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I'm trying to prove the following identity:

If $f:X \rightarrow \mathbb{R}$ is a bounded, continuous function, and $\mu$ is a borel measure in $X$, then

$\int f d\mu = \int_0^1 \mu(\{x \in X ; f(x)>t\})dt$.

(W/ loss of generality, one may consider $0\leq f\leq 1$.

This equality is used to prove one of the portmanteau theorem affirmations, as in chapter 8 of Gordan Z.'s lecture notes:

https://web.ma.utexas.edu/users/gordanz/notes/theory_of_probability_I.pdf

As he says, it is a particular case of Problem 5.23, but i'm trying to obtain it by scratch.

My attempt was: name the set $A_t=\{x \in X ; f(x)>t\} $.

Then $ \int_0^1 \mu(A_t)dt= \int_0^1 \int \mathcal{X}_{A_t} d\mu dt=\int \int_0^1 \mathcal{X}_{A_t} dt d\mu$.

Then it remains to prove that $\int_0^1 \mathcal{X}_{A_t} dt $ is equal to $f$, but I got stuck at this point.

Mateus R.
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  • I don't understand why the integral is from $0$ to $1$. Is $f(x) \in [0,1]$? – Davi Barreira Jul 14 '20 at 21:03
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    @Mateus: in Gordan's stement, $f>0$. Also, there is a typo in his expression. The integral should be over $[0,|f|\infty]$. If $|f|\infty=1$, the expression is fine. – Mittens Jul 14 '20 at 21:07
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    @DaviBarreira: these are lecture notes and prone to have typos. The OP did not realize of that. It should be $0\leq f\leq 1$. Continuity was added because in the section in question, the author is dealing with the Pormanteau theorem. As you may have already know, that assumption is not needed. – Mittens Jul 14 '20 at 21:10
  • Cheers, my friend! – Davi Barreira Jul 14 '20 at 21:11
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    @Matheus: The particular application of Fubini's theorem that will bring this puppy home is https://math.stackexchange.com/questions/460253/show-int-f-d-mu-int-limits-0-infty-mu-left-x-in-x-fxt-right – Mittens Jul 14 '20 at 21:26
  • @oliverdiaz thats right, I forgot to mention those pre requisites. Thank you for your time. – Mateus R. Jul 14 '20 at 21:32
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    @Mateus: Notice that in the application of Fubini's theorem (see the link) the condition $0\leq f\leq |f|\infty$ implies that $\mu(|f|>t)=0$ for $t\geq|f|\infty$, that is why the integration can be restricted to the interval $[0,|f|_\infty]$. – Mittens Jul 14 '20 at 21:37
  • @OliverDiaz Okay. W/ loss of generality, I think the author assumed $||f||_\infty =1$. Thank you again sir. – Mateus R. Jul 14 '20 at 21:59

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