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I want to prove the following:

If $f ≥ 0$ is measurable with respect to the measure $m$ and $\mu_f (\lambda) = m(\{x : f(x) > \lambda\})$ then $\int f dm = \int_{0}^{\infty} \mu_f (\lambda) d\lambda$ where $d\lambda$ is Lebesgue measure.

I use the fact that $f(x)= \int_{0}^{f(x)} 1 d\lambda$. Then I apply Fubini Theorem to switch integrals.

So I have : $\int_{X} f dm = \int_{X} \int_{0}^{f(x)} 1 d\lambda dm =\int_{0}^{f(x)} \int_{X} 1 dm d\lambda = \int_{0}^{f(x)} m(X) d\lambda $.

How can I replace my upper bound of the integral by infinity? Thank you

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    $\begingroup$ You did not interchange limit correctly -- the outer limit should not depend on the dummy of the inner integral. $\endgroup$ – user10354138 Nov 2 '18 at 19:08
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\begin{align} \int fdm&=\int\int_0^\infty 1\{f>\lambda\}d\lambda dm \\ &=\int_0^\infty\int 1\{f>\lambda\}dm d\lambda=\int_0^\infty m(\{f>\lambda\})d\lambda. \end{align}

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