Proof of a formula for the expectation of a product of random variables I want to prove the second task, task b) (see picture below). a) was not hard to show.

One question before I start: I am a bit confused about the notation, but $\mathbb 1(t)_{\{Y>t\}}$ is $1$ if $Y(t)>t$? and $0$ else, correct ?
Using a) I just obtain:
$\mathbb E(XY^r)=\int_0^{\infty}\mathbb E(X\cdot \mathbb{1}_{\{Y^r>t\}})dt=\int_0^{\infty}\mathbb E(X\cdot \mathbb{1}_{\{Y>t^{\frac{1}{t}}\}})dt$ but what now? I played around with some substitutions..nothing helped.
Edit: The problem is that I can't get ${Y>t}$ under my indicator function.
Thanks for any kind of help!
 A: The essential result here is $x^r = r\int_0^x t^{r-1}dr = r\int_0^\infty t^{r-1} 1_{\{t | x>t\}}(t)dt$.
Then
we have $Y^r(\omega) = r\int_0^\infty t^{r-1} 1_{\{t | Y(\omega)>t\}}(t)dt$.
Then
\begin{eqnarray}
E[X Y^r]  &=& \int X(\omega) Y^r(\omega) \mu(d\omega) \\
&=&  \int X(\omega)r\int_0^\infty t^{r-1} 1_{\{t | Y(\omega)>t\}}(t)dt \mu(d\omega) \\
&=& r\int_0^\infty t^{r-1} \int X(\omega)1_{\{t | Y(\omega)>t\}}(t) \mu(d\omega) dt \\
&=& r\int_0^\infty t^{r-1} \int X(\omega)1_{\{\omega | Y(\omega)>t\}}(\omega) \mu(d\omega) dt \\
&=& r\int_0^\infty t^{r-1} E[X 1_{\{\omega | Y(\omega)>t\}}] dt
\end{eqnarray}
Note the switch in the indicator function in the second last line
(you can see that $1_{\{t | Y(\omega)>t\}}(t) = 1_{\{\omega | Y(\omega)>t\}}(\omega)$).
Addendum: 
If you accept that $E[XY] = \int_0^\infty E[X 1_{\{\omega | Y(\omega)>t\}}] dt$, then replace $Y$ by $Y^r$ to get
$E[XY^r] = \int_0^\infty E[X 1_{\{\omega | Y^r(\omega)>t\}}] dt$. Now use the
suggested substitution $t=q^r$ to get
$E[XY^r] = r\int_0^\infty q^{r-1}E[X 1_{\{\omega | Y^r(\omega)>q^r\}}] dq$.
Now note that $1_{\{\omega | Y^r(\omega)>q^r\}} = 1_{\{\omega | Y(\omega)>q\}}$ to get the desired result.
A: Just do exactly what the question tells you to:
\begin{equation*}
\mathbb{E}[XY^r]=\int_{0}^{\infty}\mathbb{E}[X\cdot\mathbf{1}_{Y^r>t}]dt
\end{equation*}
Letting $t=q^r$ we have $dt=rq^{r-1}dq$ with the limits still $0$ and $\infty$ (since $r>0$). Hence,\begin{align*}
\mathbb{E}[XY^r]&=\int_{0}^{\infty}\mathbb{E}[X\cdot\mathbf{1}_{Y^r>q^r}]rq^{r-1}dq\\
&=\int_{0}^{\infty}\mathbb{E}[X\cdot\mathbf{1}_{Y>q}]rq^{r-1}dq,
\end{align*}
as desired.
