# Integral of shifted distribution function

Assume the integral exists. Show that for any distribution function $M$ with density function $m$ exists and $y\geq 0$, $\int_{-\infty}^{\infty} [M(x+y) - M(x)] dx = y$,

My attempt: I see that $F(x+a) - F(x) = P(x< X < x+a)$ for $X$ is a random variable. But this means the indefinite integral is just the sum of all the probabilities, so why is it $a$, but not $1$? I think I made a serious mistake somewhere in this argument.

Could someone give me some help on this problem? Would really appreciate any of your help.

• I think you need to assume the expectation of $X$ exists and is finite. Then, you might try integration by parts. – Michael Sep 15 '16 at 3:10
• A more clever way (with no assumptions on expectation of $X$ being finite) is to find an expression for $\int_{x=-\infty}^{\infty} 1_{\{X \in (x, x+y]\}}dx$ and then use a Fubini-Tonelli argument on $E[\int_{x=-\infty}^{\infty} 1_{\{X \in (x, x+y]\}}dx]$. – Michael Sep 15 '16 at 3:19
• @Michael: thank you so much for your great help! I got it. Could you please give this problem a try as well? math.stackexchange.com/questions/1927392/… – user177196 Sep 15 '16 at 4:13