### The dominated convergence theorem for the Riemann and the improper Riemann integral (Measure theory is a must – part II)

#### by Orr Shalit

(Hello students of Infi 2 – this post is for you).

In this post I will describe the dominated convergence theorem (DCT) for the Riemann and improper Riemann integrals. The previous post can serve as an introduction (a slanted one, beware) to this one. My goal is to convince that the important and useful convergence theorems in integration theory can (and therefore, needless to say, should) be taught in a first course on Riemannian integration.

The bounded convergence theorem for the Riemann integral is also known as *Arzela’s Theorem*, and this post does not contain anything new. In preparing this post I used as reference the short note “A truly elementary approach to the bounded convergence theorem”, J. W. Lewin, *The American Mathematical Monthly*. This post can be considered as a destreamlinization of that note. I think my presentation is even more “truly elementary”, since I avoid introducing inner measure. **Warning:** this post will really truly be at a very elementary level.

We will first prove the theorem on a bounded interval, and then we’ll prove it on an unbounded one.

#### 1. The DCT on a bounded interval

Let be a bounded and closed interval in the real line. In this section, when we say that a function is **integrable**, we mean that is integrable in the sense of the Riemann integral or in the equivalent sense of the Darboux integral;

**Theorem 1 (The Bounded Convergence Theorem): ***Let be a uniformly bounded sequence of Riemann integrable functions on converging pointwise to a Riemann integrable function . Then *

In other words, the theorem states that

**Remark:** In the setting of Riemann integration, we cannot drop the assumption that the limit function is Riemann integrable, since the pointwise limit of a sequence of Riemann integrable functions need not be Riemann integrable. More advanced integration theories (Lebesgue integration or Kurzweil-Henstock integration) have the nice feature that pointwise limits of functions within the class of integrable functions behave nicely.

**Proof:** It is convenient to first make a reduction to the case where all functions are non-negative, and .

**Lemma 1:** Let be a uniformly bounded sequence of non-negative converging pointwise to . Then

Suppose for the moment that we proved the lemma. If are as in the theorem, then we define . Then is a sequence of non-negative Riemann integrable functions converging pointwise to . Morevover, if for all , then for all , so is also a uniformly bounded sequence. By the lemma, we have

and the theorem follows. Thus, it suffices to prove the lemma.

**Proof of Lemma 1:** Let be the uniform bound of the sequence , and let be given. Our job is to show that there exists an integer such that for all ,

(*)

Let us assume for contradiction that there is no such . This means that (*) fails for infinitely many . We may as well discard all for those which satisfy** (*)**. Hence we assume for contradiction that** (*) **fails for all .

What could make (*) fail for a certain ? By one of the characterisations of the Riemann integral (a problem on the workshop sheet for those taking the course this semester), (*) fails for if and only if there exists a step function (and one may assume also ) such that .

Define , where is the step function from the previous paragraph. The set is a finite union of intervals, and the sum of the lengths of these intervals must be strictly bigger than , because

.

We may shrink the sets ever so slightly to obtain sets that are a union of finitely many **closed** intervals, and such that . Here and below, for a set that is a finite union of intervals, we denote by the sum of the lengths of the intervals.

We will now show that the boundedness below of the total lengths of contradicts the assumption that . Indeed, we shall prove that

**Claim: There exists such that for infinitely many . **

Once we prove the claim we are done, since from the claim it follows that for infinitely many , contradicting convergence to zero of the sequence at the point .

The claim is very easy to prove using the methods of measure theory. To prove it using the tools of a first year student is a little trickier. To prove the claim, it is convenient to formulate it afresh as a lemma.

Let us say that a set is a **simple set** if it is the disjoint union of finitely many closed intervals. We denote by the sum of the lengths of the intervals comprising the finite set, and we call it the **length** of . There are a few easy properties that the length function enjoys:

**Exercise:**

- If is simple, then is a finite union of intervals (either open or half open).
- If are simple sets and each pair intersects only at finitely many points, then .
- If are simple sets, then .
- If is simple, then .

Now we formulate the lemma from which the above claim immediately follows.

**Lemma 2:** Let be a sequence of simple sets in , and suppose that there exists such that for all . Then there exists a such that .

**Proof:**

**Step I:** We shall first prove that there exists a subsequence and a positive number such that

for all .

Suppose to the contrary that one cannot find such a subsequence and such a . Then , in particular, is not the first element of such a subsequence. This means that there are only finitely many such that . Removing these finitely many s from the sequence, we are left with a sequence (which we relabel as ) such that for all .

Now by our assumption there are only finitely many s such that for and all . We can remove these finitely many elements, and proceed inductively to eventually obtain a subsequence with the property

for all .

On the other hand, since is a subsequence of , we have for all . The last two inequalities combine to give a contradiction. Indeed, if we estimate the length the first elements of the subsequence, we obtain

,

while on the other hand

Continuing this way we obtain

and for sufficiently large this is greater than , a contradiction.

**Step II:** We will now construct a subsequence such that for all . Let and be the subsequence and positive number whose existence was proven above. Let us relabel it as , since we will need to refine it further. Put . We will find as follows.

Suppose that . We may refine the sequence further and assume that for all . Now, considering the sequence as a sequence of subsets of , we are in the same situation as we were in the beginning of Step I, so we may repeat what we did there. We therefore obtain a subsequence such that for all . We let . Obviously, , so in particular the intersection is not empty.

Now if is given as the union , then we may (after refining and perhaps renumbering the s) consider the sequence , as sequence of simple closed sets in all of which have length bounded below by some positive . We repeat what we did to find and so on. By the construction, we have indeed that for all .

**Step III:** We now find the required point that is in infinitely many of the sets in the original sequence . In fact, we will show that the subsequence constructed in the previous step has a point in its joint intersection . We have to show that . This follows from the fact that sequence of closed has the finite intersection property, together with compactness of .

I cannot resist giving the familiar bare hands proof for . Letting , we are required to show that . Since is a decreasing sequence of (simple) closed sets, this follows from Cantor’s intersection theorem, but I won’t stop now. For every , take a point . Since all of the sequence is contained in , it has a limit point . Since for all , . Thus , and the proof is (finally) complete.

#### 2. The DCT on an unbounded interval

We now prove the theorem on an unbounded interval. For definiteness, we take the interval to be . We shall say that a function is **integrable** on if it is Riemann integrable on every bounded interval and if it is integrable in the generalised sense on .

If is a sequence of function and is another function with the same domain of definition, we say that the sequence is **dominated** by if for all . It will become clear soon why it is called the **dominated** convergence theorem and not the **bounded** convergence theorem.

**Theorem 2 (the Dominated Convergence Theorem):** Let be a sequence of (generalised) integrable functions on that is dominated by an integrable function . If the sequence converges pointwise to an integrable function , then

**Remark:** One can weaken the assumption that is integrable to “ is Riemann integrable on every interval of the form “, but this is not a significant weakening, since from the assumptions and then converges absolutely by the comparison theorem (compare with the remark after Theorem 1).

**Proof:** Let , and let be such that . It follows that and for all . Since is bounded on , the sequence is uniformly bounded there. By Theorem 1, there is an such that for all ,

.

Now for all , we have

.

That completes the proof.

**Exercise:** State and prove a version of the dominated convergence theorem for a generalized Riemann integrable (i.e., not necessarily bounded) functions on a bounded interval.

please explain this part in more detail ; We may refine the sequence further and assume that for all .

By Step I all the elements of the sequence intersect with length bigger than . It follows that at least one of the intervals must be intersected by infinitely many s with length of intersection no less than . Without loss of generality, this interval can be taken to be , and putting gives what we claim (the refined sequence consists of all the infinitely many that intersect with length at least ).

Thank you very much. I find it extraordinary that this theorem can be proved

without using lebesgue’s theorem.

I’m very interested in mathematics, especially lebesgue’s theorem

and probability based on measure theory.

I have been studying mathematics by myself for almost thirty years.

Your big help warmed my heart.

Thanks!

Does this theorem become much easier to prove if the function to integrate

is a continuous function ?

Are you asking about the limit function? Then the answer is no, since heart of the matter is to prove the result for limit function equal to zero. If you assume that all functions are continuous – I don’t know. I don’t see a simplification (I’ll be happy to hear if anybody does).

Thank you very much for your answer. I assumed that all functions were

continuous. I think that the special properties of real-valued functions

defined on a bounded closed interval may make the proof easier.

I would like to rectify the second sentence; I think that the special properties

of real valued continuous functions defined on a bounded closed interval

may make the proof easier.

A friend of mine pointed out that F. Hausdorff proved if a sequence of

continuous real valued functions defined on a bounded closed interval I

1.converges pointwise on I to a continuous real valued function on I

2. is uniformly bounded on I

then the the integration on I and taking the limit of the sequence can be interchanged.

He even cited Hausdorff`s thesis (“Beweis eines Satzes von Alzela”

Math.Zeit. 26(1927), pp.135-137). But I can’t read German.

Hi. Thanks for the post. I’m a little confused by the definition of simple sets. It seems to me that the property described in Exercise 1 is not true, since the complement of a simple set would be a disjoint union of open (in [a,b]) intervals. Could it be that a simple set is meant to be a set which is a disjoint union of finitely many intervals (open or closed)? Thanks again!

Thanks Joseph,

That was a mistake in the exercise, thanks (now fixed. That part of the exercise was not important for the rest of the proof). I *do* want simple sets to be unions of closed sets, otherwise the proof won’t work – having the finite intersection property will not imply non-empty intersection.

Orr