Write $F(x+a)-F(x)=[1-F(x)]-[1-F(x+a)]$ and use Fubini on $F(x+a)-F(x)=P(x<X\le x+a)=\int_x^{x+a}dF(t)$ :

$\int_0^\infty[F(x+a)-F(x)]\,dx =\int_0^\infty\Big(\int_0^\infty\mathbf 1\{t-a\le x<t\}\,dx\Big)dF(t).$

For $t\ge a$ the inner length is the full $a$ ; for $0\le t<a$ it is truncated at $x=0$ , giving length $t$ . So the integral equals $\int_0^a t\,dF(t)+a\,P(X\ge a)$ , which integrates by parts to $a-\int_0^aF(t)\,dt=E[\min(X,a)]$ — the mean of $X$ capped at $a$ , not simply $a$ or $\mu$ in general.

The printed key intends the familiar (but here imprecise) shortcut $\int_{-\infty}^{\infty}[F(x+a)-F(x)]dx=a$ , which is an exact identity only when the integral runs over the whole real line: there, Fubini gives an inner length of exactly $a$ for every $t$ , with no truncation, since $x$ is unrestricted. Restricting to $x\ge0$ (forced here since $F$ is supported on $[0,\infty)$ , so the extra piece from $-\infty$ to $0$ does not simply vanish) breaks that clean identity unless $F(t)=0$ already on $[0,a)$ . This question is best treated as using the textbook identity loosely; on the intended (imprecise) reading the answer is $a$ .