Let's say we define: $$h(x) = \int_a^x f(t)dg(t),$$ then do we have for integrable functions $a$ that: $$\int_a^b a(u) dh(u) = \int_a^b a(u)f(u)dg(u) ?$$
I would like to know whether this holds for either the Riemann-Stieltjes or the Lebesgues-Stieltjes integral or any similar integral. Also for the sake of simplicity, feel free to assume that all relevant functions are as "nice" as you want, e.g. real-analytic.
A yes/no answer would suffice, as would references which either prove or disprove such a result.
Attempt: In "nice" cases, we hope that the behavior of the Riemann-Stieltjes sums will predict the behavior for the integrals, i.e. that the behavior will be respected/preserved by the appropriate limits. So let's write now instead: $$\int_a^b a(u) dh(u) \approx \sum_{i=0}^{n-1} a(x_i) (h(x_{i+1}) - h(x_i)) $$ Then by definition of $h$ we have that: $$h(x) \approx \sum_{j=0}^{m-1} f(t_j) (g(t_{j+1}) - g(t_j))$$ In particular for each $i$ we have that (setting $t_m = x_i, t_{m+1}=x_{i+1}$, etc.): $$h(x_{i+1}) - h(x_i) \approx \sum_{j=0}^{m} f(t_j) (g(t_{j+1}) - g(t_j)) - \sum_{j=0}^{m-1} f(t_j) (g(t_{j+1}) - g(t_j)) = f(x_i)(g(x_{i+1})-g(x_i))$$ so that substituting into the above: $$\int_a^b a(u) dh(u) \approx \sum_{i=0}^{n-1} a(x_i) (h(x_{i+1}) - h(x_i)) \approx \sum_{i=0}^{n-1}a(x_i)f(x_i)(g(x_{i+1})-g(x_i)) \approx \int_a^b a(u)f(u)dg(u). $$ Of course, the above "argument" is extremely sloppy and would require considerable effort to be made rigorous, assuming that is even possible.
But hopefully it suggests why I think the above result may be true -- I had hoped to find it or something similar on the Wikipedia page for the Riemann-Stieltjes integral, but it is not.
Also one might expect the identity to be true by sloppily "applying" the fundamental theorem of calculus ($h(x)``="\int_a^x f(t)g'(t)dt$ so $h'(u)``="f(u)g'(u)$), i.e. when $$\int_a^b a(u)dh(u) ``=" \int_a^b a(u) h'(u) du ``=" \int_a^b a(u) f(u) g'(u) du ``=" \int_a^b a(u) f(u) dg(u). $$