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

(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 $[a,b]$ be a bounded and closed interval in the real line. In this section, when we say that a function $f$ is integrable, we mean that $f$ 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 $f_n$ be a uniformly bounded sequence of Riemann integrable functions on $[a,b]$ converging pointwise to a Riemann integrable function $f$. Then

$\lim_{n\rightarrow \infty} \int_a^b f_n(t) dt = \int_a^b f(t) dt .$

In other words, the theorem states that $\lim_{n\rightarrow \infty} \int_a^b f_n(t) dt = \int_a^b \lim_{n\rightarrow \infty }f_n(t) dt .$

Remark: In the setting of Riemann integration, we cannot drop the assumption that the limit function $f$ 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 $f_n$ are non-negative, and $f = 0$.

Lemma 1: Let $g_n \geq 0$ be a uniformly bounded sequence of non-negative converging pointwise to $0$. Then

$\lim_{n \rightarrow \infty} \int_a^b g_n(t) dt = 0.$

Suppose for the moment that we proved the lemma. If $f_n$ are as in the theorem, then we define $g_n = |f-f_n|$. Then $g_n$ is a sequence of non-negative Riemann integrable functions converging pointwise to $0$. Morevover, if $|f_n| \leq M$ for all $n$, then $g_n \leq 2M$ for all $n$, so $g_n$ is also a uniformly bounded sequence. By the lemma, we have

$|\int_a^b f_n (t) dt - \int_a^b f(t) dt| \leq \int_a^b |f_n (t)- f(t)|dt \rightarrow 0,$

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

Proof of Lemma 1: Let $M$ be the uniform bound of the sequence $g_n$, and let $\epsilon > 0$ be given. Our job is to show that there exists an integer $N$ such that for all $n \geq N$,

(*) $\int_a^b g_n\leq \epsilon.$

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

What could make (*) fail for a certain $n$? By one of the characterisations of the Riemann integral (a problem on the workshop sheet for those taking the course this semester), (*) fails for $n$ if and only if there exists a step function $s_n\leq g_n$ (and one may assume also $s_n \geq 0$) such that $\int_a^b s_n > \epsilon$.

Define $E_n = \{t \in [a,b] : s_n (t) \geq \epsilon(b-a)^{-1}/2\}$, where $s_n$ is the step function from the previous paragraph. The set $E_n$ is a finite union of intervals, and the sum of the lengths $L$ of these intervals must be strictly bigger than $\delta:= \epsilon/(2M)$, because

$\int_a^b s_n = \int_{E_n} s_n + \int_{[a,b]\setminus E_n} s_n \leq L M + \epsilon/2$.

We may shrink the sets $E_n$ ever so slightly to obtain sets $F_n \subseteq E_n$ that are a union of finitely many closed intervals, and such that $|F_n| > \delta$. Here and below, for a set $F$ that is a finite union of intervals, we denote by $|F|$ the sum of the lengths of the intervals.

We will now show that the boundedness below of the total lengths of $F_n$ contradicts the assumption that $g_n \rightarrow 0$. Indeed, we shall prove that

Claim: There exists $t \in [a,b]$ such that $t \in F_n$ for infinitely many $n$

Once we prove the claim we are done, since from the claim it follows that $g_n(t) \geq s_n(t) \geq \epsilon(b-a)^{-1}/2$ for infinitely many $n$, contradicting convergence to zero of the sequence $g_n$ at the point $t$.

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 $F \subset [a,b]$ is a simple set if it is the disjoint union of finitely many closed intervals. We denote by $|F|$ the sum of the lengths of the intervals comprising the finite set, and we call it the length of $F$. There are a few easy properties that the length function enjoys:

Exercise:

1. If $F \subseteq [a,b]$ is simple, then $[a,b] \setminus F$ is a finite union of intervals (either open or half open).
2. If $F_1, \ldots, F_n$ are simple sets and each pair intersects only at finitely many points, then $|\cup_{k=1}^n F_k| = \sum_{k=1}^n|F_k|$.
3. If $E,F$ are simple sets, then $|E\cup F| = |E| + |F| - |E \cap F|$.
4. If $F \subseteq [a,b]$ is simple, then $|F| \leq b-a$.

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

Lemma 2: Let $\{F_n\}$ be a sequence of simple sets in $[a,b]$, and suppose that there exists $\delta > 0$ such that $|F_n| > \delta$ for all $n$. Then there exists a $t \in [a,b]$ such that $card \{n : t \in F_n\} = \aleph_0$.

Proof:

Step I: We shall first prove that there exists a subsequence $\{F_{n_k}\}$ and a positive number $\delta_0$ such that

$|F_{n_1} \cap F_{n_k}| > \delta_0$ for all $k$.

Suppose to the contrary that one cannot find such a subsequence and such a $\delta_0 > 0$. Then $F_1$, in particular, is not the first element of such a subsequence. This means that there are only finitely many $n >1$ such that $|F_1 \cap F_n | \geq 2^{-2}$. Removing these finitely many $F_n$s from the sequence, we are left with a sequence (which we relabel as $\{F_n\}$) such that $|F_1 \cap F_n|< 2^{-2}$ for all $n$.

Now by our assumption there are only finitely many $n$s such that $|F_j \cap F_n| \geq 2^{-3}$ for $j \leq 2$ and all $n \geq 3$. We can remove these finitely many elements, and proceed inductively to eventually obtain a subsequence $\{\tilde{F}_n\} \subseteq \{F_n\}$ with the property

$|\tilde{F}_j \cap \tilde{F}_n|< 2^{-n}$  for all $j.

On the other hand, since $\{\tilde{F}_n\}$ is a subsequence of $\{F_n\}$, we have $|\tilde{F}_n|>\delta$ for all $n$. The last two inequalities combine to give a contradiction. Indeed, if we estimate the length the first $N$ elements of the subsequence, we obtain

$|\cup_{n=1}^N \tilde{F}_n| \leq b-a$,

while on the other hand

$|\cup_{n=1}^N \tilde{F}_n| = |\tilde{F}_1| + |\cup_{n=2}^N \tilde{F}_n| - |\cup_{n=2}^N \tilde{F}_n \cap \tilde{F}_1|$

$>$   $|\tilde{F}_1| + |\cup_{n=2}^N \tilde{F}_n| - \cup_{n=2}^N 2^{-n} > |\tilde{F}_1| + |\cup_{n=2}^N \tilde{F}_n| - 1/2$

Continuing this way we obtain

$|\cup_{n=1}^N \tilde{F}_n| > \sum_{n=1}^N |\tilde{F}_n| - \sum_{k=1}^{N-1} 2^{-k} > N \delta - 1$

and for sufficiently large $N$ this is greater than $b-a$, a contradiction.

Step II:  We will now construct a subsequence $\{F'_n\} \subseteq \{F_n\}$ such that $\cap_{k=1}^n F'_k \neq \emptyset$ for all $n$. Let $\{F_{n_k}\}$ and $\delta_0$ be the subsequence and positive number whose existence was proven above. Let us relabel it as $\{F_n\}$, since we will need to refine it further. Put $F'_1 = F_1$. We will find $F'_2, F'_3, \ldots$ as follows.

Suppose that $F_1 = \cup_{k=1}^N I_k$.  We may refine the sequence $\{F_n\}$ further and assume that $| F_n \cap I_1 | > \delta_1>0$ for all $n$. Now, considering the sequence $\{I_1 \cap F_n\}_{n=2}^\infty$ as a sequence of subsets of $I_1$, 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 $\{F_{n_k}\}_{k=1}^\infty$ such that $|I_1 \cap F_{n_1} \cap F_{n_k}| > \delta_2>0$ for all $k$. We let $F'_2 = F_{n_1}$. Obviously, $|F'_1 \cap F'_2| > \delta_2 > 0$, so in particular the intersection is not empty.

Now if $F'_2 \cap I_1$ is given as the union $\cup_{k=1}^M J_k$, then we may (after refining and perhaps renumbering the $J$s) consider the sequence $\{I_1 \cap J_1 \cap F_{n_k}\}$, as sequence of simple closed sets in $I_1 \cap J_1$ all of which have length bounded below by some positive $\delta_3$. We repeat what we did to find $F'_3, F'_4$ and so on. By the construction, we have indeed that $\cap_{k=1}^n F'_k \neq \emptyset$ for all $n$.

Step III: We now find the required point $t\in [a,b]$ that is in infinitely many of the sets in the original sequence $\{F_n\}$. In fact, we will show that the subsequence $\{F'_n\}$ constructed in the previous step has a point in its joint intersection $\cap F'_n$. We have to show that $\cap_n F'_n \neq \emptyset$. This follows from the fact that sequence of closed $\{F'_n\}$ has the finite intersection property, together with compactness of $[a,b]$.

I cannot resist giving the familiar bare hands proof for $\cap_n F'_n \neq \emptyset$. Letting $H_n = \cap_{k=1}^n F'_n$, we are required to show that $\cap H_n \neq \emptyset$. Since $H_n$ is a decreasing sequence of (simple) closed sets, this follows from Cantor’s intersection theorem, but I won’t stop now. For every $k$, take a point $t_k \in H_k$. Since all of the sequence is contained in $H_1$, it has a limit point $t \in H_1$. Since $\{t_n\}_{n=k}^\infty \subset H_k$ for all $k$, $t \in H_k$. Thus $t \in \cap H_n = \cap F'_n$, 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 $[0, \infty)$. We shall say that a function is integrable on $[0,\infty)$ if it is Riemann integrable on every bounded interval and if it is integrable in the generalised sense on $[0, \infty)$.

If $f_n$ is a sequence of function and $g$ is another function with the same domain of definition, we say that the sequence $f_n$ is dominated by $g$ if $|f_n(x) | \leq g(x)$ for all $x$. 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 $f_n$ be a sequence of (generalised) integrable functions on $[0, \infty)$ that is dominated by an integrable function $g$. If the sequence $f_n$ converges pointwise to an integrable function $f$, then

$\lim_{n\rightarrow \infty}\int_0^\infty f_n(t)dt = \int_0^\infty f(t) dt.$

Remark: One can weaken the assumption that $f$ is integrable to “$f$ is Riemann integrable on every interval of the form $[0,b]$“, but this is not a significant weakening, since from the assumptions $|f| \leq |g|$ and then $\int_0^\infty f(t) dt$ converges absolutely by the comparison theorem (compare with the remark after Theorem 1).

Proof: Let $\epsilon >0$, and let $L$ be such that $\int_L^\infty g < \epsilon$. It follows that $\int_L^\infty |f| < \epsilon$ and $\int_L^\infty |f_n| < \epsilon$ for all $n$. Since $g$ is bounded on $[0,L]$, the sequence $f_n$ is uniformly bounded there. By Theorem 1, there is an $N$ such that for all $n \geq N$,

$|\int_a^L f_n - \int_a^L f| < \epsilon$.

Now for all $n \geq N$, we have

$|\int_0^\infty f_n - \int_0^\infty f|$ $\leq |\int_0^L f_n - \int_0^L f|$ $+ |\int_L^\infty f_n - \int_L^\infty f|$ $< 3 \epsilon$.

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.