Expectation of $\mathbb{E}(X^{k+1})$ I have difficulties with an old exam problem : 

Let $X$ be a positive random variable defined on a probability space $(\Omega, \mathcal{F}, \mathbf{P})$. 
  Show that
  $$\int_0^\infty t^k \mathbf{P}(X\geq t) dt = \int_0^\infty \int_{\Omega} t^k\int_{\{X(\omega)\geq t\}}dt d\mathbf{P}(\omega)$$
  Infer from this the integral expression of $\mathbb{E}(X^{k+1})$ (where $\mathbb{E}$ is the expectation)

We have Fubini theorem, which we can apply to a $\mathbb{B}(\mathbb{R})\otimes\mathcal{F}$-measurable function because the Lebesgue measure is $\sigma$-finite and $\mathbf{P}$ is also $\sigma$-finite because it is a probability. I think we can write $\mathbf{P}(X\geq t)$ as $\int_{\{ X(w)\geq t\}} d\mathbf{P}(\omega)$ but I don't know how to proceed next. Especially I don't see how to introduce the $\int_{\Omega}$.
Edit
From the comments, there must be an error in the description of the exam problem. It should have been the following :

Let $X$ be a positive random variable defined on a probability space $(\Omega, \mathcal{F}, \mathbf{P})$. 
  Show that
  $$\int_0^\infty t^k \mathbf{P}(X\geq t) dt = \int_0^\infty \int_{\Omega} t^k\mathbf{1}_{\{X(\omega)\geq t\}}dt d\mathbf{P}(\omega)$$
  Where $\mathbf{1}_{\{X(\omega)\geq t\}}$ is the characteristic function of $\{ X(\omega)\geq t\}$
Infer from this the integral expression of $\mathbb{E}(X^{k+1})$ (where $\mathbb{E}$ is the expectation)

 A: By Fubini's theorem, we have
$$\begin{align*}
\int_0^\infty t^k\mathbb P(X\geqslant t)\mathsf dt&=
 \int_0^\infty t^k \mathbb E\left[1_{\{\omega:X(\omega)\geqslant t\}} \right]\mathsf dt\\
&=\int_0^\infty t^k\int_{\Omega} 1_{\{\omega:X(\omega)\geqslant t\}}\mathsf d\mathbb P\;\mathsf dt\\
&=\int_{\Omega}\int_0^{X(\omega)}t^k \mathsf dt\; \mathsf d\mathbb P\\
&=\int_{\Omega} \frac1{k+1}X^{k+1}(\omega)\mathsf d\mathbb P(\omega)\\
&= \frac1{k+1}\mathbb E[X^{k+1}].
\end{align*}$$
Hence $$\mathbb E[X^{k+1}] = (k+1)\int_0^\infty t^k\mathbb P(X\geqslant t)\mathsf dt.$$
The crucial part here is that
$$1_{\{\omega : X(\omega) \geqslant t\}}(\omega) = 1_{\{t: t\leqslant X(\omega)\}}(t). $$
A: Since I dissed measure theory, I feel obligated to give a measure-theory answer.  This is essentially the method suggested by Stefan Hansen (using indicator function).  The correct thing we want to prove is: 
$$ \int_0^{\infty} t^k P[X\geq t]dt = \int_{\omega \in \Omega} \left[\int_0^{\infty}  1\{X(\omega) \geq t\} t^k dt\right] dP(\omega) $$
where $1\{X(\omega)\geq t\}$ is an indicator function that is 1 if $X(\omega)\geq t$, and 0 else. 
To do this, we can use the Tonelli theorem about double integrals of non-negative functions. The main steps (you can fill in details) are: 
\begin{align} 
\int_0^{\infty} t^k P[X\geq t] dt &= \int_0^{\infty} t^k \left[\int_{\omega: X(\omega)\geq t} dP(\omega)  \right] dt\\
&= \int_0^{\infty} t^k\left[ \int_{\omega\in\Omega} 1\{X(\omega)\geq t\} dP(\omega)\right]dt
\end{align} 
and then use Tonelli to switch the order of integration.
