Show that $ \int_\mathbb{R} 2F(x)dF(x)=1 $.

Question

Let $F$ be Stieltjes measure function (nondecreasing and right continuous) and $\mu$ be corresponding measure on $(\mathbb{R}, \mathcal{R})$. Assume that $F: \mathbb{R}\to [0,1]$ is continuous. Show that $$ \int_\mathbb{R} 2F(x)dF(x)=1 $$

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I found that this question is a corollary of Durrett's textbook exercise 1.7.3:

Let $F$ be continuous. Then $$ \int_{(a,b]}2 F(y)dF(y)=F^2(b)-F^2(a) $$

However, the proof relies on the previous results:

Let $F, G$ be Stieltjes measure functions and let $\mu, \nu$ be corresponding measures. Then $$\int_{(a,b]}F(y)dG(y)+\int_{(a,b]}G(y)dF(y)=F(b)G(b)-F(a)G(a)+\sum_{x\in (a,b]} \mu(\{x\})\nu(\{x\})$$

Can we prove our result without using this result?

If we just apply Durrett's result, then we take $a\to-\infty, b\to \infty$, we will prove our result.

Answer 1

Assuming the continuity of $F$, let $X$ be a random variable with distribution function $F$. Then $\int F(x)\,dF(x) = \mathbb E[F(X)]$. Since $F(X)$ is uniformly distributed in $[0,1]$, it follows $\mathbb E[F(X)] = \frac{1}{2}$, which solves the problem.

Answer 2

$$\int udv = uv - \int vdu$$

Claim:

$$\int F(x) dF(x) = F^2(b) - F^2(a) - \int F(x)dF(x)$$

Proof:

Since this is integral is Riemann integrable, we have: $$g([a,b]) = \int_{[a,b]} F(x)dF(x) = \lim_{n \rightarrow \infty} \sum_{1 \leq i \leq n-1} F(a_i) (F(a_{i+1}) - F(a_i))$$ $$ = \lim_{n \rightarrow \infty} \sum_{1 \leq i \leq n-1} F(a_i) (F(a_{i+1}) - F(a_i)) - F^2(a_{i+1}) + F^2(a_i) + F^2(a_{i+1}) - F^2(a_i)$$ $$ = F^2(b) - F^2(a) + \lim_{n \rightarrow \infty} \sum_{1 \leq i \leq n-1} F(a_i) (F(a_{i+1}) - F(a_i)) - F^2(a_{i+1}) + F^2(a_i)$$ $$ = F^2(b) - F^2(a) + \lim_{n \rightarrow \infty} \sum_{1 \leq i \leq n-1} F(a_i) F(a_{i+1}) - F^2(a_{i+1})$$ $$ = F^2(b) - F^2(a) - \lim_{n \rightarrow \infty} \sum_{1 \leq i \leq n-1} F(a_{i+1})(F(a_{i+1}) - F(a_i))$$ $$ = F^2(b) - F^2(a) - \int F(x) dF(x)$$

Rest follows since, $$\int F(x) dF(x) = \frac{1}{2} (F^2(b) - F^2(a))$$