How can I prove $\int[F(x+a)-F(x)]\,dx=a$
where $F(x)$ is the cumulative distribution function?
How can I prove $\int[F(x+a)-F(x)]\,dx=a$
where $F(x)$ is the cumulative distribution function?
Let $R, S> 0$ be large compared to $a$. Then
$\begin{align*} \int_{-R}^{S} \left[ F(x+a) - F(x) \right] \; dx &= \int_{-R}^{S} F(x+a) \; dx - \int_{-R}^{S} F(x)\; dx \\ &= \int_{-R+a}^{S+a} F(x) \; dx - \int_{-R}^{S} F(x)\; dx \\ &= \int_{S}^{S+a} F(x) \; dx - \int_{-R}^{-R+a} F(x)\; dx \\ &= \int_{0}^{a} F(x+S) \; dx - \int_{0}^{a} F(x-R)\; dx \end{align*}$
Now taking $R, S \to \infty$, Bounded Convergence Theorem shows that
$ \lim_{S\to\infty} \int_{0}^{a} F(x+S) \; dx = \int_{0}^{a} \lim_{S\to\infty} F(x+S) \; dx = a$
and
$ \lim_{R\to\infty} \int_{0}^{a} F(x-R) \; dx = \int_{0}^{a} \lim_{R\to\infty} F(x-R) \; dx = 0$
Therefore we have
$ \int_{-\infty}^{\infty} \left[ F(x+a) - F(x) \right] \; dx = a.$
We can also prove it using Fubini's theorem for non-negative functions. Let $X$ a random variable of cumulative distribution function $F$, and $(\Omega,\mathcal F,P)$ the probability space on which $X$ is defined. We have \begin{align*} \int_{\mathbb R}[F(x+a)-F(x)]dx&=\int_{\mathbb R}\int_{\Omega}\chi_{\{(u,v),u