$\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\{x\in X: f(x)>t\}) d\mu_t$ for any natural $p\ge 1$ Let $f:X\mapsto0,+\infty)$ 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 $p\in\mathbb{N}$, $p\ge 1$, $$\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\\{x\in X: f(x)>t\\}) d\mu_t$$ I know that the equality holds for $p=1$, as proved [here. That also implies that $$\int_X f^p d\mu =\int_{0,+\infty)} \mu(\\{x\in X: f(x)^p>t\\}) d\mu_t,$$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 [this, but, although related, they do not ask the same question.

Apply change of variables to the expression you have. You deduced that: $$\int_X f^p d\mu =\int_{[0,+\infty)} \mu(\\{x\in X: f(x)^p>t\\}) d\mu_t,$$ where $\mu(\\{x\in X: f(x)^p>t\\}=\mu(\\{x\in X: f(x)>t^{1/p}\\}$. Now substitute $t\mapsto t^p$ and the result follows because the derivative of this map is $pt^{p-1}$.