Consider $F$ as a distribution function of some random variable $\xi$. The problem I'm trying to solve is to find integral:
$$
\int_{-\infty}^{+\infty}F(x)dF(x)
$$
From what I see, there are two ways of solving this problem.
The first:
$\int_{-\infty}^{+\infty}F(x)dF(x)=\mathbb{E}F(\xi)=\mathbb{E}\mathbb{P}(\xi<\xi)=\mathbb{P}(\xi<\xi)=\mathbb{P}(\emptyset)=0$
The second:
$\int_{-\infty}^{+\infty}F(x)dF(x)=\frac{1}{2}\int_{-\infty}^{+\infty}dF^2(x)=\frac{1}{2}\int_{-\infty}^{+\infty}dG(x)=\frac{1}{2}.$
Where $G(x)=F(x)F(x)$ is a distribution function of some variable (not descending, $G(-\infty)=0, G(+\infty)=1$).
As for me the second answer is more satisfying. Because it's intuitively understandable ($\int x dx=\frac{x^2}{2} +C$) and probably can be proved strictly.
But what's wrong with the first way? All I can think about is that if we define $F(x)=\mathbb{P}(\xi\leq x)$ so that it was left-continuous, then $\mathbb{E}F(\xi)=\mathbb{E}\mathbb{P}(\xi\leq\xi)=\mathbb{P}(\xi=\xi)=\mathbb{P}(\Omega)=1$. And the $0.5$ is in some way the mean of $0$ and $1$.