Q:

Proof verification: $\mathbb{E}[X] = \int_0^\infty P(X > \alpha) d\alpha$

I want to show that given a probabilty density $P: \mathbb R^+ \rightarrow [0, 1]$, its expectation obeys the identity: $\mathbb{E}[X] = \int_0^\infty P(X > \alpha) d\alpha$.

We assume that the density $P$ is defined on $[0, u]$. We will get the final version by setting $u \rightarrow \infty$ (nit 1)
Begin by defining the cumulative density $C(x) = \int_0^x P(\alpha) d\alpha$.
$C'(x) = P(x) - P(0)$ from fundamental theorem of calculus. We need to assume that $P(0) = 0$ so that $C'(x) = P(x)$.  (nit 2).
This gives us $dC(x) = P(x)$.

Now compute expectation:
\begin{align*}
&\mathbb E[X] = \int_0^u xP(x)dx  \quad [UdV]\\
&\left[ \text{use product rule: }\int U dV = UV - \int V dU \right] \\
&= [xC(x)]_0^u - \int_0^u C(x) \cdot 1 dx \\
&= uC(u) -  \int_0^u C(x)dx \\
&\text{[$C(u) = 1$ since $u$ was upper bound of distribution]} \\
&= u - \int_{0}^uC(x) dx \\
&= \int_0^u 1 dx  - \int_{0}^uC(x) dx \\
&= \int_0^u [1 - C(x)] dx \\
&= \int_0^u [1 - P(X \leq x)] dx \\
&= \int_0^u P(X > x) dx \\
\blacksquare
\end{align*}
Set $u \rightarrow \infty$ to get the final desired expression.
Is this proof watertight? I'm nervous about (i) first proving it for finite $u$ and then setting the limit; (ii) The assumption that $P(0) = 0$. I believe (i) is all right since that's the definition of integral with limit infinity. As for (ii), I believe this is also okay since we are assuming something about the distribution over a set of measure zero (a single point $0$). Still, I'm nervous, so a proof verification would be very appreicated.

A:

You can fix your proof just noticing that
$$
\int_{0}^\infty tf_X(t) \mathop{}\!d t=\lim_{n\to\infty}\int_{0}^n tf_X(t) \mathop{}\!d t
$$
from the monotone convergence theorem. Hence, following your work, you get that
$$
\mathrm{E}[X]=\lim_{n\to\infty}\int_{0}^n(F_X(n)-F_X(t))\mathop{}\!d t=\int_{0}^\infty (1-F_X(t))\mathop{}\!d t 
$$
where we had used again the monotone convergence theorem as $\mathbf{1}_{[0,n]}(t)(F_X(n)-F_X(t))$ increases to $1-F_X(t)$ as $n\to\infty $.

