0.2Deficiencies of the Riemann Integral

Updated 2 Sep 2026Chapter (PDF)

This is not a section in Folland and content is drawn from a variety of places, such as Chapters 1-2 of MIRA.

We will discuss four problems with the Riemann integral.

Riemann integration does not handle unbounded functions.
Riemann integration does not handle functions with many discontinuities.
Riemann integration does not work with limits of sequences of functions.
The Riemann integral doesn't form a complete space when considering spaces of functions.

[0.2.1]Example(Riemann Integration Against Unbounded Functions)#

Consider the function $f: [0, 1] \to \bR$ defined by

f(x) = \begin{cases} 1/\sqrt{x} & 0 < x \le 1, \\ 0 & x = 0. \end{cases}

For any partition $P = x_0, \dots, x_n$ of $[0, 1]$ , note that $\sup_{[x_0, x_1]} f = \infty$ , and so $U(f, [a, b]) = \infty$ (thus $\int_0^1 f(x) \, \dd x$ is undefined). But note

\lim_{a \to 0^+} \int_a^1 f(x) \, \dd x = \lim_{a \to 0^+} [2\sqrt{x}]_{x = a}^1 = \lim_{a \to 0^+} (2 - 2\sqrt{a}) = 2,

so we'd desire the integral to equal $2$ . This function is well-behaved and has the classic notion of “area under the curve,” so it makes no sense for the integral to diverge.

[0.2.2]Example(Riemann Integration Against Discontinuous Functions)#

Consider the function $f: [0, 1] \to \bR$ defined by $f(x) = 1_\bQ$ . For any closed interval $[a, b] \subseteq [0, 1]$ , note $\inf_{[a, b]} f = 0$ and $\sup_{[a, b]} f = 1$ , since any interval in $\bR$ contains an irrational and rational number. Thus $L(f, [a, b]) = 0$ and $U(f, [a, b]) = 1$ for all partitions, meaning $f$ is not Riemann integrable. But note that $\bQ$ is countable and $[0, 1] \setminus \bQ$ is uncountable, so $f$ is zero far more often than it is $1$ . With this in mind, one might want to define the integral of $f$ to be $0$ , which is not possible with the Riemann integral.

[0.2.3]Example(Riemann Integration Against Sequences of Functions)#

Let $r_1, r_2, \dots$ be an injective sequence $\bZ_+ \to \bQ$ that contains all rational numbers in $[0, 1]$ . For $k \in \bZ_+$ , define $f_k: [0, 1] \to \bR$ by

f_k(x) = \begin{cases} 1 & \text{if } x \in \{r_1, \dots, r_k\}, \\ 0 & \text{otherwise.} \end{cases}

Recall that a function is Riemann integrable if its set of discontinuities is, at most, measure zero. The set of discontinuities for each $f_k$ is finite, and so this function is Riemann integrable. Indeed, $\int_0^1 f_k = 0$ for each $k \in \bZ_+$ . But note that $f_k \to f$ pointwise, with $f = 1_\bQ$ , which we have established is not Riemann integrable. Thus

\lim_{n \to \infty} \int_0^1 f_n(x) \, \dd x\ne \int_0^1 \lim_{n \to \infty} f_n(x) \, \dd x.

[0.2.4]Example(Space of Continuous Functions Together with Riemann Integral Norm)#

Consider the space of continuous functions $f: [0, 1] \to \bR$ , denoted $C([0, 1])$ , together with the norm

\norm{f - g}_1 = \int_0^1 |f(x) - g(x)| \, \dd x.

Ideally, we want this space to be complete with respect to the norm—we show this is not the case. Actually, we don't show it. I'm tired.

[0.2.5]Theorem(Interchanging Limit and Integral with Riemann Integration)#

Suppose $\{f_k\}$ is a sequence of Riemann integrable functions $f_k: [a, b] \to \bR$ such that $f_k \to f$ pointwise. If $f$ is Riemann integrable on $[a, b]$ , then

\int_a^b f(x) \, \dd x = \lim_{n \to \infty} \int_a^b f_k(x) \, \dd x.

This seems like it'd resolve problems the Riemann integral has with sequences of functions, but this is also not true. First of all, the proof for this theorem is unwieldy, and it immediately suggests something about our theory is not up to par. Considering the theorem further, we need that the limit function is Riemann integrable, which is undesirable. After all, we would want a sequence of Riemann integrable functions to converge pointwise to a Riemann integrable function, right? This mirrors the fact that a sequence of continuous functions need not converge pointwise to a continuous function. To fix this, we made a stronger notion of convergence (uniform convergence) that forced the limit function to be continuous.

It is true that we can repair this theorem by using uniform convergence. Indeed, if $f_n \to f$ uniformly, then

\lim_{n \to \infty} \int_a^b f_n = \int_a^b \lim_{n \to \infty} f_n.

Uniform convergence is quite a strong restriction to place on a sequence of functions and is undesirable. All the problems posed by trying to swap limit and integral, together with the aforementioned problems, motivate us to throw out Riemann integration and discover a new, better integration theory. This is the purpose of measure theory.