The following is a question from a past measure theory qual. While it might've been asked before, I cannot find a form with a similar hint or proof, so I thought to ask it.
Suppose $F$ and $G$ are of bounded variation, $F$ is continuous and $G$ is right-continuous. Let $\mu_F, \mu_G$ be the Lebesgue-Stieltjes measures corresponding to $F,G$ respectively. Show the integration by parts formula:
$$ \int_{(a,b]} Fd\mu_G = F(b)G(b) - F(a)G(a) - \int_{(a,b]} Gd\mu_F $$
Hint: Reduce to the case that $\mu_F,\mu_G$ are positive measures (and $\mu_F$ atomless), and then use a theorem on product measures.
I was able to reduce to the case of positive measures by using the decomposition of BV functions as difference of increasing functions + Jordan decomposition. However, I have no idea what `theorem on product measures' to use here. Moreover, I'm not sure if I should be thinking about $\mu_F \times \mu_G$ on $\mathbb{R} \times \mathbb{R}$, or the measure $\mu_{FG}$ induced by the product $FG$. I think it is the latter, since $\int_{(a,b]} d\mu_{FG} = F(b)G(b) - F(a)G(a)$, so it suffices to show $\int_{(a,b]} d\mu_{FG} = \int_{(a,b]} Fd\mu_G + \int_{(a,b]}Gd\mu_F$, but I'm not at all sure how to show this.
I'm familiar with the proof that shows this result by `brute force', but I don't know how to show this with the aforementioned 'theorem on product measures'.
