The Upper-Lower Sum Criterion

Why Compare Upper and Lower Sums?

A partition divides a closed interval into smaller pieces. On each piece, a bounded function has a greatest possible height in the sense of its supremum and a least possible height in the sense of its infimum. The resulting upper and lower sums bracket the values of all tagged Riemann sums for that partition.

If the upper and lower sums are close, the partition has left little room for the sum to vary with the choice of tags. The upper-lower sum criterion makes this idea exact: Riemann integrability is equivalent to being able to make that gap as small as desired by choosing a suitable partition.

Partitions and Darboux Sums

Let $f:[a,b]\to\mathbb{R}$ be bounded, where $a<b$. A partition $P$ of $[a,b]$ is a finite increasing list $a=x_0<x_1<\cdots<x_n=b$. Its $i$th subinterval is $[x_{i-1},x_i]$, with width $\Delta x_i=x_i-x_{i-1}$. On that subinterval, define $M_i=\sup\{f(x):x\in[x_{i-1},x_i]\}$ and $m_i=\inf\{f(x):x\in[x_{i-1},x_i]\}$.

The upper Darboux sum and lower Darboux sum for $P$ are

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

Boundedness ensures that every $M_i$ and $m_i$ is finite. Since $m_i\le M_i$, each subinterval contributes a nonnegative amount to the difference between the sums.

Definition: The upper-lower gap for a partition $P$ is $U(f,P)-L(f,P)$. It is the total width, measured in area units, between the upper and lower estimates supplied by that partition.

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

The quantifiers are essential. For each positive tolerance $\varepsilon$, it is enough to find a suitable partition; the partition may depend on $\varepsilon$. The criterion does not require one fixed partition to work for every tolerance. Nor does it require every partition to give a small gap.

The function must be bounded on the whole interval. Without boundedness, the finite upper and lower sums in this criterion need not be defined. The result concerns Riemann integrability on a closed, bounded interval, with upper and lower sums formed from finite partitions.

What the Gap Measures

For each subinterval, let $\omega_i=M_i-m_i$, the difference between the supremum and infimum of $f$ there. Subtract the lower sum from the upper sum and group the terms by subinterval:

\begin{aligned} U(f,P)-L(f,P) &=\sum_{i=1}^{n}M_i\Delta x_i-\sum_{i=1}^{n}m_i\Delta x_i\\ &=\sum_{i=1}^{n}(M_i-m_i)\Delta x_i\\ &=\sum_{i=1}^{n}\omega_i\Delta x_i. \end{aligned}

This identity shows why the gap is useful. A subinterval contributes more when the function varies more there, and also when that subinterval is wider. The total gap adds these contributions across the entire domain. Making the gap small means controlling the total effect of the function's variation, weighted by interval width.

There is also a direct connection to tagged sums. If a tag $\xi_i$ is chosen in each subinterval, then $m_i\le f(\xi_i)\le M_i$. Multiplying by the nonnegative width and summing gives

L(f,P)\le \sum_{i=1}^{n}f(\xi_i)\Delta x_i\le U(f,P).

Thus every tagged sum from a given partition lies between its lower and upper sums. A small gap squeezes all those sums into a narrow range. This is the practical meaning of the criterion: the upper and lower estimates can be brought arbitrarily close, so the possible accumulated values are tightly controlled.

Worked Examples

Worked Example: The Identity Function on a Uniform Partition

Let $f(x)=x$ on $[0,1]$, and divide the interval into $n$ equal subintervals. The partition points are $x_i=i/n$, so each width is $\Delta x_i=1/n$. Since $f$ is increasing, the infimum on $[x_{i-1},x_i]$ is $x_{i-1}$, and the supremum is $x_i$. Therefore,

\begin{aligned} U(f,P)-L(f,P) &=\sum_{i=1}^{n}(x_i-x_{i-1})\Delta x_i\\ &=\sum_{i=1}^{n}\left(\frac{i}{n}-\frac{i-1}{n}\right)\frac{1}{n}\\ &=\sum_{i=1}^{n}\frac{1}{n^2}\\ &=\frac{1}{n}. \end{aligned}

Given any $\varepsilon>0$, choose a positive integer $n>1/\varepsilon$. Then $1/n<\varepsilon$, so this partition has gap less than $\varepsilon$. For example, with $n=4$, the gap is $1/4$. The calculation verifies the criterion's required estimate, and the function is Riemann integrable.

As a check on the individual sums, the lower sum is $\sum_{i=1}^{n}((i-1)/n)(1/n)=(n-1)/(2n)$, while the upper sum is $\sum_{i=1}^{n}(i/n)(1/n)=(n+1)/(2n)$. Their difference is $(n+1)/(2n)-(n-1)/(2n)=1/n$, as found above.

Worked Example: The Square-Root Function on a Uniform Partition

Let $f(x)=\sqrt{x}$ on $[0,1]$, again using $x_i=i/n$. The function is increasing, so the supremum on each subinterval is $\sqrt{x_i}$, and the infimum is $\sqrt{x_{i-1}}$. Each interval has width $1/n$. Hence

\begin{aligned} U(f,P)-L(f,P) &=\frac{1}{n}\sum_{i=1}^{n} \left(\sqrt{\frac{i}{n}}-\sqrt{\frac{i-1}{n}}\right)\\ &=\frac{1}{n}\left(\sqrt{1}-\sqrt{0}\right)\\ &=\frac{1}{n}. \end{aligned}

The sum telescopes because each intermediate square-root value appears once with a plus sign and once with a minus sign. Given $\varepsilon>0$, choose an integer $n>1/\varepsilon$. The gap is then $1/n<\varepsilon$, as required. For $n=9$, the gap is $1/9$; substituting into the formula confirms this value. Thus the criterion applies, and $f$ is Riemann integrable on $[0,1]$.

This example also shows that the criterion is not restricted to functions whose upper and lower sums are especially easy to evaluate individually. Their difference may simplify because the contributions cancel in a telescoping sum.

Worked Example: A Bounded Function That Fails the Criterion

Define $g:[0,1]\to\mathbb{R}$ by setting $g(x)=1$ when $x$ is rational and $g(x)=0$ when $x$ is irrational. This function is bounded between $0$ and $1$. Every subinterval of positive width contains both rational and irrational numbers. Therefore, on every subinterval in every partition, the supremum is $1$ and the infimum is $0$.

For any partition $P$ of $[0,1]$, it follows that

U(g,P)=\sum_{i=1}^{n}1\cdot\Delta x_i =\sum_{i=1}^{n}\Delta x_i=1, \qquad L(g,P)=\sum_{i=1}^{n}0\cdot\Delta x_i=0.

Thus $U(g,P)-L(g,P)=1$ for every partition. Taking, for example, $\varepsilon=1/2$, no partition can make the gap less than $\varepsilon$. The criterion therefore says that $g$ is not Riemann integrable on $[0,1]$. Checking another partition cannot help: the calculation shows that the gap remains $1$ for all of them.

How to Use the Criterion

When applying the criterion, the task is to bound or calculate the gap and then choose a partition that meets the requested tolerance. A useful approach is to separate the reasoning into a few steps.

Check boundedness.
Confirm that the function is bounded on the closed interval, so its upper and lower sums are finite.

Describe the oscillation on each interval.
Find or estimate the difference $M_i-m_i$ for the chosen subintervals.

Compute or bound the total gap.
Use $U(f,P)-L(f,P)=\sum_{i=1}^{n}(M_i-m_i)\Delta x_i$.

Match the tolerance.
Choose the partition so the gap is strictly less than the given $\varepsilon>0$, or show that a fixed positive lower bound prevents this.

A successful calculation must end with the strict inequality required by the criterion. Showing that the gap can be made small in an informal sense is not enough; the estimate should identify how the partition controls the gap for an arbitrary positive tolerance.

A Common Quantifier Mistake

A frequent error is to test one partition and conclude that the function is integrable because its gap is small. That only establishes a useful estimate for one particular tolerance. The criterion requires a suitable partition for every positive tolerance.

For example, suppose a partition gives a gap of $0.01$. This establishes the required inequality for tolerances greater than $0.01$, but says nothing by itself about a tolerance such as $0.001$. To prove the criterion, one needs a method that can produce a sufficiently small gap no matter how small the positive tolerance is. In the identity-function example, choosing $n>1/\varepsilon$ supplies exactly that method.

The converse mistake is to demand that every partition have a small gap. That condition is not part of the criterion. A coarse partition may have a large gap even for an integrable function; what matters is whether at least one suitable partition exists for each tolerance.

Why the Criterion Matters

The upper-lower sum criterion translates integrability into a question about finite partitions and numerical bounds. It can be used without first calculating the value of an integral. Instead, it asks whether the function's upper and lower estimates can be made arbitrarily close.

The distinction between the examples is captured by the same quantity. For the increasing functions, finer partitions reduce the total gap below any prescribed tolerance. For $g$, every interval retains the same range of function values, so the total gap never decreases. The criterion therefore offers both a way to establish integrability and a way to rule it out.

Check Your Understanding

Answer these questions by referring to the definition of the upper-lower gap and its role in the criterion.

State the upper-lower sum criterion, including its boundedness condition and the quantifiers involving $\varepsilon$ and $P$.
Why is $U(f,P)-L(f,P)$ equal to $\sum_{i=1}^{n}(M_i-m_i)\Delta x_i$?
For $f(x)=x$ on $[0,1]$ with $n$ equal subintervals, what is the gap, and how can $n$ be chosen for a given $\varepsilon>0$?
Why does the rational-indicator function on $[0,1]$ fail the criterion?
Does the criterion require every partition to have a small gap? Explain the role of the word “there exists.”

The Upper-Lower Sum Criterion

What You'll Learn

Why Compare Upper and Lower Sums?

Partitions and Darboux Sums