Summary:
(1). Is the Statement below true?
(2). If so, then how to complete my proof sketched below?
(3). Is there any way to compute the integral term in the Statement more explicity?
I'm trying to prove a statement(the Statement below). I don't even know whether its true or false, though. My settings are as follows:
Let a function $\alpha:[0,\,1]\to\mathbb R$ be a monotonically increasing function. Define a linear functional $\Lambda: C([0,\,1])\to \mathbb C$ by the Riemann-Stieltjes integral $$ \Lambda f = \int_0^1 f \, d\alpha. \,\,\,\, (f\in C([0,\,1]))$$ Then by the Riesz Representation Theorem, we have a regular Borel measure $\mu$ such that $$ \Lambda f = \int_0^1 f\,d\mu$$ for all $f\in C([0,\,1])$.
What I want to prove is the following.
Statement. Let $[a,\,b]$ be a subset of $[0,\,1]$. Then we have $$\mu([a,\,b]) = \int_D\alpha ' dm +\sum_{x \in S} (\alpha(x^+)-\alpha(x^-)),$$ where $m$ is the Lebesgue measure, $D\subset [a,\,b]$ is the set on which $\alpha$ is differentiable, and $S$ is the set of all discontinuities of $\alpha$. The symbols $\alpha(x^+)$ and $\alpha(x^-)$ means the right-hand and left-hand limits of $\alpha$ at $x$, respectively.
Agagin I say, I'm not sure if this is true. But I just hope so.
Here's what I did for this problem: I partitioned the interval $[0,\,1]$ into two parts, $S$ and its complement $S^c$. Then $S$ is countable since $\alpha$ increases monotonically. Then If $s \in S$, then we have $$ \mu(\{s\}) = \alpha(s^+)-\alpha(s^-),$$ getting the sum in the statement which converges because $\alpha$ is of bounded variation.
Next, we know $\alpha$ is differentiable almost everywhere on $S^c$, so there is a subset $D$ of $[a,\,b]$ on which $\alpha$ is differentiable and $m(S^c - D) =0$. But I dont' know how to prove this.
And lastly, is there any way to compute the integral term in the Statement more explicity?