show that $E(X|G)=\int_0^{\infty}P[X>t|G]dt$ For $0\le X \in L_1$,  $X$ is a random variable on $(\Omega,B,P)$, and G is a $\sigma$-algebra and $G \subset B$, show almost surely
$$E(X|G)=\int_0^{\infty}P[X>t|G]dt$$
Here is my try:
This is about conditional expectation. By the definition of the conditional expectation $E(X|G)$ has two requirements:
$$E(X|G) \in G$$ 
And for all  $ A \in G$
$$\int_AE(X|G)dP=\int_AXdP $$
It's easy to get the second equation. That is 
$$
\begin{align}
\int_A\int_0^\infty P[X>t|G]dtdP &= \int_0^\infty \int_A P[X>t|G]dPdt \;\;(Fubini) \\ 
&= \int_0^\infty \int_A E(1_{[X>t]}|G)dPdt 
\\&=\int_0^\infty \int_{A} 1_{[X>t]}dPdt
\\&=\int_{A} \int_0^\infty  1_{[X>t]}dtdP \;\; (Fubini)
\\&=\int_A X dP
\end{align}
$$
But I tried a lot, still cannot prove the first one, because it's contained in the integral. Anyone has any idea? Thank you!
 A: I feel that the question is not well formulated but here is my answer to what I think is being asked assuming the conditional PDF exists:
$$\int_0^{z} x f(x|G)\,dx=\left.x\cdot F(x|G)\right|_0^z-\int_0^{z} F(x|G)\,dx=$$
$$ =\int_0^{z} (F(z|G)-F(x|G))dx$$ now take $z\rightarrow \infty$ to get $E[X|G]$.
Notice the convergence is monotone.
A: Let $f(t,\omega) = E[\mathbb{1}_{X>t}|\mathcal{G}](\omega)$, we are going to prove the $\mathcal{G}$-measurability of $g_a(\omega) = \int_{0}^{a}f(t,\omega)dt, \forall a>0$. Then $\int_{0}^{+\infty}f(t,\omega)dt = \lim_{a\rightarrow +\infty}g_a(\omega)$ is $\mathcal{G}$-measurable.
Firstly since $f(t,\omega)$ is monotone for t, it is Riemann integrable. So $\int_{0}^{a}f(t,\omega)dt = \lim_{n\rightarrow +\infty}\sum_{k=0}^{n-1}f(a\frac{k}{n}, \omega)\frac{a}{n}$. Note that for a fixed $t$, $f(t,\omega)$ is $\mathcal{G}$-measurable, so the finite sum is $\mathcal{G}$-measurable, and then their limit(we know it exists) is $\mathcal{G}$-measurable. Thus we have proven the $G$-measurability of $g_a(\omega)$ 
