proof check, ($ E(X^p)=\int_0^\infty px^{p-1}\mathbb{P}\{ X>x \}dx $)

Question

Problem: If $X ≥ 0,$ then for every $p \in R_+,$ $$ E(X^p)=\int_0^\infty px^{p-1}\mathbb{P}\{ X>x \}dx $$ Show this, using Fubini's theorem with the product measure $ \mathbb{P}\times Leb, $ after noting that $ X^{p}(\omega)=\int_0^\infty px^{p-1}dx=\int_0^\infty px^{p-1} \chi_{\{ X>x \}}(\omega) dx. $

Actually, I have problems with the notations. I have passed measure theory but I have problems with this probability theory. This a problem from Erhan Çinar's book.

Here is what I have done

Let $ \mu [-\infty , x]= P(X\leq x) $ so $ \mu [x , +\infty]= P(X\geq x) $ then $$ \int_0^\infty d\mu(y)=\int_x^\infty \mathbb{P}\{ X>y \}dy, $$ Now I want to write $$\int_0^\infty px^{p-1}\mathbb{P}\{ X>x \}dx=\int_0^\infty px^{p-1}(\int_x^\infty d\mu(y))dx=\int_0^\infty\int_0^y px^{p-1}\mathbb{P}\{ X>y \}dxd\mu(y)= \int_0^\infty y^{p}\mathbb{P}\{ X>y \}d\mu(y)=E(X^p)$$

Please let me know where my problem is?

Answer 1

Hint $$|X(w)|^p =\int_{0}^{|X(w)|} pt^{p-1} dt= \int_{0}^{\infty} pt^{p-1} \mathbf{1}_{\{|X|>t\}}(w)dt $$

AND

$$E(X^p)=\int_\Omega |X(w)|^p dw =\int_\Omega\int_{0}^{\infty} pt^{p-1} \mathbf{1}_{\{|X|>t\}}(w)dt~dw.=\int_{0}^{\infty} pt^{p-1} \int_\Omega\mathbf{1}_{\{|X|>t\}}(w)dw ~dt $$ and $$ \mathbb{P}\{ X>t \}= \int_\Omega \mathbf{1}_{\{|X|>t\}}(w) dw.$$ which gives the result.