Show that $\int_{\mathbb{R}} F(x) F(dx)=\frac{1}{2}$ if $F$ is a continuous distribution function If $F$ is a continuous distribution function, prove that 
\begin{align*}
\int_{\mathbb{R}} F(x) F(dx)=\frac{1}{2}
\end{align*}
What I tried
\begin{align*}
\int_{\mathbb{R}} F(x) F(dx)&=\int_{\mathbb{R}} P(X \le x) F(dx)=\int_{\mathbb{R}} E[\mathsf{1}_{X \le x}]F(dx)=\int_{\mathbb{R}}  \int_\Omega \mathsf{1}_{X \le x} dP F(dx)\\
&=  \int_\Omega  \int_{\mathbb{R}} \mathsf{1}_{X \le x} F(dx) dP 
\end{align*}
where the last step is due to Fubini's theorem.
If the above is correct what to do next? Thanks
 A: Here is the solution that I think you are looking for, recall from your other problem
$$\int_\Omega f(x) dP = \int_0^\infty P\big(\{f>t\}\big) dt,$$
observe that $\mu_F$ is a probability measure on $\mathbb{R}$, so let $\Omega := \mathbb{R}$ and $P:= \mu_F$, we get 
$$\int_{-\infty}^\infty F(x) d\mu_F = \int_0^\infty \mu_F\big(\{F>t\}\big) dt =\int_0^1 \mu_F\big(\{F>t\}\big) dt,$$
because for $t>1$, the set $\{F>t\}$ is an empty set, thus $\mu_F\big(\{F>t\}\big)=0$ . 
And for $t\in [0,1]$, I claim that $\mu_F\big(\{F>t\}\big) = 1-t$.
Here notice that since $F$ is a cdf, thus it is monotone, then the set $\{F>t\}$ is an interval $(r, \infty)$ in $\mathbb{R}$. (When $F$ is strictly increasing, then $F$ is invertable and we can take $r = F^{-1}(t)$). From the definition of L-S measure on intervals and $F$ is continuous, we have 
$$\mu_F((r, \infty)) = F(\infty) - F(r) = 1-t$$
then 
$$\int_{-\infty}^\infty F(x) d\mu_F =\int_0^1 1-t\;dt = 1/2.$$
A: You can use this result to show the following
$$
  I = \int_\Bbb R F(x)\mathrm dF(x) = F^2(x)|^\infty_{-\infty} - \int_\Bbb R F(x)\mathrm dF(x) = 1-I\implies I = \frac12
$$
Since Xiao provided another solution, I'll think it's worth telling you about a third one which is a neat probabilistic one-liner. Since $X$ is continuous, $F(X)\sim U[0,1]$ and hence your integral is an expectation of a uniformly distributed random variable on $[0,1]$ which is of course $\frac12$.
A: \begin{align*}
 \mathrm{E}[F(X)]&=\int_{\Omega}F(X(\omega))P(d\omega)\\&=\int_{{\mathbb{R}}}F(x)(P\circ X^{-1})(dx)\\&=\int_{{\mathbb{R}}}\mathrm{Pr}(X\leq x)(P\circ X^{-1})(dx)\\
 &=\int_{{\mathbb{R}}}\mathrm{E}[1(X\leq x)](P\circ X^{-1})(dx)\\
 &=\int_{{\mathbb{R}}}\int_{\Omega}1(X(\omega)\leq x)P(d\omega)(P\circ X^{-1})(dx)\\
 &=\int_{\Omega}\int_{{\mathbb{R}}}1(X(\omega)\leq x)(P\circ X^{-1})(dx)P(d\omega)\tag{Fubini’s Theorem}\\
 &=\int_{\Omega}\bigg(1-\mathrm{Pr}(x\leq X(\omega))\bigg)P(d\omega)\\
 &=1-\mathrm{E}[F(X)].
 \end{align*}
