Does the "right endpoint rule" yield the same definition of the Riemann integral?

Question

In an analysis class I'm teaching, I stated a different definition of the Riemann integral, which I thought was equivalent to the usual one. But now I'm not so sure.

Notation: let $f : [0,1] \to \mathbb{R}$ be a bounded function. A partition $\pi$ is a finite sequence $\pi = \{0 \le x_0 \le \dots \le x_n \le 1\}$. The mesh $|\pi|$ is defined as $\max_i |x_i - x_{i-1}|$. For a partition $\pi$, define:

• the upper sum $U_\pi(f) = \sum_{i=1}^n \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (x_i - x_{i-1})$

• the lower sum $L_\pi(f) = \sum_{i=1}^n \inf_{x \in [x_{i-1}, x_i]} f(x) \cdot (x_i - x_{i-1})$

• the right sum $R_\pi(f) = \sum_{i=1}^n f(x_i) (x_i - x_{i-1})$.

A standard definition of Riemann integral (see, e.g., Baby Rudin) goes as follows:

$f$ is Riemann integrable, with integral $I$, iff $\inf_\pi U_\pi(f) = \sup_\pi L_\pi(f) = I$.

The definition I gave, which here I'll call "right Riemann integrable", is this:

$f$ is right Riemann integrable, with integral $I$, iff for every $\epsilon > 0$ there exists $\delta > 0$ such that for every partition $\pi$ with $|\pi| < \delta$ we have $|R_\pi(f) - I| < \epsilon$. (In other words, $\lim_{|\pi|\to 0} R_\pi(f) = I$.)

The idea is to mimic the "right endpoint rule" definition given in introductory calculus classes.

Now it's well known that $f$ is Riemann integrable iff it is continuous almost everywhere, in which case the Riemann integral equals the Lebesgue integral.

I can prove that if $f$ is continuous almost everywhere, then it is right Riemann integrable, and the right Riemann integral equals the Lebesgue integral. Hence, every Riemann integrable function is right Riemann integrable, and in this case all integrals agree.

What I can't prove is the converse.

Is it true that every "right Riemann integrable" function is Riemann integrable?

I thought of doing something like the following: given a partition $\pi$, for each $i$ let $t_i \in [x_{i-1}, x_i]$ be a point at which the supremum $\sup_{x \in [x_{i-1}, x_i]} f(x)$ is almost attained. Then refine $\pi$ to a new partition $\pi'$ by throwing in all the $t_i$. I would like to say that $R_{\pi'}(f)$ is then close to $U_\pi(f)$. But if $t_i$ was very close to $x_{i-1}$ then the effect of throwing it in to $\pi'$ may not affect the right sum very much.

Answer 1

The book by Bartle-Sherbert employs something similar to your definition. In Theorem $7.2.1.$ he proves what he states as "Cauchy Criterion". An adaptation of the proof shows that it is equivalent to Theorem $6.6$ of Rudin. Since those are equivalences, you have that the definitions (of Rudin and Bartle-Shebert) are equivalent.

For easy reference to those not with the book, Theorem $6.6$ of Rudin is:

Theorem: $f \in \mathcal{R}(\alpha)$ on $[a,b]$ iff for every $\epsilon>0$ there exists a partition $P$ such that

$$U(P,f,\alpha)-L(P,f,\alpha) < \epsilon.$$

EDIT: As this is right now, this does not address your question. Note that the definition on Bartle-Sherbert takes in account an arbitrary "tagging". I'll leave this answer as a partial answer, since it may help somehow.

Answer 2

For bounded functions (as you assume here) it was probably first proved in D. C. Gillespie, "The Cauchy definition of a definite integral", Annals of Mathematics (2) 17 (1915-16), 61-63.