I'm struggling to find a solution for the following problem:
Let $f$ be an absolutely continuos function on [a,b], let $\mu$ be a bounded Borel measure on [a,b], and let $\Phi_\mu(t)=\mu([a,t))$ with $\Phi_\mu(a)=0$. Prove $$\int_{[a,b]}f(t)\mu(dt) = f(b)\Phi_\mu(b+)-\int_a^bf'(t)\Phi_\mu(t)dt$$, where $\Phi_\mu(b+)=\lim_{y\to b^+}\Phi_\mu(y)$
Obviously, this is some kind of integration by parts formula which per se is not hard to prove, but I have no idea how to use the general distribution function $\Phi_\mu$.
Any help is highly appreciated, thank you very much!
We have $$\Phi_{\mu}(t) = \int_{[a,t)} ~ \mu(ds) = \int_{[a,b]} 1_{[a,t)}(s) \mu(ds)$$ where $1_{[a,t)}(s) = 1$ if $s \in [a,t)$ and $0$ elsewhere. Hence, $$\int_a^b f'(t) \Phi_{\mu}(t) ~ dt = \int_a^b \int_{[a,b]} f'(t)1_{[a,t)}(s) \mu(ds) dt$$ Applying Fubini's theorem (the outer integral should be seen as a Lebesgue integral) yields $$= \int_{[a,b]} \int_a^b f'(t) 1_{[a,t)}(s) ~ dt ~ \mu(ds) = \int_{[a,b]} \int_a^b f'(t) 1_{(s,b]}(t) ~ dt ~ \mu(ds) = \int_{[a,b]} \int_s^b f'(t) ~ dt ~\mu(ds)$$ Here we used that $1_{[a,t)}(s) = 1$ iff $a \leq s < t \leq b$ which is equivalent to $1_{(s,b]}(t) = 1$. The inner integral can be computed using the fundamental theorem of calculus and we get $$= \int_{[a,b]} f(b) - f(s) ~ \mu(ds) = f(b) \Phi_{\mu}(b) - \int_a^b f(t) ~ \mu(dt)$$ which gives the desired formula.
However, I'm not quite sure why one would take the limit from above ($\Phi_{\mu}(b+)$) and not the limit from below.