How to prove the interchange of integral and expectation How to prove $\int_{0}^{\infty}{h(t)\mathbb{E}(I(X>t))dt}=\mathbb{E}(\int_{0}^{\infty}{h(t)I(X>t)dt})$.
Can I treat $h(t)$ as a constant respect to $X$? Then, directly get the result?
The point is I do not understand what $\mathbb{E}(\int_{0}^{\infty}{h(t)I(X>t)dt})$ is.
 A: The integral $\int_0^\infty h(t)I(X>t)\,dt$ is a random variable, call it $Y$. The role of the indicator random variable $I(X>t)$ is to restrict the $t$-integration to the (random) interval $(0,X)$. In other words,
$$
Y(\omega) =\int_0^{X(\omega)} h(t)\,dt,
$$
for each sample point $\omega$ in the sample space. You are then forming the expectation of $Y$.  If $h$ takes only non-negative values, then Tonelli's theorem can be used to justify the change in order of expectation and integration (in $t$).
A: Ok, it looks better now, and I think all requirements for switching the order of integration are satisfied. It seems you are assuming finite expectation of $X$. To get at your last line/question: $X$ is a random variable and thus $I(X>t)$ is a random variable (for each fixed $t$). It is either $1$ or $0$, and depends (randomly) on the value $X$ takes. Thus $\int_0^\infty h(t) I(X>t)dt$ is a random variable. The function $h(t) I(X>t)$ is going to be set to zero over a certain interval depending on what value $X$ takes, thus we don't know the value of the integral with certainty.
