Almost equivalent definitions of the Riemann–Stieltjes integral

Below, I will present two definitions of the Riemann–Stieltjes integral, the second of which is more general. My question concerns the relationship between these two definitions.

Definition 1: Let $f,g:[a,b] \to \mathbb{R}$. For a partition $P=\{x_0, x_1,x_2 \cdots x_{n-1},x_n\}$ of $[a,b]$, consider the sum $$S(P,f,g) \stackrel{\rm def}{=} \sum_{i=0}^{n-1} f(c_i) \left[g(x_{i+1}) - g(x_{i})\right]$$ where we have "sample points" $c_i \in [x_i, x_{i+1}]$.

$f$ is then said to be Riemann–Stieltjes integrable with respect to $g$ if there is a real number $L$ with the following property: for all $\epsilon>0$ there is a $\delta>0$ such that for any partition $P$ with $\text{sup}_{{0\leq i \leq n-1}}(x_{i+1} - x_i) < \delta$ and any sequence of points $\{c_i\}_{{0\leq i \leq n-1}, c_i \in [x_i, x_{i+1}]}$ we have

$$\left|S(P,f,g) - L\right| < \epsilon$$

Definition 2: We modify the above definition so that it is like this instead: for all $\epsilon>0$ there is partition $P_{\epsilon}$ such that any refinement $P' \supset P_{\epsilon}$ satisfies $$\left|S(P',f,g) - L\right| < \epsilon$$ independent of the sequence of points $\{c_i\}_{{0\leq i \leq n-1}, c_i \in [x_i, x_{i+1}]}$ we choose.

Remark: The first definition implies the second. Simply let $P_{\epsilon}$ be any partition with $\text{sup}_{{0\leq i \leq n-1}}(x_{i+1} - x_i) < \delta$. However, interestingly, the second definition does not imply the first. Take

$$g(x) = \begin{cases} 0 & x \in [0, \frac 12) \\ 1, & x \in [\frac 12, 1] \end{cases}$$

$$f(x) = \begin{cases} 0 & x \in [0, \frac 12] \\ 1, & x \in (\frac 12, 1] \end{cases}$$

as a counterexample. For this example, the integral exists and is equal to $0$ in the sense of the second definition by ensuring our chosen partition $P{_\epsilon}$ is such that $\frac 12 \in P_{\epsilon}$. This ensures $g(x_{i+1}) - g(x_i) = 0$ except in the interval $[x_k, \frac 12]$; however, this interval does not affect the sum since $f \equiv 0$ in $[x_k, \frac 12]$.

Conversely, for the first definition we needn't have $\frac 12 \in P$. $\frac 12$ may be in the interior of some subinterval $[x_i, x_{i+1}]$ (ie., $x_i < \frac 12 < x_{i+1}$). This would mean that $g(x_{i+1}) - g(x_i) = 1$, and depending on the "sample point" $c_i$ we choose in this subinterval, the sum may be $1$ or $0$. This can happen regardless of how fine the partition is, and hence the integral does not exist.

Problem:

Are there any regularity conditions we can impose on $g$ to ensure the equivalence of the above definitions? Strict monotonicity is a natural example. If that doesn't work, consider stronger conditions (e.g., $g$ is homeomorphism onto its image, or a $C^{1}$ diffeomorphism).

• just a side note: the $\delta$ such that $\sup_{0\le i \le n-1}(x_{i+1} - x_i) < \delta$ is also called the mesh of the partition. At first glance it seems that the second definition is equivalent to the first, in the same way that the different definitions of the Riemann integral are equivalent. Jul 16 '17 at 4:54
• @Masacroso Yeah I'm aware of that. I didn't explicitly use the word "mesh" in case other people did not know what it meant. And I think I showed they're not equivalent by the counterexample, unless you think it is incorrect. Jul 16 '17 at 4:59
• oh, I see... I was checking the wikipedia article. Both definitions refers to different Riemann-Stieltjes integrals, the first is the original definition and the second an integral introduced by Pollard named Generalized Riemann-Stieltjes integral. See here Jul 16 '17 at 5:09
• @Masacroso Yes I've read that. The Wikipedia page is actually what motivated this question. Jul 16 '17 at 5:12
• Good question +1. I don't think most analysis textbooks treat alternative definitions. Apostol gives both these definitions in exercise and also gives your counter-example. I don't think there is an "if and only if" condition for the equivalence of these two definitions. Jul 16 '17 at 17:37

Depending on how the Riemann-Stieltjes integral is defined (at least two ways that you mention) there are a variety of joint conditions on the integrand $f$ and integrator $g$ that guarantee existence. There is some but not complete overlap.

The Riemann-Stieltjes integral is less flexible than the Riemann integral. One impediment is that the continuity of $g$ comes into play. A function that is continuous is Riemann integrable and so too is one that is discontinuous only on a set of measure zero. Taking definition (2), if the integrator is increasing, then if $f$ is continuous, the Riemann-Stieltjes integral exists, but may not exist if $f$ is only continuous almost everywhere. That is because it is necessary that the integrand and integrator have no common points at which they are discontinuous.

Some basic relationships (of which I am aware) are:

Definition (1) holds if and only if definition (2) holds for Riemann integrals where $g(x) = x$.

Definition (1) implies definition (2) for Riemann-Stieltjes integrals when $f$ is bounded and $g$ is increasing.

Definition (2) implies definition (1) for Riemann-Stieltjes integrals when $g$ is increasing, and either $f$ or $g$ is continuous. You found a counterexample if the continuity requirement is relaxed.

To prove the third implication, first consider that $f$ is continuous and R-S integrable with respect to $g$ under definition (2). Then for any partition $P = (x_0,x_1, \ldots,x_n)$ and choice of tags we have

$$\left|S(P,f,g) - \int_a^bf \, dg\right| = \left|\sum_{j=1}^n f(\xi_j)[g(x_j) - g(x_{j-1})] - \sum_{j=1}^n \int_{x_{j-1}}^{x_j}f \, dg\right|$$

Since $f$ is continuous we can apply the integral mean value theorem to find points $\eta_j$ such that

$$\left|S(P,f,g) - \int_a^bf \, dg\right| = \left|\sum_{j=1}^n [f(\xi_j)-f(\eta_j)]\,[g(x_j) - g(x_{j-1})] \right| \\ \leqslant \sum_{j=1}^n |f(\xi_j)-f(\eta_j)|\,[g(x_j) - g(x_{j-1})].$$

By uniform continuity of $f$, for any $\epsilon >0$ there is a $\delta > 0$ such that if $\|P\| < \delta$ then $|f(\xi_j)-f(\eta_j)| < \epsilon/(g(b) - g(a))$ and

$$\left|S(P,f,g) - \int_a^bf \, dg\right| < \epsilon.$$

Proof of the implication assuming that the integrator $g$ is continuous, rather than $f$, is lengthier. In brief, we choose a partition $P' =(x_0,x_1,\ldots,x_n)$ such that the upper sum $U(P',f,g)$ and lower sum $L(P',f,g)$ are within $\epsilon/2$ of the integral. Using the uniform continuity of $g$, we find $\delta >0$ such that $|g(x) - g(y)| < \epsilon/(2nM)$ when $|x-y| < \delta$, where $M$ bounds $f$. Then a partition $P$ with $\|P\| < \delta$ is constructed through a tedious process such that

$$\int_a^b f \, dg - \epsilon < L(P',f,g) < L(P,f,g) \leqslant S(P,f,g) \leqslant U(P,f,g) < U(P',f,g) < \int_a^b f \, dg + \epsilon.$$

• That proof at the end is new for me (although it does resemble the proof that the definitions are equivalent for a Riemann integral) +1. Jul 16 '17 at 17:39

Asking that $f$ be continuous is sufficient. In fact, all you need is that $f$ be left continuous when $g$ is right-discontinuous, and vice versa. This is a rather easy exercise that can be done by considering the difference between the maximum and the minimum values attainable by choosing sample points on a given partition. As a corollary of this, $f$ must be continuous where $g$ is both left- and right-discontinuous.

Where $g$ is continuous, it can be shown fairly easily that $f$ is allowed to have finitely many discontinuities. In fact, as long as the set of discontinuities of $f$ is measure zero (in the sense of Lebesgue measure), the integral ought converge. This proof is identical to that on the Riemann integral.