$\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\{x\in X: f(x)>t\}) d\mu_t$ for any natural $p\ge 1$ [duplicate]
Solution 1:
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 $pt^{p-1}$.