Proof of the Darboux Criterion

Turning the Criterion into a Proof

The Darboux Criterion states that a bounded function on a closed interval is Riemann integrable exactly when partitions can make the difference between the upper and lower sums arbitrarily small. The previous tutorial stated the criterion and described the gap through oscillations. Here we examine the proof carefully: in one direction, we combine two separate approximations by a common refinement; in the other, a small gap forces the Darboux upper and lower integrals to agree.

Throughout, let $a<b$, and let $f:[a,b]\to\mathbb{R}$ be bounded. For a partition $P$, write $L(f,P)$ and $U(f,P)$ for its lower and upper sums. Recall that the Darboux lower integral is the supremum of all lower sums, and the Darboux upper integral is the infimum of all upper sums:

\underline{\int_a^b} f=\sup_P L(f,P), \qquad \overline{\int_a^b} f=\inf_P U(f,P).

The result we will prove is the Darboux Criterion from the previous tutorial. We will also isolate a useful tolerance estimate: two separate one-sided approximations can be combined into a single small-gap partition. This is the key step in the direction where integrability is assumed.

A Common Refinement Combines Two Estimates

Proposition (Two-Sided Tolerance Estimate): Suppose $A\in\mathbb{R}$, $\alpha>0$, and $\beta>0$. If partitions $P_L$ and $P_U$ satisfy $$ L(f,P_L)>A-\alpha, \qquad U(f,P_U)<A+\beta, $$ then there is a partition $R$ such that $$ U(f,R)-L(f,R)<\alpha+\beta. $$

Proof. By the Common Refinement Theorem, $P_L$ and $P_U$ have a common refinement $R$. By the theorem on upper and lower sums under refinement, lower sums do not decrease and upper sums do not increase. Thus

L(f,R)\geq L(f,P_L)>A-\alpha, \qquad U(f,R)\leq U(f,P_U)<A+\beta.

Subtracting the lower bound from the upper bound gives

U(f,R)-L(f,R) <(A+\beta)-(A-\alpha) =\alpha+\beta.

This proves the estimate. The two approximations need not come from the same partition initially; refinement makes them compatible without worsening either bound. $\square$

The estimate is useful beyond the integrability proof. If a lower sum is close to a target value from below and an upper sum is close to it from above, a common refinement preserves both advantages. It is important that the estimates are one-sided in the appropriate directions: increasing a lower sum and decreasing an upper sum are exactly the changes refinement guarantees.

Theorem (Darboux Criterion): A bounded function $f:[a,b]\to\mathbb{R}$ is Riemann integrable if and only if, for every $\varepsilon>0$, there is a partition $P$ of $[a,b]$ such that $$ U(f,P)-L(f,P)<\varepsilon. $$

Proof. First suppose $f$ is Riemann integrable. By definition, its Darboux lower and upper integrals agree; call their common value $I$. Fix $\varepsilon>0$. Since $I$ is the supremum of the lower sums, there is a partition $P_L$ such that

L(f,P_L)>I-\frac{\varepsilon}{2}.

Indeed, if no such partition existed, every lower sum would be at most $I-\varepsilon/2$, contradicting that their supremum is $I$. Since $I$ is also the infimum of the upper sums, there is a partition $P_U$ such that

U(f,P_U)<I+\frac{\varepsilon}{2}.

Apply the Two-Sided Tolerance Estimate with $A=I$ and $\alpha=\beta=\varepsilon/2$. It gives a partition $R$ for which $U(f,R)-L(f,R)<\varepsilon$. This proves the required condition.

Conversely, suppose that for every $\varepsilon>0$ there is a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$. By the definitions of the Darboux integrals, every lower sum is at most the lower integral, and every upper sum is at least the upper integral. The Darboux Lower Integral Does Not Exceed the Upper Integral theorem also gives that the difference of those integrals is nonnegative. Therefore, for each such partition,

0\leq \overline{\int_a^b}f-\underline{\int_a^b}f \leq U(f,P)-L(f,P) <\varepsilon.

The difference on the left is a fixed nonnegative real number, independent of $\varepsilon$. A nonnegative real number smaller than every positive $\varepsilon$ must be zero. Hence the Darboux upper and lower integrals agree, which is the definition of Riemann integrability. This proves the converse and the theorem. $\square$

A Quantitative Consequence

Proposition (Integral Lies Between the Partition Sums): If $f$ is Riemann integrable with integral $I$, then for every partition $P$, $$ L(f,P)\leq I\leq U(f,P). $$ Consequently, if $U(f,P)-L(f,P)<\varepsilon$, then both $I-L(f,P)<\varepsilon$ and $U(f,P)-I<\varepsilon$.

Proof. Since the lower integral is the supremum of lower sums, $L(f,P)\leq\underline{\int_a^b}f$. Since the upper integral is the infimum of upper sums, $\overline{\int_a^b}f\leq U(f,P)$. Integrability makes both integrals equal to $I$, proving the two-sided bound. Subtracting $L(f,P)$ from $I$, and then subtracting $I$ from $U(f,P)$, shows that each difference is nonnegative and at most $U(f,P)-L(f,P)$. Thus each is less than $\varepsilon$ whenever the gap is less than $\varepsilon$. $\square$

This consequence clarifies what a successful partition certifies: not only are the upper and lower sums close to each other, but each brackets the integral within the size of their gap. The certificate does not identify the exact integral by itself, but it controls the error of either sum.

Worked Applications of the Proof Method

Worked Example: The Square Function on a Symmetric Interval

Consider $f(x)=x^2$ on $[-1,1]$. For a positive integer $n$, divide each of $[-1,0]$ and $[0,1]$ into $n$ equal intervals. Each has length $1/n$, and $f$ is monotone on each half: it decreases from $1$ to $0$ on $[-1,0]$ and increases from $0$ to $1$ on $[0,1]$.

On each half, the oscillation on a partition interval is the absolute change in $f$ between its endpoints. The sum of those changes on $[-1,0]$ is $1-0=1$, and on $[0,1]$ it is $1-0=1$. Using the Oscillation Formula for the Darboux Gap, and the common interval length $1/n$, we obtain

U(f,P_n)-L(f,P_n) =\frac{1}{n}(1+1) =\frac{2}{n}.

For any $\varepsilon>0$, choose a positive integer $n>2/\varepsilon$. Then $2/n<\varepsilon$, so the Darboux Criterion proves that $x^2$ is Riemann integrable on $[-1,1]$. The point $0$ is included in the partition because it separates the two intervals of monotonicity.

Worked Example: The Square-Root Function

Let $f(x)=\sqrt{x}$ on $[0,1]$, and divide the interval into $n$ equal pieces. Since $f$ is increasing, the infimum and supremum on $[(i-1)/n,i/n]$ are its endpoint values. Thus the oscillation there is $\sqrt{i/n}-\sqrt{(i-1)/n}$, and the interval length is $1/n$. The gap is

U(f,P_n)-L(f,P_n) =\frac{1}{n}\sum_{i=1}^{n} \left(\sqrt{\frac{i}{n}}-\sqrt{\frac{i-1}{n}}\right) =\frac{1}{n}(1-0) =\frac{1}{n}.

The middle sum telescopes: every term $\sqrt{j/n}$ for $1\leq j<n$ appears once with a plus sign and once with a minus sign. Given $\varepsilon>0$, choose a positive integer $n>1/\varepsilon$. Then the gap is less than $\varepsilon$, and the criterion proves integrability. This example uses no derivative estimate near zero; the partition-sum argument remains valid there.

Worked Example: Thomae’s Function

Define $t:[0,1]\to\mathbb{R}$ by setting $t(x)=0$ at irrational $x$, and $t(p/q)=1/q$ when $x=p/q$ is rational and written in lowest terms with $q\geq1$. In particular, $t(0)=t(1)=1$. The function is bounded between $0$ and $1$. We construct partitions with small gap while taking care that high values at partition endpoints still count in the suprema.

Fix $\varepsilon>0$. Choose a positive integer $N$ so large that $1/(N+1)<\varepsilon/2$. Let $S_N$ be the set of rationals in $[0,1]$ whose reduced denominator is at most $N$. This is a finite set: for each denominator $q\leq N$, there are only finitely many numerators $p$ with $0\leq p\leq q$. Choose a relative neighborhood of each point of $S_N$, with the sum of the lengths of these finitely many neighborhoods less than $\varepsilon/2$. Include every neighborhood endpoint, as well as $0$ and $1$, among the partition points.

Call a partition interval exceptional if it meets $S_N$. Because the neighborhood endpoints are partition points, every exceptional interval lies within the union of the chosen neighborhoods. The total length of exceptional intervals is therefore less than $\varepsilon/2$. On each one, the oscillation is at most $1$, since all values of $t$ lie in $[0,1]$. On every nonexceptional interval, no rational in it has reduced denominator at most $N$. Its supremum is consequently at most $1/(N+1)$, and its infimum is at least zero. The total length of all nonexceptional intervals is at most $1$. The oscillation formula now yields

U(t,P)-L(t,P) <\frac{\varepsilon}{2} +\frac{1}{N+1}\cdot 1 <\varepsilon.

Thus the Darboux Criterion proves that Thomae’s function is Riemann integrable on $[0,1]$. The neighborhoods are essential: a selected rational such as $1/2$ has value $1/2$, and a partition interval with that point as an endpoint still includes its value. Making the intervals that meet the finitely many high-value rationals have small total length controls their contribution rather than incorrectly ignoring their endpoint values.

Reading the Proof Without Losing a Hypothesis

The two directions of the criterion use different properties of the Darboux integrals. When integrability is known, equality of the integrals gives one lower-sum approximation and one upper-sum approximation to the same value. The common refinement combines them. When small gaps are known, the definitions give an upper bound on the difference between the integrals, and the ability to make that bound arbitrarily small forces equality.

A common error is to use a single lower-sum approximation and a single upper-sum approximation without putting them on a common partition. Their difference does not, by itself, give the gap of either partition. The Common Refinement Theorem and refinement monotonicity are what justify the combination. Another error is to claim that a small gap for one partition proves integrability: the condition must hold for every positive tolerance, with a potentially different partition for each tolerance.

In applications, the Oscillation Formula for the Darboux Gap translates the partition problem into an estimate of oscillation times interval length. The square-root example controls this sum through telescoping changes; the Thomae example separates a small total length of exceptional intervals from uniformly small oscillations elsewhere. These are distinct ways to satisfy the same criterion.

Check Your Understanding

Use the proof and examples above to answer the following questions.

Why are two partitions needed in the integrable-to-small-gap direction of the proof?
What refinement properties ensure that a common refinement preserves the useful lower- and upper-sum estimates?
How does a bound on the Darboux gap control the difference between the upper and lower integrals?
Why does the square-root example have a telescoping sum of oscillations?
In the Thomae example, why must intervals meeting rationals of reduced denominator at most $N$ be treated separately, including when those rationals are partition points?

Proof of the Darboux Criterion

What You'll Learn

Turning the Criterion into a Proof

A Common Refinement Combines Two Estimates