Let $F$ be a probability distribution function. Prove that $\int\limits_{-\infty}^{\infty} F(x+a)-F(x)\ dx= a\ \forall a\in\Bbb{R}$ A real-valued map $F$ is called a probability distribution if

$F$ is monotonically increasing
$F$ is right continuous
$\lim\limits_{x\to\infty} F(x)=1$ and $\lim\limits_{x\to-\infty} F(x)=0$

Again I know that for such $F$, there is a probability measure $P$ on borel $\sigma$-field of $\Bbb{R}$. such that $P((a,b])=F(b)-F(a)\ \forall -\infty\le a\le b\le \infty$.
So we have to prove that $\int\limits_{-\infty}^{\infty}P((x,x+a])\ dx=a$
Here $F$ may not be absolutely continuous, so we cannot work with the probability density function.
I'm not getting any idea how to prove this result. Can anyone help me in this regard? Thanks for your help in advance.
Let $\mu$ be the probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$
induced by $F$ (i.e, $\mu(a,b]=F(b)-F(a))$. Let $m$ be the usual
Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Consider
the product measure space $(\mathbb{R}\times\mathbb{R},\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}),m\times \mu)$.
Fix $a&gt;0$ (the case that $a\leq0$ can be treated similarly). Let
$A=\{(x,y)\in\mathbb{R}^{2}\mid x&lt;y\leq a+x\}$ which is a Borel subset
of $\mathbb{R}^{2}$ (see remark below). By Fubini-Tonelli Theorem,
$$
\int\left(\int1_{A}(s,t)dm(s)\right)d\mu(t)=\int\left(\int_{A}(s,t)d\mu(t)\right)dm(s).
$$
Observe that for each $s$, $\int1_{A}(s,t)d\mu(t)=\mu\left((s,s+a]\right)=F(s+a)-F(s)$.
Therefore $\int\left(\int_{A}(s,t)d\mu(t)\right)dm(s)=\int_{-\infty}^{\infty}\left[F(s+a)-F(s)\right]ds.$
On the other hand, for each $t$, $1_{A}(s,t)=1$ iff $s&lt;t\leq s+a$
iff $s\in[t-a,t)$. Therefore, $\int1_{A}(s,t)dm(s)=m\left([t-a,t)\right)=a$.
It follows that $\int\left(\int1_{A}(s,t)dm(s)\right)d\mu(t)=\int ad\mu(t)=a$.
That is, $\int_{-\infty}^{\infty}\left[F(s+a)-F(s)\right]ds=a$.

Remark: We go to show that $A$ is a Borel set.
Let $A_{1}=\{(x,y)\in\mathbb{R}^{2}\mid x&lt;y\}$ and $A_{2}=\{(x,y)\in\mathbb{R}^{2}\mid y&gt;a+x\}$.
Note that
\begin{eqnarray*}
A_{1} &amp; = &amp; \cup_{r\in\mathbb{Q}}\{(x,y)\mid x&lt;r&lt;y\}\\
 &amp; = &amp; \cup_{r\in\mathbb{Q}}\left((-\infty,r)\times\mathbb{R}\right)\cap\left(\mathbb{\mathbb{R}}\times(r,\infty)\right)
\end{eqnarray*}
which is a countable union of measurable rectangles, so it is measurable.
Similarly, $A_{2}$ is measurable. Finally, $A=A_{1}\setminus A_{2}$,
so $A$ is measurable too.
Let $\mu$ be the probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$
induced by $F$ (i.e, $\mu(a,b]=F(b)-F(a))$. Let $m$ be the usual
Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Consider
the product measure space $(\mathbb{R}\times\mathbb{R},\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}),m\times \mu)$.
Fix $a&gt;0$ (the case that $a\leq0$ can be treated similarly). Let
$A=\{(x,y)\in\mathbb{R}^{2}\mid x&lt;y\leq a+x\}$ which is a Borel subset
of $\mathbb{R}^{2}$ (see remark below). By Fubini-Tonelli Theorem,
$$
\int\left(\int1_{A}(s,t)dm(s)\right)d\mu(t)=\int\left(\int_{A}(s,t)d\mu(t)\right)dm(s).
$$
Observe that for each $s$, $\int1_{A}(s,t)d\mu(t)=\mu\left((s,s+a]\right)=F(s+a)-F(s)$.
Therefore $\int\left(\int_{A}(s,t)d\mu(t)\right)dm(s)=\int_{-\infty}^{\infty}\left[F(s+a)-F(s)\right]ds.$
On the other hand, for each $t$, $1_{A}(s,t)=1$ iff $s&lt;t\leq s+a$
iff $s\in[t-a,t)$. Therefore, $\int1_{A}(s,t)dm(s)=m\left([t-a,t)\right)=a$.
It follows that $\int\left(\int1_{A}(s,t)dm(s)\right)d\mu(t)=\int ad\mu(t)=a$.
That is, $\int_{-\infty}^{\infty}\left[F(s+a)-F(s)\right]ds=a$.

Remark: We go to show that $A$ is a Borel set.
Let $A_{1}=\{(x,y)\in\mathbb{R}^{2}\mid x&lt;y\}$ and $A_{2}=\{(x,y)\in\mathbb{R}^{2}\mid y&gt;a+x\}$.
Note that
\begin{eqnarray*}
A_{1} &amp; = &amp; \cup_{r\in\mathbb{Q}}\{(x,y)\mid x&lt;r&lt;y\}\\
 &amp; = &amp; \cup_{r\in\mathbb{Q}}\left((-\infty,r)\times\mathbb{R}\right)\cap\left(\mathbb{\mathbb{R}}\times(r,\infty)\right)
\end{eqnarray*}
which is a countable union of measurable rectangles, so it is measurable.
Similarly, $A_{2}$ is measurable. Finally, $A=A_{1}\setminus A_{2}$,
so $A$ is measurable too.