Prove that $\int_{-\infty}^{\infty} (F(x+a)-F(x))dx=a$ for all $a>0$.

Question

This is a qualifying exam problem. Suppose $F$ is a distribution function of a Borel measure $\mu$ with $\mu (\Bbb R)=1$. Prove that $\int_{-\infty}^{\infty} (F(x+a)-F(x))dx=a$ for all $a>0$.

Answer 1

Fubini. Or rather Tonelli. We have $$F(x+a)-F(x)=\mu([x,x+a)).$$And $$\int\mu([x,x+a))dx =\int\int\chi_{[x,x+a)}(t)\,d\mu(t)dx =\int\int\chi_{[x,x+a)}(t)\,dxd\mu(t)=\int a\,d\mu=a.$$

Answer 2

By Fubini's theorem, \begin{align*} & \int_{-\infty}^\infty(F(x + a) - F(x)) dx \\ = & \int_{-\infty}^\infty\left[\int_{-\infty}^{x + a}\mu(dt) - \int_{-\infty}^x\mu(dt)\right]dx \\ = & \int_{-\infty}^\infty\left[\int_{-\infty}^\infty I_{(x, x + a]}(t)\mu(dt)\right]dx \\ = & \int_{-\infty}^\infty\left[\int_{-\infty}^\infty I_{(x, x + a]}(t)dx\right]\mu(dt) \\ = & \int_{-\infty}^\infty\left[\int_{t - a}^{t} 1 dx\right]\mu(dt) \\ = & a\int_\mathbb{R} \mu(dt) \\ = & a. \end{align*}