d\mu=\mu(\{x\in X\,|\,f(x)\ne 0\})$.
In the cases where $f\in L^{p_0}$ for some $0
$\lim_{p\rightarrow 0}\int_X|f|^p\,d\mu=\mu(\{x\in X\,|\,f(x)\ne 0\})$
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real-analysis
functional-analysis
lebesgue-integral
lp-spaces
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0Surely $X$ is a finite measure space? – 2017-01-12
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1If $f$ is in no $L^p$, the LHS is infinite hence you simply have to check that $\mu(f\ne0)=\infty$. Now, $f$ is bounded, hence... – 2017-01-12
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1@OpenBall No. $ $ – 2017-01-12
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0Sorry, somehow saw that as $\left(\int |f|\right)^p$. – 2017-01-12
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1@Did : Oh, it was very simple after all. Thank you! – 2017-01-12
1 Answers
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We may assume that $f$ is not in $L^p$ for any $0