# Why does the monotone convergence theorem not apply on Riemann integrals?

I had just learned in measure theory class about the monotone convergence theorem in this version:

For every monotonically increasing sequence of functions $$f_n$$ from measurable space $$X$$ to $$[0, \infty]$$, $$\text{if}\quad \lim_{n\to \infty}f_n = f, \quad\text{then}\quad \lim_{n\to \infty}\int f_n \, \mathrm{d}\mu = \int f \,\mathrm{d}\mu .$$

I tried to find out why this theorem apply only for a Lebesgue integral, but I didn't find a counter example for Riemann integrals, so I would appreciate your help.

(I guess that $$f$$ might not be integrable in some cases, but I want a concrete example.)

Riemann integrable functions (on a compact interval) are also Lebesgue integrable and the two integrals coincide. So the theorem is surely valid for Riemann integrals also.

However the pointwise increasing limit of a sequence of Riemann integrable functions need not be Riemann integrable. Let $$(r_n)$$ be an ennumeration of the rationals in $$[0,1]$$, and let $$f_n$$ be as follows:

$$f_n(x) = \begin{cases} 1 & \text{if x \in \{ r_0, r_1, \dots, r_{n-1} \}} \\ 0 & \text{if x \in \{ r_n, r_{n+1}, \dots \}} \\ 0 & \text{if x is irrational} \\ \end{cases}$$

Then the limit function is nowhere continuous, hence not Riemann integrable.

• I think the problem is more the other way round. If you start with a sequence of Riemann integrable functions $(f_n)_{n\in\mathbb{N}}$ that converges point-wise to $f$. Is $f$ again Riemann integrable? And what about the szenario where you are not on a compact interval? Nov 29, 2019 at 9:53
• @NathanaelSkrepek Thank you. I have given a counter-example for the domain $[0,1]$. Nov 29, 2019 at 10:02

Here is a version of the Monotone convergence theorem for Riemann integrals that can be proved without referring to measure theory:

Theorem. Let $$\{f_n\}$$ be a nondecreasing sequence of Riemann integrable functions on $$[a,b]$$ converging pointwise to a Riemann integrable function $$f$$ on $$[a,b]$$. Then $$\lim_{n\to \infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx.$$

An elementary proof is given in this paper.

• thanks! so I guess that I'm looking for an concrete example of monotonic series of positive riemann integrable functions whose limit is not riemann integrable. Nov 29, 2019 at 9:59
• @AmitKeinan I have given such an example in my answer. Nov 29, 2019 at 10:02
• Look at this question.
– user140541
Nov 29, 2019 at 10:03

I spent about two weeks working on an elementary proof of this problem earlier this year.

There is an elementary proof of something slightly stronger: Let $$f_n$$ tend pointwise to $$0$$ on $$[0,1]$$, with $$f_n$$ Riemann integrable, and for all $$x$$ sup $$|f_n(x)| \leq 1$$. (This has the obvious generalisation to an arbitrary interval, with convergence pointwise to a Riemann integrable function and $$f_n$$ uniformly bounded).

I would be happy to write up my proof after my exams are finished this summer :)

This is a slightly harder version of a problem given on this problem set for second year undergraduates at the university of cambridge, although that also allows the assumption that the $$f_n$$ are continuous.