Tail sum for expectation In Pitman's Probability, the tail sum formula for expectation is introduced for a nonnegative (0,1,...) discrete random variable $X$:
$$E(X) = \sum_{i=0}^\infty P(X > i).$$


*

*I wonder if there is a similar formula for nonnegative continuous
random variable $X$:
$$E(X) = \int_0^\infty P(X > x) dx?$$
If no, are there some conditions for it to hold? And how can it be
proved?
Here is my thought:
If the cdf $F$ of $X$ is bijective, then $X=F^{-1}(U)$ for some
random variable $U$ uniformly distributed over $[0,1)$. So $$E(X) =
    \int_0^1 F^{-1}(u) du.$$
To prove the tail sum formula, it suffices to prove $$\int_0^1
    F^{-1}(u) du = \int_0^\infty P(X > x) dx.$$ But I am stuck here. 
What's more, is the condition that the cdf $F$ of $X$ is bijective
really necessary for tail sum formula to hold?

*Can tail sum formula be generalized to a random variable that is not
necessarily nonnegative?


Thanks!
 A: First we prove that for $X \geq 0$
$$EX= \int_{0}^{\infty}P(X > t)dt.$$
We apply Fubini's theorem. Our product measure will be a product of the distribution of $X$ and Lebesgue measure. 
So $$\int_{0}^{\infty}P(X>t)dt = \int_0^{\infty}\left( \int_{t}^{\infty} \nu_X(ds)\right)dt=\int_{\mathbb{R}}\left( \int_{0}^{\infty}\textbf{1}_{[0,\infty)}(t)\textbf{1}_{(t,\infty)}(s)\nu_X(ds)\right)dt=$$
$$=\int_{0}^{\infty}\left(\int_0^s dt \right) \nu_X(ds)= \int_{0}^{\infty}s \nu_X(ds)=\int_{\mathbb{R}}s\nu_X(ds)=EX$$
But, you want to have $EX=\int_{0}^{\infty}P(X \geq t)dt.$
Applying what we have already proved, it is enough to show that $\int_{0}^{\infty}P(X=t)dt=0$.
A: @Did gave a great answer, and while I'm thinking about it I want to record a similar answer using slightly different notation in the case where the cumulative density function $F$ is differentiable. 
In this case,
\begin{align}
\mathbb E(X) &= \int_0^\infty x F'(x) \, dx \\
&= \int_0^\infty \int_0^\infty [x \geq t] \, dt \,F'(x) \, dx \\
&= \int_0^\infty \int_0^\infty [x \geq t] F'(x) \, dt \, dx \\
&= \int_0^\infty \int_0^\infty [x \geq t] F'(x) \, dx \, dt \quad \text{(Fubini's theorem)}\\
&= \int_0^\infty \int_t^\infty F'(x) \, dx \, dt \\
&= \int_0^\infty 1 - F(t) \, dt.
\end{align}
(The quantity $[x \geq t]$ is equal to $1$ if $x \geq t$ and $0$ otherwise.)
A: This is Fubini theorem for nonnegative functions/sums. If $x$ is a nonnegative integer,

$$
x=\sum_{i=0}^{+\infty}[x\gt i].
$$

Likewise, if $x$ is a nonnegative real number,

$$
x=\int_{0}^{+\infty}[x\gt t]\mathrm dt,\qquad x=\int_{0}^{+\infty}[x\geqslant t]\mathrm dt.
$$

Then one integrates both sides of the relevant identity with respect to the distribution $\mathrm P_X$ of $X$ and one uses Fubini theorem to change the order of the summation/integral and of the expectation. 
For example, the second identity yields
$$
\mathbb E(X)=\int_\Omega X\ \mathrm d\mathbb P=\int_\Omega \int_{0}^{+\infty}[X\gt t]\mathrm dt\ \mathrm d\mathbb P=\int_{0}^{+\infty}\int_\Omega [X\gt t]\mathrm d\mathbb P\ \mathrm dt
$$ that is,

$$\mathbb E(X)=\int_{0}^{+\infty}\mathbb P(X\gt t)\ \mathrm dt.
$$

Likewise, the first identity yields
$$
\mathbb E(X)=\int_\Omega X\ \mathrm d\mathbb P=\int_\Omega\ \sum\limits_{i=0}^{+\infty}[X\gt i]\ \mathrm d\mathbb P=\sum\limits_{i=0}^{+\infty}\ \int_\Omega[X\gt i]\ \mathrm d\mathbb P
$$ that is,

$$\mathbb E(X)=\sum\limits_{i=0}^{+\infty}\mathbb P(X\gt i).
$$

A: This is integration by parts. Take $X\geq 0$, and call its distribution function $F$. Let $g$ be an increasing differentiable function with $g(0)=0$.
$$\begin{align*}
\mathbb{E}[g(X)]&=\int_0^\infty g(t) dF(t) \\
              &= \int_0^\infty -g(t) d(1-F(t)) \\
&= [-g(t)(1-F(t))]^\infty_0 - \int_0^\infty 1-F(t)d(-g(t)) \\
&= \int_0^\infty g'(t)\mathbb{P}[X>t]dt
\end{align*}$$
This reduces to what you want when $g(X)=X$.
One way we could compute $\mathbb{E}[X]$ for general $X$ would be to compute $\mathbb{E}[X^+]$ and $\mathbb{E}[X^-]$ in this way and then take the difference, where $X^+=\max(X,0)$ and $X^-=\max(-X,0)$.
Justification of integration by parts: The integration by parts works when $g(X)$ is integrable, since, by dominated convergence,
$$\begin{align}
\limsup_x g(x)\mathbb{P}[X>x] &\le \limsup_x\mathbb{E}[g(X)\mathbf{1}(X>x)]\\
                              &= \mathbb{E}[\limsup_x g(X)\mathbf{1}(X>x)]\\
                              &= 0.
\end{align}$$
