How can I prove $\int[F(x+a)-F(x)]\,dx=a$ How can I prove $$\int[F(x+a)-F(x)]\,dx=a$$
where $F(x)$ is the cumulative distribution function?
 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&ltv\leq v+a\}}(x,X(\omega))dP(\omega)dx\\\
&=\int_{\Omega}\int_{\mathbb R}\chi_{\{(u,v),u&ltv\leq v+a\}}(x,X(\omega))dxdP(\omega)\\\
&=\int_{\Omega}\int_{X(\omega)-a}^{X(\omega)}dxdP(\omega)\\\
&=\int_{\Omega}adP(\omega)\\\
&=a.
\end{align*}
A: 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
$$ \int_{-\infty}^{\infty} \left[ F(x+a) - F(x) \right] \; dx = a.$$
