Darboux Upper and Lower Integrals

From Sums for One Partition to Values for All Partitions

In “Upper and Lower Sums,” each partition gave a lower sum and an upper sum for a bounded function. Refining a partition can raise its lower sum and lower its upper sum, while the Comparison Across Partitions theorem ensures that every lower sum is at most every upper sum. We now collect the lower sums and upper sums across all partitions of the interval. Their supremum and infimum are the Darboux lower and upper integrals.

Throughout this tutorial, let $a<b$ and let $f:[a,b]\to\mathbb{R}$ be bounded. Write $\mathcal{P}$ for the collection of all partitions of $[a,b]$. The lower and upper sums $L(f,P)$ and $U(f,P)$, for $P\in\mathcal{P}$, are the sums defined in the previous tutorial.

Definition: The Darboux lower integral and Darboux upper integral of $f$ on $[a,b]$ are, respectively, $$ \underline{\int_a^b} f=\sup\{L(f,P):P\in\mathcal{P}\}, \qquad \overline{\int_a^b} f=\inf\{U(f,P):P\in\mathcal{P}\}. $$ The first is the supremum of all lower sums; the second is the infimum of all upper sums.

The definitions involve every partition, not a limiting process along one selected sequence of partitions. A particular sequence can help calculate a Darboux integral, but it must be used with bounds that show no other partition can give a better lower or upper sum. We will see this in the calculation for $f(x)=x$.

These Two Values Are Finite and Ordered

Because $f$ is bounded, there are real numbers $m$ and $M$ such that $m\leq f(x)\leq M$ for all $x\in[a,b]$. On each subinterval of any partition, its infimum and supremum are therefore also between $m$ and $M$. The subinterval lengths are positive and sum to $b-a$, so every lower sum and upper sum lies between $m(b-a)$ and $M(b-a)$. In particular, the sets used in the definition are nonempty and bounded, and their supremum and infimum are finite real numbers.

Theorem (Darboux Lower Integral Does Not Exceed the Upper Integral): For every bounded $f:[a,b]\to\mathbb{R}$, $$ \underline{\int_a^b} f\leq\overline{\int_a^b} f. $$

Proof. By the Comparison Across Partitions theorem from “Upper and Lower Sums,” if $P,Q\in\mathcal{P}$, then $L(f,P)\leq U(f,Q)$. Fix a partition $P$. Since $L(f,P)$ is at most $U(f,Q)$ for every $Q$, it is a lower bound for the set of all upper sums. It follows that

L(f,P)\leq \inf_{Q\in\mathcal{P}}U(f,Q) =\overline{\int_a^b}f.

This holds for every $P$. Thus the upper integral is an upper bound for the set of all lower sums, and

\underline{\int_a^b}f =\sup_{P\in\mathcal{P}}L(f,P) \leq\overline{\int_a^b}f.

This proves the assertion. $\square$

The same global bounds also give

m(b-a)\leq\underline{\int_a^b}f \leq\overline{\int_a^b}f\leq M(b-a).

The first inequality follows because every lower sum is at least $m(b-a)$, so its supremum is too. The last follows because every upper sum is at most $M(b-a)$, so its infimum is too. These bounds locate the two values even when they are not equal.

Approximating the Supremum and Infimum

The definitions as a supremum and an infimum have a practical consequence: each Darboux integral can be approached by an appropriate partition sum. The approximation need not be attained. In particular, there may be no single partition whose lower sum equals the lower integral, or whose upper sum equals the upper integral.

Theorem (Approximation by Partition Sums): For every $\varepsilon>0$, there are partitions $P_L,P_U\in\mathcal{P}$ such that $$ L(f,P_L)>\underline{\int_a^b}f-\varepsilon, \qquad U(f,P_U)<\overline{\int_a^b}f+\varepsilon. $$

Proof. Let $\alpha=\underline{\int_a^b}f$. If there were no partition $P_L$ with $L(f,P_L)>\alpha-\varepsilon$, then every lower sum would satisfy $L(f,P)\leq\alpha-\varepsilon$. This would make $\alpha-\varepsilon$ an upper bound for all lower sums, contradicting $\alpha=\sup_{P\in\mathcal{P}}L(f,P)$. Thus such a $P_L$ exists.

Now let $\beta=\overline{\int_a^b}f$. If there were no partition $P_U$ with $U(f,P_U)<\beta+\varepsilon$, then every upper sum would satisfy $U(f,P)\geq\beta+\varepsilon$. This would make $\beta+\varepsilon$ a lower bound for all upper sums, contradicting $\beta=\inf_{P\in\mathcal{P}}U(f,P)$. Hence the required $P_U$ exists as well. $\square$

The two partitions supplied by this theorem need not be the same. The infimum may be approached best by one partition, while the supremum is approached best by another. If a single partition is needed to control both sums, a common refinement can be used: refinement does not decrease the lower sum or increase the upper sum, by the Upper and Lower Sums Under Refinement theorem.

Calculating Darboux Integrals

Worked Example: A Constant Function

Let $f(x)=c$ on $[a,b]$, where $c$ is any real number. On every subinterval, the infimum and supremum are both $c$. For any partition $P=\{x_0,\ldots,x_n\}$,

L(f,P)=U(f,P) =\sum_{i=1}^{n}c(x_i-x_{i-1}) =c\sum_{i=1}^{n}(x_i-x_{i-1}) =c(b-a).

Every lower sum has the same value, so its supremum is $c(b-a)$; every upper sum has the same value, so its infimum is also $c(b-a)$. Therefore

\underline{\int_a^b}f=\overline{\int_a^b}f=c(b-a).

This calculation also works when $c<0$: the sums are signed quantities, and the telescoping sum of the subinterval lengths remains $b-a$.

Worked Example: The Identity Function on the Unit Interval

Let $f(x)=x$ on $[0,1]$, and take any partition $P=\{0=x_0<x_1<\cdots<x_n=1\}$. On $[x_{i-1},x_i]$, the infimum is $x_{i-1}$ and the supremum is $x_i$. Set $\Delta_i=x_i-x_{i-1}$, so $\Delta_i>0$. The identity

x_i\Delta_i =\frac{x_i^2-x_{i-1}^2}{2}+\frac{\Delta_i^2}{2}

follows by substituting $\Delta_i=x_i-x_{i-1}$: its right-hand side is $\frac{(x_i-x_{i-1})(x_i+x_{i-1})}{2}+\frac{(x_i-x_{i-1})^2}{2}=x_i(x_i-x_{i-1})$. Likewise,

x_{i-1}\Delta_i =\frac{x_i^2-x_{i-1}^2}{2}-\frac{\Delta_i^2}{2}.

Since the differences of squares telescope and $x_n^2-x_0^2=1$, adding these identities gives

U(f,P)=\frac12+\frac12\sum_{i=1}^n\Delta_i^2, \qquad L(f,P)=\frac12-\frac12\sum_{i=1}^n\Delta_i^2.

Thus every upper sum is at least $\frac12$, and every lower sum is at most $\frac12$. For the uniform partition into $n$ equal subintervals, every $\Delta_i=1/n$, so $\sum_{i=1}^n\Delta_i^2=1/n$. Its sums are

U(f,P)=\frac12+\frac{1}{2n}, \qquad L(f,P)=\frac12-\frac{1}{2n}.

Given any $\varepsilon>0$, choose $n$ large enough that $1/(2n)<\varepsilon$. The upper sums then approach $\frac12$ from above, and the lower sums approach it from below. Together with the bounds for every partition, this proves

\underline{\int_0^1}x\,dx =\overline{\int_0^1}x\,dx =\frac12.

Worked Example: A Function Nonzero at One Point

Define $f:[0,1]\to\mathbb{R}$ by $f(1/2)=1$ and $f(x)=0$ for $x\ne1/2$. Every subinterval of positive length contains a point other than $1/2$, where $f$ is zero. Therefore its infimum is zero, whether or not it contains $1/2$, and every lower sum is zero. Hence

\underline{\int_0^1}f=0.

For any partition, all upper sums are nonnegative. To make one as small as desired, choose $0<\delta<1/2$ and use the partition $P_\delta=\{0,1/2-\delta,1/2,1/2+\delta,1\}$. Only the two subintervals touching $1/2$ have a positive supremum; their suprema are both $1$, and each has length $\delta$. Thus

U(f,P_\delta)=1\cdot\delta+1\cdot\delta=2\delta.

Given $\varepsilon>0$, choose $\delta<\min(1/2,\varepsilon/2)$. Then $0\leq U(f,P_\delta)<\varepsilon$. Since all upper sums are nonnegative and can be made arbitrarily small, their infimum is zero. Therefore

\underline{\int_0^1}f=\overline{\int_0^1}f=0.

The function’s value at the single exceptional point affects upper sums for intervals containing that point, but those intervals can be made short. The definition takes the infimum over all partitions, so the upper integral records the limiting effect of such increasingly localized intervals.

What Equality Means—and What It Does Not Yet Say

For any bounded function, the lower integral is at most the upper integral. The examples show that equality can occur, but equality is not automatic from boundedness alone. When they are equal, the common value is the Riemann integral under the Darboux criterion; when they differ, no single value can serve as both the supremum of lower sums and the infimum of upper sums. The fact that upper and lower integrals are always defined should not be confused with the claim that every bounded function is integrable.

The point-value example highlights another useful distinction. A function can change at a point without changing either Darboux integral, because partitions can isolate that point in intervals of arbitrarily small total length. By contrast, the rational-irrational function from “Upper and Lower Sums” has lower sum $0$ and upper sum $1$ for every partition, so its lower integral is $0$ and its upper integral is $1$. There is no partition that makes the two kinds of sums approach one another.

A common calculation error is to find sums along a convenient sequence of partitions and treat their limits as the Darboux integrals without proving bounds for arbitrary partitions. The identity-function example avoided this: it established that every upper sum is at least $\frac12$ and every lower sum is at most $\frac12$, then used uniform partitions to approach those bounds. Both parts are needed to identify the infimum and supremum exactly.

The approximation theorem provides a general starting point for later comparisons: partitions can be chosen to bring lower sums close to the lower integral and upper sums close to the upper integral. The next tutorial develops further properties of these two values and the conditions under which they coincide.

Check Your Understanding

Use the definitions and proofs in this tutorial to answer the following questions.

Which collection of partition sums defines the Darboux lower integral, and which defines the upper integral?
Why are both Darboux integrals finite for a bounded function on a finite interval?
How does the Comparison Across Partitions theorem imply that the lower integral is at most the upper integral?
For $f(x)=x$ on $[0,1]$, what formulas do an arbitrary partition’s upper and lower sums satisfy in terms of its subinterval lengths?
For the function that is $1$ at $1/2$ and $0$ elsewhere on $[0,1]$, why can its upper sums be made arbitrarily small?