Proof verification: $E\space H(X)=\int_{-\infty}^{\infty}h(y)P(X\geq y)dy$ 
Let $H(x)=\int_{(-\infty,x]}h(y)dy$ with $h(y)\geq 0$, then
  $$E\space H(X)=\int_{-\infty}^{\infty}h(y)P(X\geq y)dy$$

Solution: $$\int_{-\infty}^{\infty}h(y)P(X\geq y)dy=\int_{-\infty}^{\infty}h(y)\int_{\Omega}1_{(X\geq y)}dPdy
\\=\int_{\Omega}\int_{-\infty}^{\infty}h(y)1_{(X\geq y)}dydP=\int_{\Omega}\int_{-\infty}^{X}h(y)dydP
\\=\int_{\Omega}H(X)dP=E\space H(X)$$
Something is wrong, isn't? I didn't use the fact that $h(y)\geq 0$. Anyhelp appreciated.
Got the answer: Fubini's theorem.
 A: Indeed Fubini. 
Allowed because the integrand is non-negative.
Your solution is okay. 
$$\begin{aligned}\int h\left(y\right)P\left(X\geq y\right)dy & =\int h\left(y\right)\int1_{\left\{ X\geq y\right\} }\left(\omega\right)P\left(d\omega\right)dy\\
 & =\int\int h\left(y\right)1_{\left\{ X\geq y\right\} }\left(\omega\right)dyP\left(d\omega\right)\\
 & =\int\int_{\left(-\infty.X\left(\omega\right)\right]}h\left(y\right)dyP\left(d\omega\right)\\
 & =\int H\left(X\left(\omega\right)\right)P\left(d\omega\right)\\
 & =\mathbb{E}H\left(X\right)
\end{aligned}
$$

Alternative (backwards and using CDF of $X$)
$$\begin{aligned}\mathbb{E}H\left(X\right) & =\int H\left(x\right)dF_{X}\left(x\right)\\
 & =\int\int1_{\left(-\infty.x\right]}\left(y\right)h\left(y\right)dydF_{X}\left(x\right)\\
 & =\int\int1_{\left(-\infty.x\right]}\left(y\right)h\left(y\right)dF_{X}\left(x\right)dy\\
 & =\int h\left(y\right)\int1_{\left[y,\infty\right)}\left(x\right)dF_{X}\left(x\right)dy\\
 & =\int h\left(y\right)P\left(X\geq y\right)dy
\end{aligned}
$$
