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?

Proof:

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

$$ \int_{-\infty}^{\infty} \left[ F(x+a) - F(x) \right] \; dx = a.$$

Proof 2:

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<v\leq v+a\}}(x,X(\omega))dP(\omega)dx\\\ &=\int_{\Omega}\int_{\mathbb R}\chi_{\{(u,v),u<v\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*}