When is a bounded, real-valued function Riemann integrable? I ask this question for a few reasons. The first is because I assigned a couple exercises on the my students’ take-home final exam introducing ‘measure zero’ and the types of discontinuities of functions. The second is that I was reading the statement of Weyl equidistribution theorem, where the condition is Riemann integrable, as opposed to Lebesgue integrable. In fact, the theorem is false if you use the latter. The third and last is that I have known the answer for some time, but I have never worked through the proof in full detail. Well, I plan to change that with today’s post.
Theorem 1. (Lebesgue) A bounded function is Riemann integrable if and only if the set of discontinuities of is of measure zero.
It turns out that a bounded function is Riemann integrable if and only if it is continuous and its discontinuities form a set of measure zero. The French mathematician Henri Lebesgue proved this result over a century ago. It is important to not confuse this ‘Lebesgue’s theorem’ with other results about the Lebesgue integral, a more powerful integral based on measure theory. However, the limitations of the Riemann integral shown by Lebesgue motivate the study of his integral.
Indeed, we would think that a function zero almost everywhere (a.e.) would be Riemann integrable with integral zero. This is not the case. Consider the Dirichlet function
which is zero a.e. However, is not Riemann integrable on any compact interval . Given any partition , the upper Riemann sum
is while the lower Riemann sum
is by the density of the rationals in the reals.
Enough with the motivation, let’s get down to the details. To prove Lebesgue’s theorem, we will need to define the oscillation of a bounded function at a point , which quantifies how much varies in an arbitrarily small interval about . Precisely, the oscillation of at the point is
Note that the supremum is well-defined since is bounded. The oscillation of at a point is a useful notion, since it compactly encodes the continuity of a function.
Lemma 2. is continuous at a point if and only if .
which implies by the triangle inequality that
Letting , we see that . Since was arbitrary, we conclude that .
The other direction of the proof essentially follows from retracing our steps above, and I leave it as an exercise to the reader.
Lemma 3. Let be a bounded function. Then for any ,
Proof. Since is compact, it suffices to show that is a closed subset of , or is open in . Let , so that . Set . By definition of oscillation, there exists a constant such that implies that
By the triangle inequality,
So, for any , we can take sufficiently small so that and therefore
Let and be the supremum and infimum of , respectively, on . Denote the set of discontinuities of by , and note that
which can be represented as the countable union
First, suppose that is Riemann integrable. Then, for any given, there exists a partition
If , for some and , then
Hence, if , then
Multiplying both sides of the inequality by , we see that . The intervals , for , form an open cover of , which is equal in measure to . Hence,
which implies that . Since the countable union of null sets is again a null set, we conclude that .
Now suppose that . Since , we see that , for all . Hence, for be given, there exists a countable collection of open intervals such that
The preceding lemma implies that is compact, hence there exists a finite subcover of . If , then . So for such an , there exists such that for all , . The collection of intervals
forms an open cover of , which is compact. Hence, there exists a finite subcover
This collection of intervals together with the intervals form an open cover of . We want to use this open cover to obtain a partition of such that . In particular, we want the partition to have the property that any interval is contained entirely in an element of the open cover, for which we will need Lebesgue’s number lemma. We postpone the proof of the lemma until the end in favor of completing the proof of Theorem 1.
Lemma 4. (Lebesgue) If is a compact metric space and is an open cover of , then there exists a such that, for every subset with , there exists some such that .
Divide into two sets and , where
For points in , the oscillation can be ‘big’, but the length of the interval is very small. For points in , the oscillation is sufficiently ‘small’, and the total length of the intervals is at most .
Proof of Lebesgue’s Number Lemma. We may assume that , otherwise any works. Since is compact, there exist finitely many elements in such that . For each , define . Define a real-valued function on by
I claim that for all . Indeed, if , then we can take sufficiently small so that the open ball . Hence, and .
is continuous and is compact, so by the extreme value theorem, attains its minimum on . Suppose we are given a set such that . For any point , the ball contains the set , by definition of diameter. If denotes the index of a set satisfying , then
which implies that .