Let $g:\mathbb R\to\mathbb R$ be bounded, nondecreasing and right-continuous and $\mu_g$ denote the Lebesgue-Stieltjes measure on $\mathcal B([0,\infty))$ associated with $g$.
Moreover, let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$. We know that $g$ is differentiable $\lambda$-almost everywhere.
However, are we able to show that $$\int f\:\mu_g=\int fg'\:{\rm d}\lambda\tag1$$ for all $f\in\mathcal L^1(\mu_g)$? If not, what do we need to assume to make this claim true?
I know that the claim is true if $g$ is continuously differentiable (everywhere), but does it hold in general?
EDIT: It is known that if $h:\mathbb R\to\mathbb R$ is locally Lebesgue integrable, then $$r(t):=\int_{-\infty}^th(s)\:{\rm d}s\;\;\;\text{for }t\ge0$$ is differentiable Lebesgue almost everywhere with $r'=h$. So, maybe I'm missing something, but the claim should be true when we impose the stronger assumption $$g(b)-g(a)=\int_a^bh(t)\:{\rm d}t\;\;\;\text{for all }b\ge a\ge0\tag1$$ for some locally Lebesgue integrable $h:\mathbb R\to\mathbb R$. Since $\mathcal B(\mathbb R)$ is generated by $\{(a,b]:a\le b\}$, we should easily see that $$\mu_g=h\lambda$$ from which $(1)$ immediately follows (noting that, necessarily, $h=g'$ Lebesgue almost everywhere).
Best Answer
If $g$ is the Cantor function then $g'=0$ almost evrywhere so the right hand side is $0$ for any $f$. Take $f=\chi_{[0,1]}$ to get a contradiction.