Expectation of Integral

Question

For $a>0$ and a r.v. $X$, is it true that $\int_0^\infty a^{p-1} \mathbb{P}[{{X}\ge a}] \operatorname{d}a = \mathbb{E}[{\int_0^{{X}}a^{p-1}\operatorname{d}a}]$? If so, why?

Answer 1

$\int_0^\infty a^{p-1} \mathbb{P}[{{X}\ge a}] \operatorname{d}a = \int_0^\infty a^{p-1} \int_{\mathbb{R}}I_{[x\ge a]}d \mathbb{P}(x) \operatorname{d}a $ and by Fubini's theorem $$ \int_0^\infty a^{p-1} \int_{\mathbb{R}}I_{[x\ge a]}d \mathbb{P}(x) \operatorname{d}a =\int_{\mathbb{R}} \int_0^\infty a^{p-1} I_{[x\ge a]}\operatorname{d}a d \mathbb{P}(x) = \mathbb{E}\left[{\int_0^{{X}}a^{p-1}\operatorname{d}a}\right].$$