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Let $f: \mathbf{R}^d \rightarrow [0,\infty]$ be a Lebesgue integrable function. Prove $$\lim_{\alpha \rightarrow \infty} \alpha m(\{x:f(x)>\alpha\})=0.$$ Hint: For $\epsilon>0$ take $g$ simple such that $0 \leq g \leq f$ and $\int(f-g)dx<\epsilon$. Then note that $g<\alpha/2$ if $\alpha$ is large and for such $\alpha$ compare $\{f>\alpha\}$ with $\{f-g>\alpha/2\}$.

So this was a problem from a midterm and I get the argument that follows from the hint but no matter how many manipulations of Markov's Inequality I do, I can't prove the hint so I'm pretty sure that I'm missing some kind of intuition here.

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We have $\{f\gt \alpha\}\subset \{f-g\gt \alpha/2\}\cup\{ g\gt \alpha/2\}$. It can be seen looking at the inverse inclusion for complements. Since $g$ takes a finite numbers of values, it is bounded. So take $\alpha\geqslant 2\sup g$ in order to get for such $\alpha$, the inclusion $\{f\gt\alpha\}\subset \{f-g\gt \alpha/2\}$. Now we can use classical Markov's inequality.

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  • $\begingroup$ wow, that seems so obvious that I'm kicking myself now, I even knew that simple functions were bounded. Thanks. $\endgroup$ Oct 26, 2013 at 19:25

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