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


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

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


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

where $M(g) = \sup_{u \in [a,b]} g(u)$, $M_k(\alpha) = \sup_{u \in [x_{k-1},x_k]} \alpha(u)$ 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

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

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 $U(P,\alpha,g) - L(P,\alpha,g) < \epsilon/M(g)$ and

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


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

  • $\begingroup$ Wow, this is a really well-reasoned and both coherently and logically organized answer! (Hopefully that doesn't sound weird, those are just properties of mathematical writing which I highly admire since I'm generally not able to achieve them.) I really appreciate you taking the time to write and post this, and I hope/expect that other people will appreciate it as well. The strength and clarity of the result as presented here definitely exceeds what my expectations were at the time of writing the question. $\endgroup$ – Chill2Macht Dec 12 '17 at 17:29
  • 1
    $\begingroup$ @Chill2Macht: You're very welcome. Thanks for those comments. That this holds under weak assumptions seems not to be readily found in books as you discovered. However, it can be quite useful. Here is an example of an application $\endgroup$ – RRL Dec 12 '17 at 18:47

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