$\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\{x\in X: f(x)>t\}) d\mu_t$ for any natural $p\ge 1$ Let $f:X\mapsto[0,+\infty)$ be a non-negative measurable function defined on the space $X$, endowed with the complete $\sigma$-additive, $\sigma$-finite, measure $\mu$ defined on the $\sigma$-algebra of the measurable subsets of $X$.  I read that, for $p\in\mathbb{N}$, $p\ge 1$, $$\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\{x\in X: f(x)&gt;t\}) d\mu_t$$  I know that the equality holds for $p=1$, as proved . That also implies that $$\int_X f^p d\mu =\int_{[0,+\infty)} \mu(\{x\in X: f(x)^p&gt;t\}) d\mu_t,$$but I am not able to use this result, nor induction, to prove the desired result. How could we prove it? I heartily thank any answerer!  EDIT: This question has been marked as a duplicate of , but, although related, they do not ask the same question.
Proof of $\int_{[0,\infty)}pt^{p-1}\mu(\{x:|f(x)|\geq t\})d\mu(t)=\int_{[0,\infty)}\mu(\{x:|f(x)|^p\geq s\})d\mu(s)$ Let $({\Bbb R},{\mathcal A},\mu)$ be the measure space where ${\mu}$ is the Lebesgue measure.  Assume that $\int_{\Bbb R}|f|^pd\mu&lt;\infty$ ($p\geq1$). There is an exercise for proving that       $$\int_{[0,\infty)}pt^{p-1}\mu(\{x:|f(x)|\geq t\})d\mu(t)=\int_{[0,\infty)}\mu(\{x:|f(x)|^p\geq s\})d\mu(s).$$   If one formally let $s=t^p$, then $ds=pt^{p-1}dt$ and one would immediately get the equality above. However, how can I justify it rigorously in the sense of Lebesgue integration?    [Edited] What puzzles me is that I have the "change of variable" theorem for Riemann integration, but all the integrations here are in the Lebesgue sense. Unless one has shown that the integrands are also Riemann integrable, how can I directly use "change of variable"?
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