Do Riemann-Stieltjes integrals "iterate"?

Question

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 $$\int_a^b a(u)dh(u) ``=" \int_a^b a(u) h'(u) du ``=" \int_a^b a(u) f(u) g'(u) du ``=" \int_a^b a(u) f(u) dg(u). $$

Answer 1

You are using $a$ to denote both the lower integration limit and the function. To avoid confusion let your function $a$ be written as $\alpha$.

This is true given only that $\alpha$ and $f$ are Riemann-Stieltjes integrable with respect to $g$ on $[a,b]$ and $g$ is increasing. This can be generalized if $g$ has bounded variation as well. Note that R-S integrability implies that $\alpha$ and $f$ are also bounded and that the product $\alpha f$ is R-S integrable with respect to $g$.

Take a partition $P = (x_0, x_1, \ldots, x_n)$ of $[a,b]$. Any corresponding Riemann-Stieltjes with tags $\xi_k \in [x_{k-1},x_k]$ can be written as

$$S(P,\alpha,h) = \sum_{k=1}^n \alpha(\xi_k)[h(x_k)-h(x_{k-1})] = \sum_{k=1}^n \alpha(\xi_k)\int_{x_{k-1}}^{x_k}f(u) \, dg(u)\\ = \sum_{k=1}^n \int_{x_{k-1}}^{x_k}\alpha(\xi_k)f(u) \, dg(u). $$

We also have

$$\int_{a}^{b}\alpha(u)f(u) \, dg(u) = \sum_{k=1}^n \int_{x_{k-1}}^{x_k}\alpha(u)f(u) \, dg(u).$$

Thus,

$$\tag{*}\left|S(P,\alpha,h) - \int_a^b \alpha(u)f(u) \, dg(u)\right| = \left|\sum_{k=1}^n \int_{x_{k-1}}^{x_k}[\alpha(\xi_k)-\alpha(u)]f(u) \, dg(u)\right| \\ \leqslant \sum_{k=1}^n \int_{x_{k-1}}^{x_k}|\alpha(\xi_k)-\alpha(u)||f(u)| \, dg(u) \\ \leqslant M(f) \sum_{k=1}^n \int_{x_{k-1}}^{x_k}(M_k(\alpha) - m_k(\alpha)) \, dg(u), $$

where $M(f) = \sup_{u \in [a,b]} |f(u)|$, $M_k(\alpha) = \sup_{u \in [x_{k-1},x_k]} \alpha(u)$ and $m_k(\alpha) = \inf_{u \in [x_{k-1},x_k]} \alpha(u).$

Note that the RHS of (*) can be written in terms of upper and lower Riemann-Stieltjes sums as

$$M(f)\sum_{k=1}^n \int_{x_{k-1}}^{x_k}(M_k(\alpha) - m_k(\alpha)) \, dg(u) = \sum_{k=1}^n \int_{x_{k-1}}^{x_k}(M_k(\alpha) - m_k(\alpha)) [g(x_k) - g(x_{k-1})] \\ = M(f)(U(P,\alpha,g) - L(P,\alpha,g)).$$

Since $\alpha$ is R-S integrable with respect to $\alpha$ it follows that for any $\epsilon >0$ there is a partition $P_\epsilon$ such that if $P$ is a refinement then $U(P,\alpha,g) - L(P,\alpha,g) < \epsilon/M(f)$ and

$$\left|S(P,\alpha,h) - \int_a^b \alpha(u)f(u) \, dg(u)\right|< \epsilon.$$

Therefore,

$$\int_a^b \alpha(u) \, dh(u)= \int_a^b \alpha(u)f(u) \, dg(u).$$

Answer 2

I stumbled over this question and present here a shift of framework which supplies a nice proof (in the frame work where $g$ is increasing and $f$ is non-negative such that $h$ is non-decreasing as well): First let me introduce some notation: To any non-decreasing function $g$. we can associate a locally finite measure (exactly its Lebesgue Stiltjes measure) and we denote it by $\mu_g$. Thus reformulated like this the question becomes. For $h(x)=\int_c^x f(y) d\mu_g(y)$ when does $$\int_c^x a(y) d\mu_h(y)=\int_c^x a(y)f(y) d\mu_g(y)$$ hold for all $\mu_h$-integrable $a$. The latter condition is equivalent to saying that $f$ is a density of $\mu_h$ w.r.t. $\mu_g$. So the question boils down to showing that $f=\frac{d\mu_h}{d\mu_g}$ under your definition of $h$. What finishes this proof now is the fact that we have (under assumed continuity of $h$ which hinges on $\mu_g$ having no atoms) that $$\mu_h((a,b])=h(b)-h(a)=\int_c^b f(y) d\mu_g(y)-\int_c^a f(y) d\mu_g(y)\\ =\int_c^b f(y) d\mu_g(y)+\int^c_a f(y) d\mu_g(y) =\int_a^b f(y) d\mu_g(y)$$ and this suffice to show that $f$ is the Radon-Nikodym derivative as $\{(a,b]|a\leq b\in\mathbb{R}\}$ is a generator of the Borell $\sigma$-algebra of $\mathbb{R}$.