How to Tell Whether a Function Is Integrable

Whether a function is “integrable” depends on which integral you mean. In a first real-analysis course, the usual question is whether a function has a proper Riemann integral on a closed, bounded interval $[a,b]$. For that setting, there is a precise test, along with several quicker sufficient tests.

The Riemann integrability test

A bounded function $f:[a,b]\to\mathbb{R}$ is Riemann integrable if and only if the set of points where $f$ is discontinuous has Lebesgue measure zero. This is often called the Lebesgue criterion for Riemann integrability.

The criterion has two parts:

1. Boundedness: There must be some finite $M$ such that $|f(x)|\le M$ for every $x\in[a,b]$.
2. Small discontinuity set: The discontinuities can be numerous, even dense, but together they must have measure zero.

In particular, every continuous function on $[a,b]$ is Riemann integrable. So is every bounded function with only finitely or countably many discontinuities. Countability is sufficient because a countable set has measure zero, but it is not necessary.

Fast sufficient tests

When you do not need a necessary-and-sufficient test, check whether one of these familiar conditions applies:

• Continuous: If $f$ is continuous on $[a,b]$, it is Riemann integrable.
• Piecewise continuous: A bounded function with finitely many jumps or other discontinuities is Riemann integrable.
• Monotone: Every bounded monotone function on a closed interval is Riemann integrable, even though it may have discontinuities.
• Continuous except on a measure-zero set: This is the full criterion once boundedness has been checked.

These are sufficient tests, not a complete list. If none applies, the function may still be integrable; examine its discontinuities or use upper and lower sums.

A worked example: continuous function

Consider $f(x)=x^2$ on $[0,1]$. It is continuous there, so it is Riemann integrable. To see the integral using equal-width Riemann sums, divide the interval into $n$ pieces of width $1/n$ and use right endpoints:

$$\begin{aligned} \sum_{k=1}^{n} f\left(\frac{k}{n}\right)\frac{1}{n} &=\sum_{k=1}^{n}\left(\frac{k}{n}\right)^2\frac{1}{n}\\ &=\frac{1}{n^3}\sum_{k=1}^{n} k^2\\ &=\frac{n(n+1)(2n+1)}{6n^3}\\ &\longrightarrow \frac{1}{3}. \end{aligned}$$

Thus $\int_0^1 x^2\,dx=1/3$. The continuity test establishes integrability; the Riemann sums calculate the value.

A bounded function that is not Riemann integrable

Define $f(x)=1$ for rational $x$ in $[0,1]$ and $f(x)=0$ for irrational $x$. The function is bounded, but it is discontinuous at every point: every interval contains both rational and irrational numbers, so values of $0$ and $1$ occur arbitrarily close to each point.

Its discontinuity set is all of $[0,1]$, which has positive measure. Therefore it is not Riemann integrable. In terms of Darboux sums, every subinterval contains both kinds of numbers, so every upper sum is $1$ and every lower sum is $0$; their gap never shrinks.

Compare this with a bounded function that is discontinuous only at a countable set: such a function is Riemann integrable. Having discontinuities everywhere is a different situation from having many discontinuities on a measure-zero set.

What if the function is unbounded?

An unbounded function on $[a,b]$ is not Riemann integrable in the proper sense described above. It may nevertheless have a convergent improper integral. For example, $f(x)=1/\sqrt{x}$ is unbounded near $0$, but

$$\int_0^1 \frac{1}{\sqrt{x}}\,dx =\lim_{\varepsilon\to0^+}\int_{\varepsilon}^1 x^{-1/2}\,dx =\lim_{\varepsilon\to0^+}2(1-\sqrt{\varepsilon})=2.$$

That is an improper integral, not a proper Riemann integral of a bounded function on the whole interval.

How to use upper and lower sums

For a partition $P$ of $[a,b]$, form the lower sum by taking the infimum of $f$ on each subinterval and the upper sum by taking its supremum. A bounded function is Riemann integrable exactly when partitions can make the difference between the upper and lower sums arbitrarily small.

This is useful when the discontinuity criterion is hard to apply directly. If you can bound the total contribution from intervals containing troublesome behavior so that it becomes as small as you like, the function is integrable. If the upper and lower sums remain separated by a fixed positive amount, it is not.

Common mistakes

• Ignoring the interval: A function can be integrable on one interval but not another, or have an improper integral at an endpoint.
• Treating any discontinuity as fatal: A bounded function with finitely many, countably many, or even some uncountable measure-zero sets of discontinuities can be Riemann integrable.
• Checking boundedness alone: Boundedness is necessary for a proper Riemann integral on a compact interval, but it is not sufficient, as the rational/irrational example shows.
• Mixing integral definitions: A function may fail to be properly Riemann integrable but still have an improper integral or a Lebesgue integral.

If you mean the Lebesgue integral

For a measurable function on a finite interval, being bounded is not required for Lebesgue integrability. The usual condition is absolute integrability: $\int_a^b |f(x)|\,dx<\infty$. For example, $1/\sqrt{x}$ is Lebesgue integrable on $[0,1]$ despite being unbounded. On a finite interval, every bounded measurable function is Lebesgue integrable.