Integral of $-\alpha\int_{x_l}^1F(x)dF(x)$?, $F(x)$ continuous with support $[x_l,1]$

Question

where $x_l$ such that $supp(F(x)) =[x_l,1]$ (supp means support)

Also, $F(x)$ is continuous

I don't know how to deal with the $F(x)$ in the integral.

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From the first line we know that $F(x_l)=0$ and $F(1)=1$, so if there is some way to write out the integral in terms of those two things then we know the answer. But I can't think of a way to write is a such (without there being a constant?), and I can't think of another approach.

Answer 1

$$-\alpha\int_{x_l}^1F(x)dF(x)$$

$$ = -\alpha \frac{F(x)^2}{2}|_{x_l}^{1}$$

$$ = - \frac{\alpha}{2} F(x)^2|_{x_l}^{1}$$

$$ = - \frac{\alpha}{2} [(1)^2 - (x_l)^2]$$

Answer 2

In the sense of Riemann-Stieltjes integral, we can provide a direct proof:

Claim. If $f : [a, b] \to \mathbb{R}$ is continuous and monotone increasing, then $$ \int_{a}^{b} f(x) \, df(x) = \left[ \frac{f(x)^2}{2} \right]_{a}^{b} $$ as in Riemann-Stieltjes integral sense.

Indeed, let $\Pi = \{a = x_0 < \cdots < x_n = b \}$ be any partition of $[a, b]$. Then the upper sum $U(\Pi, f)$ and the lower sum $L(\Pi, f)$ is given by

\begin{align*} U(\Pi, f) &= \sum_{i=1}^{n} f(x_i)[f(x_i) - f(x_{i-1})], \\ L(\Pi, f) &= \sum_{i=1}^{n} f(x_{i-1})[f(x_i) - f(x_{i-1})]. \end{align*}

Now notice that

\begin{align*} U(\Pi, f) - L(\Pi, f) &= \sum_{i=1}^{n} [f(x_i) - f(x_{i-1})]^2 \\ &\leq [f(b) - f(a)]\max_{1\leq i\leq n} [f(x_i) - f(x_{i-1})] \xrightarrow[\|\Pi\|\to0]{} 0 \end{align*}

by the uniform continuity of $f$ on $[a, b]$ and that

$$ U(\Pi, f) + L(\Pi, f) = \sum_{i=1}^{n} [f(x_i)^2 - f(x_{i-1})^2] = \left[ f(x)^2 \right]_{a}^{b}. $$

Therefore the desired Riemann-Stieltjes integral exists with value $\frac{1}{2}[f(b)^2 - f(a)^2]$.

One can prove a more general statement in a similar spirit:

Proposition. (Integration by parts) If $f, g$ are functions on $[a, b]$ such that $\int_{a}^{b}f(x) \, dg(x)$ exists as Riemann-Stieltjes integral, then $\int_{a}^{b}g(x) \, df(x)$ also exists and $$ \int_{a}^{b}f(x) \, dg(x) = [f(x)g(x)]_{a}^{b} - \int_{a}^{b}g(x) \, df(x). $$

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An overkill. In the sense of Lebesgue-Stieltjes integral, we can provide another proof.

Claim. If $f : [a, b] \to \mathbb{R}$ is right-continuous and monotone increasing, then $$ \int_{a}^{b} f(x) \, df(x) = \left[ \frac{f(x)^2}{2} \right]_{a}^{b} + \frac{1}{2}\sum_{x \in (a,b]} [f(x) - f(x^-)]^2, $$ as in Lebesgue-Stieltjes integral sense, where $f(x^-) = \lim_{y \uparrow x} f(y)$ is the left-limit and hence $f(x) - f(x^-)$ is the jump size of $f$ at $x$.

Indeed, there exists a Stieltjes measure $\mu$ such that $f(x) = \mu([a,x])$ and hence

\begin{align*} 2\int_{a}^{b} f(x) \, df(x) &= \int_{[a,b]} \int_{[a,x]} \mu(dy)\mu(dx) + \int_{[a,b]} \int_{[a,y]} \mu(dx)\mu(dy) \\ &= \int \left( \mathbf{1}_{\{ a \leq y \leq x \leq b\}} + \mathbf{1}_{\{ a \leq x \leq y \leq b\}} \right) \, \mu^{\otimes 2}(dx,dy) \\ &= \int \left( \mathbf{1}_{[a,b]^2} + \mathbf{1}_{\{ a \leq y = x \leq b\}} \right) \, \mu^{\otimes 2}(dx,dy) \\ &= \mu([a,b])^2 + \int \left( \sum_{x \in [a, b]} \mu(\{x\}) \mathbf{1}_{\{y=x\}} \right) \, \mu(dy) \\ &= \mu([a,b])^2 + \sum_{x\in[a,b]} \mu(\{x\})^2. \end{align*}

Rewriting everything in terms of $f$ yields the desired equality.