$$ \newcommand{\RR}{\mathbb{R}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\FF}{\mathbb{F}} \renewcommand{\epsilon}{\varepsilon} % ALTERNATE VERSIONS % \newcommand{\uppersum}[1]{{\textstyle\sum^+_{#1}}} % \newcommand{\lowersum}[1]{{\textstyle\sum^-_{#1}}} % \newcommand{\upperint}[1]{{\textstyle\smallint^+_{#1}}} % \newcommand{\lowerint}[1]{{\textstyle\smallint^-_{#1}}} % \newcommand{\rsum}[1]{{\textstyle\sum_{#1}}} \newcommand{\uppersum}[1]{U_{#1}} \newcommand{\lowersum}[1]{L_{#1}} \newcommand{\upperint}[1]{U_{#1}} \newcommand{\lowerint}[1]{L_{#1}} \newcommand{\rsum}[1]{{\textstyle\sum_{#1}}} % extra auxiliary and additional topic/proof \newcommand{\extopic}{\bigstar} \newcommand{\auxtopic}{\blacklozenge} \newcommand{\additional}{\oplus} \newcommand{\partitions}[1]{\mathcal{P}_{#1}} \newcommand{\sampleset}[1]{\mathcal{S}_{#1}} \newcommand{\erf}{\operatorname{erf}} $$

29 Construction

Highlights of this Chapter: We discuss the difficulty of constructing an integral, and then provide a definition (the Darboux integral) of a potential contender. We then prove that the Darboux integral satisfies our axioms of integration, and thus is truly an integral.

29.1 $\bigstar$ Failure of the ‘Calculus Integral’

The example below shows this is actually a difficult problem to answer: one might try to define the integral using a right endpoint Riemann sum (as one would in a calculus course): from this definition one can prove that all continuous functions are integrable, but then when one goes to try and verify the axioms, one finds this is actually not an integral at all!

Definition 29.1 (The “Calculus Integral”) Let $f$ be a function defined on the interval $[a,b]$, and $N$ a natural number. With $\Delta = (b-a)/N$ we define the (right endpoint) Riemann sum for $f$ with $N$ subintervals is \[\sum_{i=1}^n f(a+i\Delta)\Delta\]

Such a function $f$ is Calculus - integrable if the limit of its Riemann sums exists as the number of subintervals goes to infinity. In this case, the Calculus Integral is defined as the limiting value: \[\int_{[a,b]}^{\textrm{Calc}} f(x)dx=\lim_{N\to\infty}\sum_{i=1}^N f(a+i\Delta x)\Delta x\]

It turns out that while this definition seems unproblematic when applied to elementary functions seen in a calculus course, it has some rather surprising behavior in general: and taking it as our definition would destroy some of the familiar pillars of integration theory!

To find the trouble, we need to look away from the well behaved functions, and investigate the integrability of some monsters. Here we’ll look at the characteristic function of the rationals.

\[ \chi(x)=\begin{cases} 1 & x\in\QQ \\ 0 & x\not\in\QQ \end{cases} \]

Example 29.1 Let $\chi$ be the above function, equal to $1$ on the rationals and $0$ on the irrationals. Then $f$ is Calculus - Integrable on every interval of the form $[0,a]$ but \[ a\in\QQ\implies \int_{[0,a]}^{\textrm{Calc}} \chi dx=a \] \[ a\not\in\QQ\implies \int_{[0,a]}^\textrm{Calc} \chi dx=0 \]

In fact, its worse than this! As a natural extension of the above, one can show the following:

Exercise 29.1 The function $\chi$ is Calculus-Integrable on any closed interval in $\RR$, and the resulting value is:

The length of the interval, when both endpoints are rational.
Zero, when one endpoint is rational and the other irrational

This has a very important consequence to our theory: our proposed definition of the integral violates the subdivision rule.

Exercise 29.2 The subdivison rule \[ \int_{[a,b]}^{\textrm{Calc}} f=\int_{[a,c]}^{\textrm{Calc}} f+\int_{[c,b]}^{\textrm{Calc}} f \] is false for the integral as defined in ?def-calc-Riemann-Integral.
Hint: look at the interval $[0,2]$, and note $0<\sqrt{2}<2$.

Thus, the function defined by this construction does not satisfy the axioms of integration, and does not define an integral! Apparently, we have our work cut out for us.

29.2 The Darboux Integral

The failure of the “Calculus Class Integral” above can be traced back to the fact that it relies on specific partitions, and specific means of sampling those partitions. These specific choices cause the pathologies of the function $\chi_{\mathbb{Q}}$ to remain invisible to the limiting procedure. Below we propose a more careful limiting procedure that avoids making such specific choices: this procedure will notice the pathological behavior of functions like $\chi_{\mathbb{Q}}$ and rule them non-integrable; avoiding the problem we just encountered.

Definition 29.2 (Partitions) A partition of the interval $I=[a,b]$ is a finite ordered set $P=\{t_0,t_1,\ldots,t_n\}$ with $a=t_0<t_1<\ldots<t_{n-1}<t_N=b$.

$N$ is called the length of the partition
We write $P_i=[t_i,t_{i+1}]$ for the $i^{th}$ interval of $P$, and $|P_i|=(t_{i+1}-t_i)$ for its width.
The maxwidth of $P$ is the maximal width of the $P$’s intervals, $\mathrm{maxwidth}(P)=\max_{0\leq i< N}\{|P_i|\}$.
The set of all partitions on a fixed interval $I$ is denoted $\mathcal{P}_I$.\[\partitions{I}=\{P: P\textrm{ is a partition of } I\}\]

Definition 29.3 (Upper and Lower Sums) Let $f$ be a function, and $P$ a partition of the closed interval $I$. For each segment $P_i=[t_{i},t_{i+1}]$, we define \[m_i=\inf_{x\in P_{i}}\{f(x)\}\hspace{1cm}M_i=\sup_{x\in P_{i}}\{f(x)\}\]

We then define the upper sum $\uppersum{I}(f,P)$ and the lower sum $\lowersum{I}(f,P)$ as

\[\lowersum{I}(f,P)=\sum_{0\leq i < N} m_i |P_i|\] \[\uppersum{I}(f,P)=\sum_{0\leq i < N} M_i |P_i|\]s

Definition 29.4 (Upper and Lower Integrals) Let $f$ be a function on the closed interval $I$. Then we define the upper integral $\upperint{I}(f)$ and the lower integral $\lowerint{I}(f)$ as \[U(f)=\inf_{P\in\partitions{I}}\{\uppersum{I}(f,P)\}\] \[L(f)=\sup_{P\in\partitions{I}}\{\lowersum{I}(f,P)\}\]

Definition 29.5 (Darboux Integral) Let $f$ be a function on the closed interval $I$. Then $f$ is Darboux-Integrable on $I$ if $U(f)=L(f)$, and we define the integral to be this common value: \[\int_{[a,b]}f = U(f)=L(f)\]

Exercise 29.3 Prove the characteristic function of the rationals is not Darboux integrable on $[0,1]$.

29.3 Working with Partitions

The goal of this section is to prove the seemingly obvious fact $\lowerint{I}(f)\leq \upperint{I}(f)$. This takes more work than it seems at first because of the definitions of $\lowerint{I}(f)$ as a supremum and $\upperint{I}(f)$ as an infimum, but proves an invaluable tool in analyzing integrability.

Proof. Here we give the argument for lower sums, the analogous case for upper sums is asked in Exercise 29.4. Since $P\subset Q$ and both $P, Q$ are finite sets we know $Q$ contains finitely many more points than $P$. Here we will show that if $Q$ contains exactly one more point than $P$, that the claim holds; the general case follows by induction.

In this case we may write $Q = P\cup \{z\}$, where $z$ lies within the partition $P_k=[t_k,t_{k+1}]$. Thus, $Q_k=[t_k,c]$ for the left half after subdivision, and $Q_{k+1}=[c,t_{k+1}]$ for the right half. Outside of $P_k$, the two partitions are identical, so their difference is given only by the difference of their values on $P_k$: \[\lowersum{I}(f,Q)-\lowersum{I}(f,P)=\] \[\left(\inf_{x\in Q_k}\{f(x)\}\, |Q_k|+\inf_{x\in Q_{k+1}}\{f(x)\}\, |Q_{k+1}|\right)-\left(\inf_{x\in P_k}\{f(x)\}\, |P_k|\right)\]

Since both $Q_k$ and $Q_{k+1}$ are subsets of $P_k$, the infimum over each of them is at its smallest the infimum over the whole set. This implies

\[\begin{align*} &\inf_{x\in Q_k}\{f(x)\}\, |Q_k|+\inf_{x\in Q_{k+1}}\{f(x)\}\, |Q_{k+1}|\\ &\geq \inf_{x\in P_k}\{f(x)\} |Q_k|+\inf_{x\in P_k}\{f(x)\}|Q_{k+1}\\ &= \inf_{x\in P_k}\{f(x)\}\left(|Q_k|+|Q_{k+1}\right)\\ &= \inf_{x\in P_k}\{f(x)\}|P_k| \end{align*}\]

Thus, the first term in the difference above is bigger than the second, so the overall difference is positive. Thus $\lowersum{I}(f,Q)-\lowersum{I}(f,P)\geq 0$ and so as claimed, \[\lowersum{I}(f,Q)\geq \lowersum{I}(f,P)\]

Exercise 29.4 Following the structure above, prove that if $Q$ refines $P$, that \[\uppersum{I}(f,Q)\leq \uppersum{I}(f,P)\]

Proposition 29.2 (Upper/Lower Sum Inequality) Lower sums are always smaller than upper sums, independent of partition. That is, if $P,Q$ be two arbitrary partitions of a closed interval $I$, for any bounded function $f$, \[\lowersum{I}(f,P)\leq \uppersum{I}(f,Q)\]

Proof. Let $P$ and $Q$ be two arbitrary partitions of the interval $I$, and consider the partition $P\cup Q$. This contains both $P$ and $Q$ as subsets, so is a common refinement of both.

Using our previous work, this implies \[L(f,P)\leq L(f,P\cup Q)\hspace{1cm} U(f,P\cup Q)\leq U(f,Q)\]

We also know that for the partition $P\cup Q$ itself, \[L(f,P\cup Q)\leq U(f,P\cup Q)\]

Taken together these produce the the string of inequalities

\[L(f,P)\leq L(f,P\cup Q)\leq U(f,P\cup Q)\leq U(f,Q)\]

From which immediately follows that $L(f,P)\leq U(f,Q)$, as desired.

Proposition 29.3 (Upper/Lower Integral Inequality) Let $I$ be any closed interval and $f$ a bounded function on $I$. Then the lower integral is less than or equal to the upper integral, \[\lowerint{I}f\leq \upperint{I}f.\]

Proof. Recall that $U(f)$ is the infimum over all partitions of the upper sums.
Let $P$ be an arbitrary partition. By ?prp-upper-lower-on-different-partitions we know the upper sum with respect to any partition whatsoever is greater than or equal to $L(f,P)$, so $L(f,P)$ is a lower bound for the set of all upper sums.

Thus, the infimum of the upper sums - the greatest of all lower bounds - must be at greater or equal to this specific lower bound, \[L(f,P)\leq \inf_{Q\in\mathcal{P}}\{U(f,Q)\}=U(f)\]

But this holds for every partition $P$. That means this number $U(f)$ is actually an upper bound for the set of all $L(f,P)$. And so, it must be greater than or equal to the least upper bound $L(f)$: \[L(f)\leq U(f)\]

Corollary 29.1 To show a function $f$ is integrable, it suffices to show that $U(f)\leq L(f)$.

(To see this, recall in general that $\lowerint{I}(f)\leq \upperint{I}(f)$ from ?prp-lower-int-leq-upper-int. So, if $\upperint{I}f\leq\lowerint{I}f$ then in fact they are equal, which is the definition of $f$ being integrable.)

29.4 Integrability Criteria

Here we prove a very useful condition to test if a function is integrable, by finding sufficient partitions.

Theorem 29.1 Let $f$ be a bounded function on a closed interval $I$. Then $f$ is integrable if for every $\epsilon>0$ there exists a partition $P$ of $I$ such that \[U(f,P)-L(f,P)<\epsilon\]

Here we prove one direction of this theorem, namely that if such partitions exist for all $\epsilon>0$ then $f$ is integrable. We prove the converse below.

Proof. Let $\epsilon>0$, and assume there is a partition $P$ with \[\uppersum{I}(f,P)-\lowersum{I}(f,P)<\epsilon\] Then, recalling $\lowersum{I}(f,P)\leq \lowerint{I}(f)$ and $\upperint{I}(f)\leq \uppersum{I}(f,P)$ by definition, we chain these together with $\lowerint{I}(f)\leq \upperint{I}(f)$ to get

\[\lowersum{I}(f,P)\leq \lowerint{I}(f)\leq \upperint{I}(f)\leq \uppersum{I}(f,P)\]

Thus, the interval $[\lowerint{I}(f), \upperint{I}(f)]$ is contained within the interval $[\lowersum{I}(f,P),\uppersum{I}(f,P)]$ which has length $<\epsilon$. Thus its length must also be less than $\epsilon$:

\[0\leq \upperint{I}(f)-\lowerint{I}(f)\leq \epsilon\]

But $\epsilon$ was arbitrary! Thus the only possibility is that $\upperint{I}(f)-\lowerint{I}(f)=0$, and so the two are equal, meaning $f$ is integrable as claimed.

Now we prove the second direction of ?thm-epsilon-integrability: the proof is reminiscent of the triangle inequality, though without absolute values (as we know terms of the form $U-L$ are always nonnegative already)

Proof. Assume that $f$ is integrable, so $\lowerint{I}(f)=\upperint{I}(f)$. Since $\upperint{I}(f)$ is the greatest lower bound of all the upper sums, for any $\epsilon>0$, $\upperint{I}(f)+\frac{\epsilon}{2}$ is not a lower bound: that is, there must be some partition $P_1$ where \[\uppersum{I}(f,P_1)<\upperint{I}(f)+\frac{\epsilon}{2}\]

Similarly, since $\lowerint{I}(f)$ is the least upper bound of the lower sums, there must be some partition $P_2$ with \[\lowersum{I}(f,P_2)>\lowerint{I}(f)-\frac{\epsilon}{2}\]

Now, define $P=P_1\cup P_2$ to be the common refinement of these two partitions, and observe that

\[\begin{align*} \uppersum{I}(f,P)-\lowersum{I}(f,P) &\leq \uppersum{I}(f,P_1)-\lowersum{I}(f,P_2)\\ &< \left(\upperint{I}(f)+\frac{\epsilon}{2}\right)-\left(\lowerint{I}-\frac{\epsilon}{2}\right)\\ &= \upperint{I}(f)-\lowerint{I}(f)+\epsilon\\ &=\epsilon \end{align*}\] Where the last inequality uses $\lowerint{I}(f)=\upperint{I}(f)$. Thus, for our arbitrary $\epsilon$ we found a partition on which the upper and lower sums differ by less than that, as claimed.

And finally, we provide an even stronger theorem than $\epsilon$-integrability, that lets us prove a function is integrable and calculate the resulting value, by taking the limit of carefully chosen sequences of partitions. More precisely, we want to consider any sequence of partitions that’s getting finer and finer:

Definition 29.7 (Shrinking Partitions) A sequence $P_n\in\partitions{I}$ of partitions is said to be shrinking if the corresponding sequence of max-widths converges to $0$.

We often abbreviate the phrase $P_n$ is a shrinking sequence of partitions by $P_n\to 0$.

Theorem 29.2 (Integrability & Shrinking Partitions) Let $f$ be a function on the interval $I$, and assume that $P_n, P_n^\prime$ are two sequences of shrinking partitions such that \[\lim \lowersum{I}(f,P_n)= \lim \uppersum{I}(f,P_n^\prime)\] Then, $f$ is integrable on $I$ and $\int_I f$ is equal to this common value.

Proof.

Proof. Call this common limiting value $X$. As $\lowerint{I}f$ is defined as a supremum over all lower sums

\[\begin{align*} \lim \lowersum{I}(f,P_n) &\leq \sup_{\{n\in\NN\}}\{\lowersum{I}(f,P_n)\}\\ &\leq \sup_{P\in\partitions{I}}\{\lowersum{I}(f,P)\}\\ &=\lowerint{I}(f) \end{align*}\]

Similiarly, as $\upperint{I}(f)$ is the infimum over all upper sums, we have \[\lim \uppersum{I}(f,P_n^\prime)\geq \upperint{I}(f)\]

By ?prp-lower-int-leq-upper-int we know $\lowerint{I}(f)\leq \upperint{I}(f)$, which allows us to string these inequalities together:

\[\lim\lowersum{I}(f,P_n)\leq \lowerint{I}(f)\leq \upperint{I}(f)\leq \lim\uppersum{I}(f,P_n^\prime)\]

Under the assumption that these two limits are equal, all four quantities in this sequence must be equal, and in particular $\lowerint{I}(f)=\upperint{I}(f)$. Thus $f$ is integrable, and its value coincides with the limit of either of these sequences of shrinking partitions, as claimed.

29.5 Verification of Axioms

Proposition 29.4 (Integrability of Constants) Let $f(x)=k$ be a constant function, and $[a,b]$ an interval. Then $k$ is Darboux integrable on $[a,b]$ and \[\int_{[a,b]}k = k(b-a)\]

Proof. For any partition $P$, we have \[M_i=\sup_{x\in P_i}\{f(x)\}=k=\inf{x\in P_i}\{f(x)\}=m_i\] as $f$ is constant. Thus, \[U(f,P)=\sum_{P_i\in P} M_i|P_i|=k\sum_{P_i\in P}|P_i|=k(b-a)\] \[L(f,P)=\sum_{P_i\in P} m_i|P_i|=k\sum_{P_i\in P}|P_i|=k(b-a)\] The upper and lower sums are constant, independent of partition, and so their respective infima/suprema are also constant, equal to this same value. Thus $k$ is integrable, and the integral is also this common value \[\int_{[a,b]}k = k(b-a)\]

Proposition 29.5 (Integration and Inequalities) Let $f,g$ be Darboux integrable functions on $[a,b]$ and assume that $f(x)\leq g(x)$ for all $x\in[a,b]$. Then \[\int_{[a,b]}f\leq \int_{[a,b]}g\]

Proof. The constraint $f\leq g$ implies that on any partition $P$ we have \[L(f,P)\leq L(g,P)\] Or, equivalently $L(g,P)-L(f,P)\geq 0$. Taking the supremum over all $P$ of this set of nonnegative numbers yields a nonnegative number, so

\[\sup_{P\in\partitions{[a,b]}}\{L(g,P)-L(f,P)\}\geq 0\] \[L(g)-L(f)\geq 0\,\implies L(f)\leq L(g)\] But since we’ve assumed $f$ and $g$ are integrable we know that $L(f)=U(f)=\int_{a,b}f$ and $L(g)=U(g)=\int_{[a,b]}g$. Thus

\[\int_{[a,b]}f \leq \int_{[a,b]}g \]

Proposition 29.6 (Integration and Subdivision) Let $[a,b]$ be an interval and $c\in(a,b)$. Then a function $f$ defined on $[a,b]$ is Darboux-integrable on this interval if and only if it is Darboux integrable on both $[a,c]$ and $[c,b]$. Furthermore, when defined these three integrals satisfy the identity \[\int_{[a,b]}f = \int_{[a,c]}f+\int_{[c,b]}f\]

Proof. First, assume that $f$ is integrable on $[a,b]$. By ?thm-epsilon-integrability, this means for any $\epsilon>0$ there exists a partition $P$ where $U(f,P)-L(f,P)<\epsilon$. Now consider the refinement $P_c=P\cup \{c\}$. By the refinement lemma, \[L(f,P)\leq L(f,P_c)\leq U(f,P_c)\leq U(f,P)\] Thus $U(f,P_c)-L(f,P_c)<\epsilon$ as well. Next we take this partition and divide it into partitions of each subinterval $P_1=P_c\cup [a,c]$ and $P_2= P_c\cup [c,b]$. By simply re-grouping the finite sums, we see \[L(f,P_c)=L(f,P_1)+L(f,P_2)\hspace{1cm}U(f,P_c)=U(f,P_1)+U(f,P_2)\]

And, by the definitions of upper and lower sums, for each we know $U(f,P_i)-L(f,P_i)\geq 0$. All that remains to insure the integrability of $f$ on $[a,c]$ and $[c,b]$ is to show that these differences are individually less than $\epsilon$. But this is immediate, as for example,

\[\begin{align*} U(f,P_1)-L(f,P_1)&\leq U(f,P_1)-L(f,P_1) + (U(f,P_2)-L(f,P_2))\\ &= (U(f,P_1)+U(f,P_2))-(L(f,P_1)+L(f,P_2))\\ &=U(f,P_c)-L(f,P_c)\\ &\leq \epsilon \end{align*}\]

and the same argument applies to $U(f,P_2)-L(f,P_2)$.

Next we assume integrability on the two subintervals, and prove integrability on the whole interval.

Proof. Let $\epsilon>0$ and by our integrability assumptions choose partitions $P_1$ of $[a,c]$ and $P_2$ of $[c,b]$ such that \[U(f,P_i)-L(f,P_i)\leq \frac{\epsilon}{2}\hspace{1cm}i\in\{1,2\}\] Now, their union $P=P_1\cup P_2$ is a partition of $[a,b]$, and re-grouping the finite sums, we see \[L(f,P)=L(f,P_1)+L(f,P_2)\hspace{1cm}U(f,P)=U(f,P_1)+U(f,P_2)\]

Thus,

\[\begin{align*} U(f,P)-L(f,P) &= (U(f,P_1)+U(f,P_2))-(L(f,P_1)+L(f,P_2))\\ &= (U(f,P_1)-L(f,P_1))+ (U(f,P_2)-L(f,P_2))\\ &\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon \end{align*}\]

So, we see that integrability on $[a,b]$ is equivalent to integrability on $[a,c]$ and $[c,b]$. Finally, we need to show in the case where all three integrals are defined, the subdivision equality actually holds.

Proof. Let $P$ be any partition of the interval $[a,b]$ and define the usual suspects: \[P_c=P\cup\{c\}\hspace{1cm}P_1=P_c\cup [a,c]\hspace{1cm}P_2=P_c\cup[c,b]\] We need three pieces of data. First, the inequalities relating integrals to upper and lower sums \[L(f,P_1)\leq \int_{[a,c]}f\leq U(f,P_1)\hspace{0.75cm}L(f,P_2)\leq \int_{[c,b]}f\leq U(f,P_2)\] Second, the inequalities of refinements: \[L(f,P)\leq L(f,P_c)\leq U(f,P_c)\leq U(f,P)\] and third, the relationships between $P_1,P_2$ and $P_c$: \[L(f,P_c)=L(f,P_1)+L(f,P_2)\hspace{1cm}U(f,P_c)=U(f,P_1)+U(f,P_2)\]

Putting all of these together, we get both lower and upper estimates for the sum of the integrals over the subdivision:

\[L(f,P)\leq L(f,P_c)=L(f,P_1)+L(f,P_2)\leq \int_{[a,c]}f+\int_{[c,b]}f\] \[\int_{[a,c]}f+\int_{[c,b]}f\leq U(f,P_1)+U(f,P_2)= U(f,P_c)\leq U(f,P)\]

And concatenating these inequalities gives the overall bound, for any arbitrary partition $P$:

\[L(f,P)\leq \int_{[a,c]}f+\int_{[c,b]}f \leq U(f,P)\]

Thus, the sum of these integrals lies between the upper and lower sum of $f$ on $[a,b]$ for every partition. As $f$ is integrable, we know there is a single number with this property, and that number is by definition the integral. Thus \[\int_{[a,b]}f = \int_{[a,c]}f+\int_{[c,b]}f\]

Phew! We’ve successfully verified all three axioms for the Darboux integral. Taken together, these prove that our construction really is an integral!

Corollary 29.2 The equality of upper and lower sums satisfies the axioms of integration, and thus the Darboux Integral really does define an integral.

29.1 \(\bigstar\) Failure of the ‘Calculus Integral’

29.2 The Darboux Integral

29.3 Working with Partitions

29.4 Integrability Criteria

29.5 Verification of Axioms