$$$$

29  Construction

29.1 Definition

Here we give some preliminary definitions leading to the construction of an integral that we will then try to prove satisfies the axioms. The main difficulty is to put down a strict criterion that determines when a function is integrable, and when it is not.

The idea here - due to Darboux - is to try to overestimate and underestimate the true value of an integral by increasingly precise estimates. If the two estimates coincide as they get better and better we say the function is integrable. If they do not, we say it is not.

We build the estimates as bar-graph-like functions; similar to the familiar Riemann Sums from calculus.

Definition 29.1 (Partition) A partition of the interval $I=[a,b]$ is a finite ordered set $P=\{t_0,t_1,\dots ,t_n\}$ with $a=t_0<t_1<\dots <t_{n-1}<t_N=b$.

• $N$ 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, $\mathrm{maxwidth}(P)=\underset{0\le i<N}{max}\{|P_i|\}$.
• The set of all partitions on a fixed interval $I$ is denoted ${\mathcal{P}}_I$.$${\mathcal{P}}_I=\{P:P\text{ is a partition of }I\}$$

Definition 29.2 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 $$m_i=\underset{x\in P_i}{inf}\{f(x)\}{M}_i=\underset{x\in P_i}{sup}\{f(x)\}$$

We then define the upper sum $U_I(f,P)$ and the lower sum $L_I(f,P)$ as

$$L_I(f,P)=\sum _{0\le i<N}{m}_i|P_i|$$ $$U_I(f,P)=\sum _{0\le i<N}{M}_i|P_i|$$

Definition 29.3 Let $f$ be a function on the closed interval $I$. Then we define the upper integral $U_I(f)$ and the lower integral $L_I(f)$ as $$U(f)=\underset{P\in {\mathcal{P}}_I}{inf}\{U_I(f,P)\}$$ $$L(f)=\underset{P\in {\mathcal{P}}_I}{sup}\{L_I(f,P)\}$$

Definition 29.4 (The 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: $${\int }_{[a,b]}fdx=U(f)=L(f)$$

29.2 Partitions

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

Definition 29.5 A partition $Q$ is a refinement of a partition $P$ if $Q$ contains all the points of $P$ (that is, $P\subset Q$).

Proposition 29.1 (Refinement Lemma) If $Q$ is a refinement of the partition $P$ on a closed interval $I$, then for any bounded function $f$ the following inequalities hold $$L_I(f,P)\le L_I(f,Q)\le U_I(f,Q)\le U_I(f,P)$$

Proof. Here we give the argument for lower sums, the analogous case for upper sums is asked in Exercise 29.1. 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 $P_k$: $$L_I(f,Q)-L_I(f,P)=$$ $$(\underset{x\in Q_k}{inf}\{f(x)\}|Q_k|+\underset{x\in Q_{k+1}}{inf}\{f(x)\}|Q_{k+1}|)-(\underset{x\in P_k}{inf}\{f(x)\}|P_k|)$$

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

$$\begin{array}{rl}& \underset{x\in Q_k}{inf}\{f(x)\}|Q_k|+\underset{x\in Q_{k+1}}{inf}\{f(x)\}|Q_{k+1}|\\ & \ge \underset{x\in P_k}{inf}\{f(x)\}|Q_k|+\underset{x\in P_k}{inf}\{f(x)\}|Q_{k+1}\\ & =\underset{x\in P_k}{inf}\{f(x)\}(|Q_k|+|Q_{k+1})\\ & =\underset{x\in P_k}{inf}\{f(x)\}|P_k|\end{array}$$

Thus, the first term in the difference above is bigger than the second, so the overall difference is positive. Thus $L_I(f,Q)-L_I(f,P)\ge 0$ and so as claimed, $$L_I(f,Q)\ge L_I(f,P)$$

Exercise 29.1 Following the structure above, prove that if $Q$ refines $P$, that $$U_I(f,Q)\le U_I(f,P)$$

Proposition 29.2 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 $f$, $$L_I(f,P)\le U_I(f,Q)$$

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 $$L(f,P)\le L(f,P\cup Q)U(f,P\cup Q)\le U(f,Q)$$

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

Taken together these produce the the string of inequalities

$$L(f,P)\le L(f,P\cup Q)\le U(f,P\cup Q)\le U(f,Q)$$

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

Proposition 29.3 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, $$L_{I}f\le U_{I}f.$$

Proof. Recall that $U(f)$ is the infimum over all partitions of the upper sums.
Let $P$ be an arbitrary partition. By Proposition 29.2 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, $$L(f,P)\le \underset{Q\in \mathcal{P}}{inf}\{U(f,Q)\}=U(f)$$

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 $L(f)$: $$L(f)\le U(f)$$

Corollary 29.1 To show integrability it is enough to prove $U_{I}f\le L_{I}f$.

Proof. We know in general that $L_I(f)\le U_I(f)$ from Proposition 29.3. So, if $U_{I}f\le L_{I}f$ then in fact they are equal, which is the definition of $f$ being integrable.

29.3 Integrability Criteria

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

Theorem 29.1 (Darboux Integrability Criterion) Let $f$ be a bounded function on a closed interval $I$. Then $f$ is integrable if and only if for all $ϵ>0$ there exists a partition $P$ such that $$U_I(f,P)-L_I(f,P)<ϵ$$

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

Proof. Let $ϵ>0$, and assume there is a partition $P$ with $$U_I(f,P)-L_I(f,P)<ϵ$$ Then, recalling $L_I(f,P)\le L_I(f)$ and $U_I(f)\le U_I(f,P)$ by definition, we chain these together with $L_I(f)\le U_I(f)$ to get

$$L_I(f,P)\le L_I(f)\le U_I(f)\le U_I(f,P)$$

Thus, the interval $[L_I(f),U_I(f)]$ is contained within the interval $[L_I(f,P),U_I(f,P)]$ which has length $<ϵ$. Thus its length must also be less than $ϵ$:

$$0\le U_I(f)-L_I(f)\le ϵ$$

But $ϵ$ was arbitrary! Thus the only possibility is that $U_I(f)-L_I(f)=0$, and so the two are equal, meaning $f$ is integrable as claimed.

Now we prove the second direction of Theorem 29.1: 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 $L_I(f)=U_I(f)$. Since $U_I(f)$ is the greatest lower bound of all the upper sums, for any $ϵ>0$, $U_I(f)+\frac{ϵ}{2}$ is not a lower bound: that is, there must be some partition $P_1$ where $$U_I(f,P_1)<U_I(f)+\frac{ϵ}{2}$$

Similarly, since $L_I(f)$ is the least upper bound of the lower sums, there must be some partition $P_2$ with $$L_I(f,P_2)>L_I(f)-\frac{ϵ}{2}$$

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

$$\begin{array}{rl}{U}_I(f,P)-L_I(f,P)& \le U_I(f,P_1)-L_I(f,P_2)\\ & <(U_I(f)+\frac{ϵ}{2})-(L_I-\frac{ϵ}{2})\\ & =U_I(f)-L_I(f)+ϵ\\ & =ϵ\end{array}$$ Where the last inequality uses $L_I(f)=U_I(f)$. Thus, for our arbitrary $ϵ$ 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 $ϵ$-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 29.6 (Shrinking Paritions) A sequence $P_n\in {\mathcal{P}}_I$ 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 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 $$lim{L}_I(f,P_n)=lim{U}_I(f,P_n^{\prime })$$ Then, $f$ is integrable on $I$ and ${\int }_{I}fdx$ is equal to this common value.

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

$$\begin{array}{rl}lim{L}_I(f,P_n)& \le \underset{\{n\in \mathbb{N}\}}{sup}\{L_I(f,P_n)\}\\ & \le \underset{P\in {\mathcal{P}}_I}{sup}\{L_I(f,P)\}\\ & =L_I(f)\end{array}$$

Similiarly, as $U_I(f)$ is the infimum over all upper sums, we have $$lim{U}_I(f,P_n^{\prime })\ge U_I(f)$$

By Proposition 29.3 we know $L_I(f)\le U_I(f)$, which allows us to string these inequalities together:

$$lim{L}_I(f,P_n)\le L_I(f)\le U_I(f)\le lim{U}_I(f,P_n^{\prime })$$

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

29.4 Riemann Sums

There is an alternative (and equivalent) construction of this integral which predates the construction above. Originally due to Riemann, this alternative version can be more complicated to work with, but has certain advantages: it makes a clear path from the general theory to numerical integration, and occasionally allows alternative, more algebraic proofs of various integral identities, avoiding discussions of suprema and infima.

We can compute an integral via Riemann sums, when the integral exists. To write down the definition we need to talk of sample points for a partition.

Definition 29.7 (Sample Set) Let $P$ be a partition of $[a,b]$. Then an sample set for $P$ is a set $S=\{s_1,\dots ,s_n\}\subset [a,b]$ of points with $s_i\subset P_i$ for all $i$.

The set of all sample sets for a fixed partition $P$ is denoted ${\mathcal{S}}_P$. $${\mathcal{S}}_P=\{S\mid S\text{ is a sample set for }P\}$$

Given a partition and a set of sample points, we can define a Riemann sum

Definition 29.8 If $P$ is a partition of an interval $I$, $S$ is an evaluation set for $P$, and $f$ is a function defined on $I$, the *Riemann sum of $f$ with respect to $(P,S)$ is $$\sum _I(f,P,S):=\sum _{i=1}^{n}f(s_i)|P_i|$$

Definition 29.9 (The Riemann Integral) Let $f$ be a function defined on the closed interval $[a,b]$. Then $f$ is Riemann-integrable on $[a,b]$ if for every sequence of $P_n$ shrinking partitions, and for every choice of sample sets $S_n$ for these partitions, $lim\sum _I(f,P_n,S_n)$ exists, and is independent of the choice of sequences $P_n,S_n$. In this case, we write $${\int }_I^{\text{Riem.}}f=lim\sum _I(f,P_n,S_n)$$

Theorem 29.3 Let $f$ be a function on a closed interval $I$. Then $f$ is Darboux integrable on $I$ if and only if $f$ is Riemann-Integrable on $I$, and the two integrals agree $${\int }_{I}fdx={\int }_I^{\text{Riem.}}fdx$$

Proof (Riemann $⟹$ Darboux).

Proof (Darboux $⟹$ Riemann).

Corollary 29.2 If $f$ is integrable, then ${\int }_{[a,b]}fdx$ can by computed via Riemann Sum: choosing any sequence $P_n$ of shrinking partitions, and any sequence $S_n$ of sample points for each partition, $${\int }_{[a,b]}fdx=\underset{n}{lim}\sum _{[a,b]}(f,P_n,S_n)$$