Let us define a partition $a=x_0<x_1<...<x_n=b$ of interval $[a,b]$ and let us define the Riemann-Stieltjes integral $\int_a^b fd\Phi$ of a bounded function $f:[a,b]\to\mathbb{C}$, or $f:[a,b]\to\mathbb{R}$, as the limit, as the maximum of the leght of the partition intervals approaches 0, which must be independent from the partition chosen, of the Riemann-Stieltjes sum$$\sum_{i=1}^n f(\xi_i)[\Phi(x_i)-\Phi(x_{i-1})]$$where $\xi_i\in[x_{i-1},x_i)$ and $\Phi:[a,b]\to\mathbb{R}$ is a function of bounded variation and continuous from the left. This is Kolmogorov and Fomin's definition in Элементы теории функций и функционального анализа (p. 362 here).

In the English translation Introductory Real Analysis the definition allows to chose $\xi_i$ at the right endpoint: $\xi_i=x_i$.

The left-continuity of $\Phi$ make me suppose that such defintions are the same, intuitively. Is that so, and how can it be proved?


1 Answer 1


It appears that the definitions of integral are equivalent for $f$, $\Phi$ real with $f$ bounded and $\Phi$ non-decreasing on $[a,b]$. That's probably good enough for your purposes. The proof comes down to showing that the set of discontinuities $E$ of $f$ is of $\Phi$-measure zero; in other words, for every $\epsilon > 0$, there exists countable number of relatively open subintervals $I_i$ of $[a,b)$ for which $E \subseteq \bigcup_{i}I_i$ with $\sum_i |\Phi(I_i)| < \epsilon$, where $|\Phi(I)|=\Phi(b')-\Phi(a')$ if the endpoints of $I$ are $a'$, $b'$. The proof that $f$ is Riemann-Stieltjes integrable with respect to $\Phi$ if the set of discontinuities of $f$ is of $\Phi$-measure zero is too much to duplicate for this post.

Preliminaries: Assume $f$, $\Phi$ are real and bounded, with $\Phi$ non-decreasing. Suppose $f$ is KR integrable with respect to $\Phi$. The variation of $f$ over an interval $I\subseteq[a,b)$ is defined to be $$ \omega(f,I) = \sup_{x\in I}f(x)-\inf_{x\in I}f(x). $$ The variation of $f$ at a point $x$ is defined to be $\omega(f,x)=\inf_{x\in I}\omega(f,I)$, where the infimum is taken over all relatively open subintervals of $[a,b)$. The function $f$ is continuous at $x$ iff $\omega(f,x)=0$. So the set $E$ of discontinuities of $f$ can be written as $$ E = \bigcup_{m=1}^{\infty}E_{m},\;\;\; E_{m}=\{ x \in [a,b) : \omega(f,x) \ge 1/m\}. $$ If $I$ is a relatively open subinterval of $[a,b)$ containing $x \in E_{m}$, then $\omega(f,I) \ge 1/m$.

Lemma: Suppose $f$ is KR integrable with respect to $\Phi$. Let $m$ be a positive integer. For every $\epsilon > 0$, there exists a finite number $\{ I_{j}\}_{j=1}^{n}$ of relatively open subintervals of $[a,b)$ with $$ E_{m} \subseteq \bigcup_{j=1}^{n}I_{j},\;\;\;\sum_{j=1}^{n}|\Phi(I_j)| < \epsilon. $$

Proof: Assuming $f$ is integrable with respect to $\Phi$, then, for every $\epsilon > 0$ there exists $\delta > 0$ such that $|K_{\mathcal{P}}(f,\Phi)-K_{\mathcal{Q}}(f,\Phi)| < \epsilon/(2m+1)$ whenever $\|\mathcal{P}\| < \delta$ and $\|\mathcal{Q}\| < \delta$. Let $\mathcal{P}$ and $\mathcal{Q}$ have the same subdivision points $a=t_{0} < t_{1} < t_{2} < \cdots < t_{n}=b$, but with different augmentation points $t_{k}^{\star}$ and $t_{k}^{\star\star}$. Then it follows that $$ \sum_{k=1}^{n}\omega(f,I_{k})|\Phi(I_{k})| \le \frac{\epsilon}{2m+1}, \;\;\; I_{k} = [t_{k-1},t_{k}). $$ Let $S_{m}$ be the set of $k$ for which $(t_{k-1},t_{k})\cap E_{m} \ne \emptyset$; include $1$ in $S_{m}$ if $a \in S_{m}$. Then, $$ \frac{1}{m}\sum_{k \in S_{m}}|\Phi(I_{k})| \le \sum_{k\in S_{m}}\omega(f,I_{k})|\Phi(I_{k})| \le \frac{\epsilon}{2m+1}, \\ \implies \sum_{k\in S_{m}}|\Phi(I_{k})| < \frac{\epsilon}{2}. $$ Every point of $E_{m}$ is in $\bigcup_{k \in S_{m}}(t_{k-1},t_{k})$ except for the points of $E_{m}$ which are interior partition points of $\mathcal{P}$ (note: include the relatively open interval $[a,t_{1})$ in the union if $1 \in S_{m}$).

To cover any partition points of $\mathcal{P}$ in $(a,b)$ which may not be in the above union, choose another partition $\mathcal{Q}$ with $\|\mathcal{Q}\| < \delta$ with the remaining partition points of $\mathcal{P}$ in the interiors of intervals of $\mathcal{Q}$. The same argument as above applies to $\mathcal{Q}$ and, so, after combining relatively open intervals of $\mathcal{P}$ containing $E_{m}$ with intervals of $\mathcal{Q}$, it follows that $E_{m}$ is fully contained in relatively open intervals with total $\Phi$-measure less than $\epsilon$, which completes the proof. $\;\;\Box$

  • $\begingroup$ Great proof. I like the care for details such as "include 1 in $S_m$ if $a\in S_m$" very much! Every bounded variation $\Phi$ is the difference of two non-decreasing functions and the complex case follows by decomposing the functions into real and imaginary parts... $\endgroup$ Dec 9, 2014 at 8:32
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    $\begingroup$ @DavideZena : Actually, $f$ could be unbounded on an interval where $\Phi$ is constant. Yes, it's just a technicality, but it's just easier to assume $f$ is bounded than to discuss such issues at length. $\endgroup$ Dec 9, 2014 at 8:53
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    $\begingroup$ In the case $f$ is unbounded, do you think the Riemann-Stieltjes integral may not be the same according to the two definitions? Thanks again! $\endgroup$ Dec 9, 2014 at 9:49
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    $\begingroup$ @DavideZena : For Riemann, you can deduce boundedness of $f$ because, regardless of partition, you can choose augmentation points so that the Riemann sum ranges over an infinite set of values. For RS, you can basically do the same thing, except for the complication of intervals where $\Phi$ is constant. So everyone just assumes bounded there, too, regardless of the definition of RS. $\endgroup$ Dec 9, 2014 at 14:45
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    $\begingroup$ @DavideZena : For Stieltjes integrals, you don't want integrability ruined because the function $f$ is unbounded on an interval where the integral should be $0$ because $\Phi$ is constant on that interval. Boundedness is an essential assumption for Stieltjes integrals, whereas it is deduced for Riemann integrals. $\endgroup$ Dec 9, 2014 at 15:02

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