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Let $(X, M, \mu)$ a complete measure space. Let $f:X\to[0,\infty)$ be a measurable function and integrable function. Let $\psi(t)=\mu\{x \in X : f(x) \geq t\})$. Prove that

$$\int_X\!f\,\mathrm{d}\mu=\int_0^\infty\!\psi(t)\,\mathrm{d}t.$$

Now, we can see an answer here : $\int_X g \, d\mu =\int_0^\infty\mu(x:g(x)>t) \, dt$

But the subtle difference is : In the link above $\psi(t)$ is defined for $x \in X$ such that $f(x) \gt t$ and not $f(x) \geq t$. Also it does not make use of complete measure hypothesis (I have an intuition this is useful for $f(x) =0$).

The question is : This subtle difference between the exercises makes the answer different?

Thanks.

UDAC
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  • Is $\int_0^\infty \chi_{{g(x) > t}} , dt $ different from $\int_0^\infty \chi_{{g(x) \ge t}} , dt ,?$ – Kurt G. Aug 07 '23 at 13:42
  • @KurtG. Is that a rethorical question? – UDAC Aug 07 '23 at 19:02
  • Perhaps it was. More than that it was a hint to go through all steps in the other post to see if the proof with $\ge$ yields the same result as with $>,.$ Too late. The given answer below got that hint. – Kurt G. Aug 07 '23 at 19:04

1 Answers1

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You just need to notice that $$\int_0^\infty \mu(x: g(x)=t) dt=0$$ This follows from Fubini's theorem: $$\int_0^\infty \mu(x: g(x)=t) dt=\int_0^\infty \int_X \chi_{\{g(x)=t\}} d\mu(x) dt=\int_X \int_0^\infty \chi_{\{g(x)=t\}} dtd\mu(x)=0$$ with the additional hypothesis that $\mu$ is $\sigma$-finite.

Dark Magician
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  • What if the measure is not sigma finite? – UDAC Aug 07 '23 at 18:06
  • @UDAC . The question in OP was if $\ge$ is a subtle difference to $>,.$ The formula in the post you found assumes $\sigma$-finite and so should you from the beginning. If the measure is not $\sigma$-finite I'd rather hear your own thoughts on those problems. – Kurt G. Aug 07 '23 at 19:11