Calculate the integral for a general distribution function

$F$ is a distribution function .Take any $a \in \mathbb{R}$. Then evaluate : $\displaystyle\int_{\mathbb{R}} (F(x+a)-F(x)) dx$

As it is a general distribution function , I cannot employ densities to solve this problem . Any insights or a complete solution will be very appreciated .

Solution 1:

$\int (F(x+a)-F(x))dx=\int \int I_{\{x< X\leq x+a\}}dPdx$ where $X$ is a random variable with distribution $F$. By Fubini's Theorem this can be written as $\int \int I_{\{X-a \leq x< X\}}dxdP=\int a dP=a$.

Solution 2:

Answer that leaves out a random variable.

If $a\geq0$ then $F\left(x+a\right)-F\left(x\right)$ can be recognized as integral $\int\mathbf{1}_{\left(x,x+a\right]}\left(y\right)dF\left(y\right)$.

Applying Fubini we find: $$\int F\left(x+a\right)-F\left(x\right)dx=\int\int\mathbf{1}_{\left(x,x+a\right]}\left(y\right)dF\left(y\right)dx=$$$$\int\int\mathbf{1}_{\left(x,x+a\right]}\left(y\right)dxdF\left(y\right)=\int\int\mathbf{1}_{\left[y-a,y\right)}\left(x\right)dxdF\left(y\right)=\int adF\left(y\right)=a$$

If $a<0$ then: $$\int F\left(x+a\right)-F\left(x\right)dx=-\int F\left(x\right)-F\left(x-\left(-a\right)\right)dx=-\int F\left(x+\left(-a\right)\right)-F\left(x\right)dx=-(-a)=a$$where the second equality is based on the fact that the Lebesgue measure is translation invariant.