The Darboux Criterion

From Agreement of Integrals to a Test Using Partitions

Riemann integrability was defined in the previous tutorial by equality of the Darboux lower and upper integrals. That definition describes the desired outcome, but it does not immediately tell us how to verify it from a particular partition. The Darboux Criterion supplies this practical test: a bounded function is integrable exactly when some partition makes the difference between its upper and lower sums as small as desired.

Throughout, let $a<b$, and let $f:[a,b]\to\mathbb{R}$ be bounded. A partition is written $P=\{x_0,x_1,\ldots,x_n\}$, where $a=x_0<x_1<\cdots<x_n=b$. On each subinterval $[x_{i-1},x_i]$, the lower sum uses the infimum of $f$, and the upper sum uses its supremum. We use the notation $L(f,P)$ and $U(f,P)$ from Upper and Lower Sums.

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 $$ U(f,P)-L(f,P)<\varepsilon. $$

The criterion concerns the existence of a suitable partition for each requested accuracy. It does not say that every partition has a small gap, or that one fixed partition works for every positive $\varepsilon$. The choices of partition may become finer or may place points strategically where the function varies most.

What Makes the Gap Between the Sums?

For a partition $P=\{x_0,\ldots,x_n\}$, define the oscillation of $f$ on its $i$th subinterval by $\omega_i=\sup_{[x_{i-1},x_i]}f-\inf_{[x_{i-1},x_i]}f$. Because $f$ is bounded, these quantities are finite and nonnegative. Each subinterval contributes its oscillation multiplied by its length to the difference between the upper and lower sums.

Proposition (Oscillation Formula for the Darboux Gap): For every partition $P=\{x_0,\ldots,x_n\}$ of $[a,b]$, $$ U(f,P)-L(f,P) =\sum_{i=1}^{n}\omega_i(x_i-x_{i-1}). $$

Proof. Write $M_i=\sup_{[x_{i-1},x_i]}f$ and $m_i=\inf_{[x_{i-1},x_i]}f$. By the definitions of the upper and lower sums,

U(f,P)=\sum_{i=1}^{n}M_i(x_i-x_{i-1}), \qquad L(f,P)=\sum_{i=1}^{n}m_i(x_i-x_{i-1}).

Subtract the second finite sum from the first, term by term. Since $\omega_i=M_i-m_i$, the result is

U(f,P)-L(f,P) =\sum_{i=1}^{n}(M_i-m_i)(x_i-x_{i-1}) =\sum_{i=1}^{n}\omega_i(x_i-x_{i-1}).

This proves the formula. $\square$

The formula helps explain how a partition can have a small gap. Subintervals where $f$ varies substantially must have small total length, while intervals where $f$ varies little can be longer. In particular, a large oscillation on one interval need not prevent a small total gap if that interval is sufficiently short.

Proof of the Criterion

Two facts established in Upper and Lower Integrals will be used here. For every partition $P$, the lower sum is at most the Darboux lower integral, and the upper sum is at least the Darboux upper integral. Also, approximation by partition sums allows the lower and upper integrals to be approached by lower and upper sums, respectively. The common-refinement result ensures that two partitions can be replaced by one partition refining both.

Proof of the Darboux Criterion. First suppose $f$ is Riemann integrable, and write its common lower and upper integral as $I$. Fix $\varepsilon>0$. By approximation by partition sums, choose partitions $P_L$ and $P_U$ such that

L(f,P_L)>I-\frac{\varepsilon}{2}, \qquad U(f,P_U)<I+\frac{\varepsilon}{2}.

Take a common refinement $R$ of $P_L$ and $P_U$. Under refinement, lower sums do not decrease and upper sums do not increase. Therefore

L(f,R)\geq L(f,P_L), \qquad U(f,R)\leq U(f,P_U).

Combining these inequalities gives

U(f,R)-L(f,R) \leq U(f,P_U)-L(f,P_L) <\left(I+\frac{\varepsilon}{2}\right) -\left(I-\frac{\varepsilon}{2}\right) =\varepsilon.

Thus a partition with gap less than $\varepsilon$ exists.

Conversely, suppose that for every $\varepsilon>0$ there is a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$. The partition-sum bounds for the Darboux integrals imply

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

The first inequality follows from the Darboux Lower Integral Does Not Exceed the Upper Integral theorem. Since this holds for every positive $\varepsilon$, the nonnegative difference of the integrals must be zero. The upper and lower integrals agree, so $f$ is Riemann integrable by the definition in the previous tutorial. This proves both directions. $\square$

Worked Applications of the Criterion

Worked Example: A Linear Function on an Interval

Consider $f(x)=3x+2$ on $[-1,2]$. This function is increasing, so on each subinterval of a uniform partition its infimum and supremum are its values at the left and right endpoints. Divide $[-1,2]$ into $n$ equal pieces, each of length $3/n$. On every piece, the change in $f$ is $3(3/n)=9/n$. Hence each piece contributes $(9/n)(3/n)=27/n^2$ to the Darboux gap, and

U(f,P_n)-L(f,P_n) =n\frac{27}{n^2} =\frac{27}{n}.

Given any $\varepsilon>0$, choose a positive integer $n>27/\varepsilon$. Then $27/n<\varepsilon$, so the Darboux Criterion proves that $f$ is integrable on $[-1,2]$. The calculation uses $f(2)-f(-1)=8-(-1)=9$; the total gap is also the subinterval length $3/n$ times this total change, namely $27/n$.

Worked Example: A Function with One Jump

Define $g:[0,1]\to\mathbb{R}$ by $g(x)=0$ when $x<2/5$, and $g(x)=1$ when $x\geq2/5$. We construct a partition that isolates the only interval where the values can vary. Choose $0<\delta<2/5$ and take $P=\{0,2/5-\delta,2/5,1\}$.

On $[0,2/5-\delta]$, $g$ is constantly zero, so the contribution to the gap is zero. On $[2/5,1]$, it is constantly one, so that contribution is also zero. On $[2/5-\delta,2/5]$, the infimum is zero and the supremum is one: the endpoint $2/5$ is included and has value one. This interval has length $\delta$, so

U(g,P)-L(g,P) =(1-0)\delta =\delta.

For any $\varepsilon>0$, choose $\delta$ with $0<\delta<\min\{2/5,\varepsilon\}$. The resulting gap is less than $\varepsilon$, and the criterion proves that $g$ is integrable. The partition point at the jump is not enough by itself to eliminate the gap, because the interval ending there includes the value at the jump. Instead, the interval containing the variation is made arbitrarily short.

Worked Example: The Indicator of the Rational Numbers

On $[0,1]$, define $h(x)=1$ when $x$ is rational and $h(x)=0$ when $x$ is irrational. Every subinterval of positive length contains both rational and irrational numbers, by the density of each set in the real line. Therefore, on every subinterval of every partition, the supremum of $h$ is one and its infimum is zero.

For any partition $P=\{x_0,\ldots,x_n\}$, the upper sum is the sum of the subinterval lengths, while the lower sum is zero:

U(h,P)-L(h,P) =\sum_{i=1}^{n}(1-0)(x_i-x_{i-1}) =\sum_{i=1}^{n}(x_i-x_{i-1}) =1.

No partition can make this gap smaller than, for example, $1/2$. The Darboux Criterion therefore shows that $h$ is not Riemann integrable on $[0,1]$. The decisive point is not simply that the function has many discontinuities; it is that every interval retains the full oscillation, so refinement never reduces the total gap.

How to Use the Criterion Carefully

The Darboux Criterion is useful because it separates the task of proving integrability into a concrete estimate: construct partitions and control the sum of oscillation times interval length. For a monotone function, changes across adjacent intervals can often be controlled by the total change of the function. For a function with a small number of jumps, one can isolate the intervals containing the jumps and make their lengths small. The oscillation formula makes both strategies precise.

A common pitfall is to calculate a small gap for one partition and treat that as a proof of integrability without explaining how to make the gap smaller than an arbitrary positive tolerance. The criterion requires a partition for each $\varepsilon>0$. Another pitfall is to assume that uniform partitions are always the best choice. The jump example shows why a partition adapted to the location of variation can be more effective.

Finally, boundedness matters throughout: the suprema and infima used in the sums must be finite. The criterion characterizes Riemann integrability for bounded functions on a closed, bounded interval; it does not extend the definition here to unbounded functions.

Check Your Understanding

Use the criterion and the arguments in this tutorial to answer the following questions.

What must be true about the difference between upper and lower sums for every positive tolerance?
Why does taking a common refinement help combine a lower-sum approximation and an upper-sum approximation?
In the oscillation formula, what two factors determine each subinterval’s contribution to the Darboux gap?
Why does isolating the jump in a short interval prove integrability of the step function?
Why can no partition prove integrability of the rational-indicator function using the Darboux Criterion?

What You'll Learn