$$$$

28  Properties

28.1 Integrability

With our minimal notion of integration, we now confront the question of what functions are actually integrable. We know it can’t be all (as we’ve seen the Characteristic function of the rationals is one non-example), but it is a very large class of functions - containing all the reasonable functions we wil need! We prove the most important result first, that all continuous functions are integrable, and then continue to show how the similar method generalizes to related cases.

Theorem 28.1 (Continuous functions are Integrable) Every continuous function on a closed interval is Darboux integrable.

Proof. Let $f$ be continuous on the interval $[a,b]$ and choose $\epsilon >0$. We will prove integrability by finding a partition $P$ such that $U(f,P)-L(f,P)<\epsilon$.

As $f$ is continuous it is bounded (by the extreme value theorem), so the upper and lower sums are defined for all partitions. It is also uniformly continuous (as $[a,b]$ is a closed interval), so we can find a $\delta$ such that $$|x-y|<\delta ⟹|f(x)-f(y)|<\frac{\epsilon }{b-a}$$

Now, choose a partition $P$ of $[a,b]$ where the width of each interval is less than $\delta$. Comparing upper and lower sums on this interval, $$U(f,P)-L(f,P)=\sum _{P_i\in P}{M}_i|P_i|-\sum _{P_i\in P}{m}_i|P_i|=\sum _{P_i\in P}[M_i-m_i]|P_i|$$

Since $|P_i|<\delta$, we know that for any $x,y\in P_i$ the values $f(x),f(y)$ differ by less than $\epsilon /(b-a)$. Thus the difference of between the infimum and supremum over this interval must be less than or equal to this bound:

$$M_i-m_i\le \frac{\epsilon }{b-a}$$

Using this to bound our sum, we see

$$U(f,P)-L(f,P)=\sum _{P_i\in P}[M_i-m_i]|P_i|\le \frac{\epsilon }{b-a}\sum _{P_i\in P}|P_i|$$ $$=\frac{\epsilon }{b-a}(b-a)=\epsilon$$

Thus, $f$ is integrable!

But the Darboux integral allows us to integrate even more things than the continuous functions. For example, it is quite straightforward to prove that all monotone functions are integrable (even those with many discontinuities!)

Theorem 28.2 (Monotone functions are Integrable) Every monotone bounded function on a closed interval is integrable.

Proof. Without loss of generality let $f$ be monotone increasing and bounded on the interval $[a,b]$ and choose $\epsilon >0$. We will prove integrability by finding a partition $P$ such that $U(f,P)-L(f,P)<\epsilon$.

Let $B=f(b)-f(a)$ be the difference between values of $f$ at the endpoints. If $B=0$ then $f$ is constant, and we already know constant functions are integrable so we are done.

Otherwise, let $P$ be an arbitrary evenly spaced partition of widths $\mathrm{\Delta }=\epsilon /B$, we consider the difference $U(f,P)-L(f,P)$:

$$U(f,P)-L(f,P)=\sum _{P_i\in P}{M}_i|P_i|-\sum _{P_i\in P}{m}_i|P_i|$$ $$=\sum _{P_i\in P}[M_i-m_i]|P_i|=\mathrm{\Delta }\sum _{P_i\in P}[M_i-m_i]$$

Since $f$ is increasing, its supremum on each interval occurs on the right, and its infimum on the left. That is, if $P_i=[t_{i-1},t_i]$ we have $$m_i=f(t_{i-1})M_i=f(t_i)$$ Plugging this into the above gives a telescoping sum! $$U(f,P)-L(f,P)=\mathrm{\Delta }\sum _{1\le i\le n}[f(t_i)-f(t_{i-1})]=\mathrm{\Delta }[f(t_n)-f(t_0)]$$ But $t_0=a$ and $t_n=b$ are the endpoints of our partition, and so this equals

$$=\mathrm{\Delta }[f(b)-f(a)]=\frac{\epsilon }{f(b)-f(a)}[f(b)-f(a)]=\epsilon$$

And, inductively its straightforward to show (via subdivision) that if the domain of a function can be partitioned into finitely many intervals on which it is integrable, than its integrable on the entire thing. Thus, for example piecewise continuous functions are Darboux Integrable. The precise statement and theorem is below.s

Definition 28.1 (Piecewise Integrable Functions) A function $f$ defined on a domain $I$ is piecewise integrable if $I$ is the disjoint of a finite sequence of intervals $I=I_1\cup I_2\cup \dots \cup I_n$, and $f$ restricted to each interval is integrable.

Proposition 28.1 (Piecewise Integrable $⟹$ Integrable) If $f$ is piecewise integrable, then it is integrable.

Proof. We begin with the case that $f$ is piecewise integrable on two subintervals, $[a,c]$ and $[c,b]$ of the interval $[a,b]$. Then the subdivision axiom immediately implies that $f$ is in fact integrable on the entire interval.

Now, assume for induction that all functions that are piecewise integrable on intervals with $\le n$ subdivisions are actually integrable, and let $f$ be a piecewise integrable function on a union of $n+1$ intervals $$[a,b]=I_1\cup I_2\cup \cdots \cup I_n\cup I_{n+1}$$

Set $J$ equal to the union of the first $n$, so that $[a,b]=J\cup I_{n+1}$. Then when restricted to $J$, the function $f$ is piecewise integrable on $n$ intervals, so its integrable by assumption. And so, $f$ is integrable on both $J$ and $I_{n+1}$, so its piecewise integrable with two intervals, and hence integrable as claimed.

Because all continuous functions and all monotone functions are integrable, we have the following useful corollary covering most functions usually seen in a calculus course.

Corollary 28.1 All piecewise continuous and piecewise monotone functions with finitely many pieces are integrable.

28.2 $★$ Linearity

We can continue using the definition of the Darboux integral to discover some familiar properties: proving that if a given function is integrable, so is any constant multiple, and if two functions are integrable, so is their sum. While straightforward, these proofs are a bit tedious as we must again dig into the definitions of the upper and lower sums. Happily, this section is optional to us as we will quickly have much easier proofs of these facts available to us for continuous functions, once we’ve proven the fundamental theorem of calculus.

Theorem 28.3 (Integrability of Constant Multiples) Let $f$ be an integrable function a closed interval $I$, and $c\in \mathbb{R}$. Then the function $cf$ is also integrable on $I$, and furthermore $${\int }_{I}cf=c{\int }_{I}f$$

We separate into cases depending on the sign of $c$. Below we complete $c\ge 0$, and leave $c<0$ as an exercise.

Proof ($c=0$). When $c=0$ the function $cf$ is identically the zero function. Thus by Proposition 26.4 $${\int }_{I}cf={\int }_{I}0=0|I|=0$$ This is equal to $c{\int }_{I}f=0{\int }_{I}f=0$, so we’ve proven $c{\int }_{I}f={\int }_{I}cf$ as required.

Proof ($c>0$). For $c>0$, note that on any interval $J$ we have $$\underset{x\in J}{inf}\{cf(x)\}=c\underset{x\in J}{inf}\{f(x)\}\underset{x\in J}{sup}\{cf(x)\}=c\underset{x\in J}{sup}\{f(x)\}$$ Thus for any partition $P$, $$L(cf,P)=\sum _i\underset{x\in P_i}{inf}\{cf(x)\}|P_i|=c\sum _i\underset{x\in P_i}{inf}f(x)|P_i|=cL(f,P)$$ $$U(cf,P)=\sum _i\underset{x\in P_i}{sup}\{cf(x)\}|P_i|=c\sum _i\underset{x\in P_i}{sup}f(x)|P_i|=cU(f,P)$$

Let $P_n$ be any sequence of shrinking partitions: since $f$ is integrable we know $limU(f,P_n)=limL(f,P_n)={\int }_{I}f$. Computing with the limit laws

$$limL(cf,P_n)=limcL(f,P_n)=climL(f,P_n)=c{\int }_{I}f$$ $$limU(cf,P_n)=limcU(f,P_n)=climU(f,P_n)=c{\int }_{I}f$$

Thus the upper and lower sums are equal in the limit, so $cf$ is integrable and its integral is equal to their common value $c{\int }_{I}f$.

Exercise 28.1 Prove the $c=-1$ case: if $f$ is integrable on $I$ then so is $-f$ and ${\int }_I(-f)=-{\int }_{I}f$. Hint: what does multiplying by $-1$ do to $m=inf$ and $M=sup$ on each partition? What does it do to the sums $U(f,P)$ and $L(f,P)$?

Combine this with the $c>0$ case above to prove the analogous result for any negative constant multiple.

Theorem 28.4 (Integrability of Sums) Let $f,g$ be integrable functions on a closed interval $I$. Then their sum $f+g$ is also integrable on $I$. Furthermore, its integral is the sum of the integrals of $f$ and $g$: $${\int }_I(f+g)={\int }_{I}f+{\int }_{I}g$$

Proof. The key inequality bounding sums of functions on an arbitrary interval $J$ is $$\underset{x\in J}{inf}\{f(x)\}+\underset{x\in J}{inf}\{g(x)\}\le \underset{x\in J}{inf}\{f(x)+g(x)\}\le \underset{x\in J}{sup}\{f(x)+g(x)\}\le \underset{x\in J}{sup}\{f(x)\}+\underset{x\in J}{sup}\{g(x)\}$$

Given an arbitrary partition $P$, summing over the subintervals $P_i$ yields

$$L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P)\le U(f,P)+U(g,P)$$

By assumption $f$ and $g$ are both integrable, so we may select a sequence $P_n$ of shrinking partitions such that $$limL(f,P_n)=limU(f,P_n)={\int }_{I}flimL(g,P_n)=limU(g,P_n)={\int }_{I}g$$

Taking the limit of the above inequalities along this sequence of partitions yields

$${\int }_{I}f+{\int }_{I}g\le limL(f+g,P)\le limU(f+g,P)\le {\int }_{I}f+{\int }_{I}g$$

Thus by the squeeze theorem these limits are equal; so $f+g$ is integrable, and its integral equals their common value ${\int }_{I}f+{\int }_{I}g$.

Each of these theorems does two things: it proves something about the space of integrable functions and also about how the integral behaves on this space. Below we rephrase the conclusion of these theorems in the terminology of linear algebra - a result so important it deserves the moniker of “Theorem” itself.

Theorem 28.5 (Linearity of the Riemann/Darboux Integral) For each interval $[a,b]\subset \mathbb{R}$, the set $\mathcal{I}([a,b])$ of Riemann integrable functions forms a Vector Subspace of the set of all functions $[a,b]\to \mathbb{R}$. On this subspace, the Riemann integral defines a linear map

$${\int }_{[a,b]}:\mathcal{I}([a,b])\to \mathbb{R}$$

28.3 $★$ Dominated Convergence for Integrals

Being able to move limits in and out of sums proved to be an incredibly useful skill in our work with functions defined as sums. Similarly, being able to move limits in and out of integrals proves to be a very useful property for reasoning with calculus. So we pause here to prove a version of the Dominated Convergence Theorem for our minimal integral, the Darboux Integral. This theorem is often called the Arzela Bounded Convergence Theorem, and was first proved by Arzela in 1885.

Theorem 28.6 (Dominated Convergence for the Darboux Integral) Let $\{f_n\}$ be a sequence of Darboux integrable functions on a closed interval $I$, and assume that the functions $f_n$ converge pointwise to a Riemann integrable function $f$. Then if there exists some $M$ where $|f_n(x)|<M$ for all $x\in I$, the order of integration and limit may be interchanged: $$lim{\int }_I{f}_n={\int }_{I}f$$

We prove this result in steps; doing a special case $f=0$ first, and then using the special case to argue the general case. Our proof follows the beautifuly short paper A Concise, Elementary Proof of Arzela’s Bounded Convergence Theorem

Proposition 28.2 Let $f_n$ be a sequence of integrable functions on $[0,1]$, with $f_n(x)\in [0,1]$ and $f_n(x)\to 0$ for all $x\in [0,1]$. Then the limit of their integrals also tends to zero: $$\underset{n}{lim}{\int }_{[0,1]}{f}_n=0$$

Proof. We prove the contrapositive, and show that if $lim{\int }_{[0,1]}{f}_n\ne 0$ the functions must not actually tend to zero, at least at some point. Since Let ${\mathcal{[}\mathcal{0}\mathcal{,}\mathcal{1}\mathcal{]}}_n={\int }_{[0,1]}{f}_n$ be the sequence of integrals. Since this is not tending to 0, we can pick a bound $2\epsilon$ (written this way for convenience), and for every $N$ find some $n>N$ where $|{\mathcal{I}}_n|>2\epsilon$. Potentially passing to this subsequence (and negating), we can assume without loss of generality we can simply assume ${\mathcal{I}}_n>2\epsilon$ for all $n$.

For each $n$, we can define a sequence $U_n$ of open subintervals of $I$ of length at least $\epsilon$, where $f_n$ is always greater than $\epsilon$ on $U_n$. To see this, recall our upper and lower estimates: since the integral is greater than $2\epsilon$, we can find a partition $P$ on which the lower sum is at least $2\epsilon$ (if we could not, then all lower sums would be $<2\epsilon$, and so the supremum over lower sums would be $\le 2\epsilon$. But this is the value of our integral!) Call a rectangle short if its height is less than $\epsilon$, and tall if its height is $\ge \epsilon$. The collection of all short rectangles have a total area less than $\epsilon$ (as their bases together are a subset of the unit interval, of length $1$). Since the total area of all the rectangles is $\ge 2\epsilon$, there must also be tall rectangles (to make up for the extra at least $\epsilon$ of area required). And, since $f_n(x)\le 1$ for all $x$, this $\epsilon$ of area requires at least $\epsilon$ of base. So take $U_n$ to be these (open) bases, then $f_n(x)>\epsilon$ for all $x\in U_n$ and $U_n$’s total length is at least $\epsilon$ as required.

Recall our overall goal is to find an $x$ where $f_n(x)\nrightarrow 0$. If there is an $x$ which lies in infinitely many of these sets $U_n$ we are done, because then for these infinitely many values of $n$ we know $f_n(x)>\epsilon$, so the whole sequence cant possibly be converging to zero! So, we’ve reduced our question about integration to a question about intervals. But we will make one more reduction - for each $N$ build the set $V_N={\cup }_{k\ge N}{U}_k$; the union of all the U’s past a given point. Since each of the sets $U_n$ contains intervals of total length at least $\epsilon$, the set $V_N$ also has total length at least $\epsilon$, but now the $V_N$ sets are nested: each $V_N$ contains $V_{N+1}$. So, seeking a point in infinitely many of the $U$s is the same as seeking a point in the intersection of the $V$s, and all we need to do is show the intersection of all the $V_N$ is nonempty. This is a quite intuitive statement, and indeed a straighforward proof:

Theorem: the intersection of a nested collection of intervals, each of total length $>\epsilon$ has total length $\ge \epsilon$. (though the proof goes a bit outside of the techniques we are focusing on here so we omit it, and refer the interested reader to the argument in A Concise, Elementary Proof of Arzela’s Bounded Convergence Theorem).

Having found a point in the intersection of $V_N$ gives us a point in infinitely many $U_n$, which means a point $x$ where $f_n(x)\nrightarrow 0$, finishing the proof of the contrapositive.

Exercise 28.2 Extend the above proof to apply to sequences of functions $f_n:I\to [0,1]$ where $I$ is any interval (not just unit length).

Now onto the general case.

Proof. Let $f_n$ be a sequence of Darboux integrable functions on an interval $I$, which converge pointwise to $f$. Consider the functions $f_n(x)-f(x)$: since $f_n$ and $f$ are bounded by $M$ in absolute value, their difference is bounded by $2M$, so set $$g_n(x)=\frac{1}{2M}(f_n(x)-f(x))$$ Then the $g_n$ are a sequence of functions with range $[0,1]$, which are integrable and converge pointwise to zero on $I$. Thus we can apply the special case above to conclude $$\underset{n\to \mathrm{\infty }}{lim}{\int }_I\frac{1}{2M}(f_n-f)=0$$

From here, we can apply the linearity of the integral for each fixed $n$ to conclude $$\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{2M}{\int }_I{f}_n-\frac{1}{2M}{\int }_{I}f)=0$$ Since the second integral (of $f$) is constant, we can apply limit laws to conclude the limit of ${\int }_I{f}_n$ exists, and distribute the limits: $$\frac{1}{2M}\underset{n}{lim}{\int }_I{f}_n-\frac{1}{2M}{\int }_{I}f=0$$

Multiplying by $2M$ to clear this constant factor and adding ${\int }_{I}f$ to both sides yields $$\underset{n}{lim}{\int }_I{f}_n={\int }_{I}f$$

as desired.

Just as we used dominated convergence for series to prove the continuity of power series, we can use dominated convergence for integrals to prove the integrability of power series. However, we will not pause to do so, as thanks to the fundamental theorem of calculus, we will soon have the technology for a much quicker proof.