Darboux Sums and Partitions

Partitions as Finite Bounds

On a compact interval, partitions replace a function by finite collections of local upper and lower bounds. The gap between these bounds records how much uncertainty remains at a given partition. Tagged sums provide a related approximation: on each subinterval, a chosen point supplies the function value used in the sum. We will define both constructions and use them to compute integrals of constant, step, and affine functions.

Throughout, let $a<b$ be real numbers and let $f:[a,b]\to\mathbb{R}$ be bounded. Boundedness ensures that every supremum and infimum below is finite. We consider the Riemann integral on this compact interval; unbounded functions and improper integrals are not included in this definition.

Definition: A partition of $[a,b]$ is a finite increasing list $P=\{x_0,x_1,\dots,x_n\}$ with $a=x_0<x_1<\cdots<x_n=b$. Its subintervals are $[x_{i-1},x_i]$, their lengths are $\Delta x_i=x_i-x_{i-1}$, and its mesh is $|P|=\max_{1\le i\le n}\Delta x_i$. A tagged partition is a partition together with choices $\xi_i\in[x_{i-1},x_i]$. Its tagged Riemann sum is $S(f;P,\xi)=\sum_{i=1}^n f(\xi_i)\Delta x_i$.

Tags may be endpoints or interior points. The sum uses the function value at the tag, whereas the Darboux sums use the full range of the function on each closed subinterval. In particular, if a function jumps at a partition point, its value there can affect the supremum or infimum on an adjacent subinterval.

Definition: For a bounded real-valued function $f$ on $[a,b]$, its lower and upper Darboux sums for $P$ are $$ L(f,P)=\sum_{i=1}^n \left(\inf_{x\in[x_{i-1},x_i]} f(x)\right)\Delta x_i, \qquad U(f,P)=\sum_{i=1}^n \left(\sup_{x\in[x_{i-1},x_i]} f(x)\right)\Delta x_i. $$ The lower integral is $\underline{\int_a^b} f=\sup_P L(f,P)$, and the upper integral is $\overline{\int_a^b} f=\inf_P U(f,P)$, where the supremum and infimum range over all partitions of $[a,b]$. The function is Riemann integrable if these two quantities are equal; their common value is denoted $\int_a^b f(x)\,dx$.

For every partition $P$, $L(f,P)\le U(f,P)$. More generally, every tagged sum for $P$ lies between its lower and upper sums, since on each subinterval the infimum is at most $f(\xi_i)$, which is at most the supremum. These inequalities hold whether $f$ takes positive, negative, or both positive and negative values.

Refinement and Tagged Sums

A partition $Q$ refines $P$ if it contains every point of $P$. Refinement can only raise a lower sum and lower an upper sum. This fact makes the lower and upper integrals compatible definitions rather than unrelated bounds.

Proposition. If $Q$ refines $P$, then $L(f,P)\le L(f,Q)$ and $U(f,Q)\le U(f,P)$. Consequently, $\underline{\int_a^b} f\le\overline{\int_a^b} f$.

Proof. Consider one subinterval $J=[x_{i-1},x_i]$ of $P$, divided by $Q$ into subintervals $J_1,\dots,J_k$. The infimum of $f$ over $J$ is at most its infimum over each $J_j$, because $J_j\subseteq J$. Multiplying these inequalities by the positive lengths and adding gives \[ \left(\inf_J f\right)|J|\le\sum_{j=1}^k\left(\inf_{J_j}f\right)|J_j|, \] since $\sum_j|J_j|=|J|$. The same reasoning with suprema gives $\sum_j(\sup_{J_j}f)|J_j|\le(\sup_J f)|J|$. Summing over the subintervals of $P$ proves both refinement inequalities.

For any two partitions $P_1,P_2$, their union is a common refinement $R$. The inequalities just proved, and the fact that a lower sum never exceeds the upper sum for the same partition, give $L(f,P_1)\le L(f,R)\le U(f,R)\le U(f,P_2)$. Thus every lower sum is at most every upper sum. Taking the supremum over $P_1$ and then the infimum over $P_2$ proves $\underline{\int_a^b} f\le\overline{\int_a^b} f$. $\square$

Theorem: If a bounded real-valued function $f$ on $[a,b]$ is Riemann integrable with value $I$, then for every $\varepsilon>0$ there is a $\delta>0$ such that every tagged partition $(P,\xi)$ with $|P|<\delta$ satisfies $|S(f;P,\xi)-I|<\varepsilon$.

Proof. Write $B=\sup_{x\in[a,b]}|f(x)|$, which is finite, and choose a partition $P_0$ with $r$ subintervals such that $U(f,P_0)-L(f,P_0)<\varepsilon/2$. Such a partition exists because the upper and lower integrals both equal $I$: one can choose an upper sum less than $I+\varepsilon/4$ and a lower sum greater than $I-\varepsilon/4$, then take their common refinement. Refinement preserves the required inequalities.

Let $Q$ be any partition with mesh less than $\delta$. Among its subintervals, call an interval crossing if its interior contains an interior partition point of $P_0$. Each crossing interval can be associated with one such point, and distinct crossing intervals have distinct associated points. There are at most $r-1$ crossing intervals, so their total length is at most $(r-1)\delta$. On each, the supremum minus the infimum of $f$ is at most $2B$, so their total contribution to $U(f,Q)-L(f,Q)$ is at most $2B(r-1)\delta$.

Every non-crossing interval of $Q$ lies within a subinterval $J$ of $P_0$. Its supremum minus infimum is therefore at most $\sup_J f-\inf_J f$. Adding the contributions of all such intervals within each $J$ gives at most $(\sup_J f-\inf_J f)|J|$. Summing over $P_0$ bounds their total contribution by $U(f,P_0)-L(f,P_0)$. Choose $\delta>0$ so that $2B(r-1)\delta<\varepsilon/2$ if $B(r-1)>0$; if $B(r-1)=0$, any positive $\delta$ suffices. It follows that $U(f,Q)-L(f,Q)<\varepsilon$.

The common Darboux value $I$ lies between $L(f,Q)$ and $U(f,Q)$, and every tagged sum for $Q$ lies in the same interval. Hence $|S(f;Q,\xi)-I|\le U(f,Q)-L(f,Q)<\varepsilon$. This proves the assertion for every choice of tags. $\square$

Three Basic Computations

Worked Example: A Negative Constant Function

Let $f(x)=-2$ on $[-1,3]$. For any partition, the supremum and infimum on each subinterval both equal $-2$. The sum of the subinterval lengths is $3-(-1)=4$, so for every partition $P$, $$ L(f,P)=U(f,P)=\sum_{i=1}^n(-2)\Delta x_i=(-2)\cdot4=-8. $$ Every tagged sum has the same value, since each tag also gives $f(\xi_i)=-2$. Thus the lower integral, upper integral, and Riemann integral are all $-8$. Check: for the partition $\{-1,0,3\}$, the sum is $(-2)(0-(-1))+(-2)(3-0)=(-2)(1)+(-2)(3)=-8$, as required.

Worked Example: A Step Function with One Jump

On $[0,2]$, set $f(x)=1$ for $0\le x<1$ and $f(x)=3$ for $1\le x\le2$. Use a partition that isolates a short interval around the jump. For $0<\delta<1$, take $P_\delta=\{0,1-\delta,1+\delta,2\}$. On the first interval the function is constantly $1$; on the middle interval its infimum is $1$ and supremum is $3$; on the last interval it is constantly $3$. Therefore $$ L(f,P_\delta)=(1-\delta)\cdot1+(2\delta)\cdot1+(1-\delta)\cdot3=4-2\delta, $$ and $$ U(f,P_\delta)=(1-\delta)\cdot1+(2\delta)\cdot3+(1-\delta)\cdot3=4+2\delta. $$

For example, $\delta=1/4$ gives lengths $3/4,1/2,3/4$, so the lower sum is $3/4+1/2+9/4=7/2$, and the upper sum is $3/4+3/2+9/4=9/2$. As $\delta$ tends to zero, the displayed lower sums tend to $4$ from below and the upper sums tend to $4$ from above. The lower integral is at least $4$, because it is at least every lower sum in this sequence; the upper integral is at most $4$, because it is at most every upper sum in the sequence. Since the lower integral never exceeds the upper integral, both equal $4$, and $f$ is Riemann integrable with integral $4$.

For the partition with $\delta=1/4$, choose the midpoint of the middle interval as its tag. That midpoint is $1$, where $f(1)=3$. The tagged sum is $(3/4)\cdot1+(1/2)\cdot3+(3/4)\cdot3=3/4+3/2+9/4=9/2$, equal to the upper sum. Choosing a tag in the middle interval strictly below $1$ instead gives $3/4+(1/2)\cdot1+9/4=7/2$, the lower sum. Both sums are within $1/2$ of the integral $4$.

Worked Example: An Increasing Affine Function

Let $f(x)=2x+1$ on $[-1,2]$, and divide the interval into $n$ equal subintervals. The $i$-th partition point is $x_i=-1+3i/n$, and each subinterval has length $3/n$. Since $f$ is increasing, its infimum and supremum on that subinterval are its left- and right-endpoint values. Substitution gives $f(x_i)=-1+6i/n$, so $$ L(f,P_n)=\frac{3}{n}\sum_{i=1}^n\left(-1+\frac{6(i-1)}{n}\right) =\frac{3}{n}\left(-n+\frac{6}{n}\cdot\frac{n(n-1)}2\right) =6-\frac9n. $$

Likewise, $$ U(f,P_n)=\frac{3}{n}\sum_{i=1}^n\left(-1+\frac{6i}{n}\right) =\frac{3}{n}\left(-n+\frac{6}{n}\cdot\frac{n(n+1)}2\right) =6+\frac9n. $$

These lower and upper sums tend to $6$, so the lower integral is at least $6$, the upper integral is at most $6$, and their ordering forces them both to equal $6$. As an independent check, an antiderivative is $x^2+x$, and $(2^2+2)-((-1)^2+(-1))=6-0=6$. For $n=3$, the formulas give $L=6-9/3=3$ and $U=6+9/3=9$; their midpoint is $6$. More generally, the midpoint of each subinterval gives a tagged sum of $6$: for an affine function, the midpoint value is the average of its endpoint values, so the midpoint sum is $(L(f,P_n)+U(f,P_n))/2=6$. This also checks that the bounds approach the computed integral from opposite sides.

A Common Pitfall: Endpoint Sums Are Not Automatically Upper Sums

A right-endpoint sum samples the function at one point per subinterval. It is not an upper sum unless those samples are in fact the suprema on their subintervals. For instance, let $g(x)=3-2x$ on $[0,1]$, and use the partition $\{0,1/2,1\}$. The right-endpoint sum is $$ \frac12g(1/2)+\frac12g(1)=\frac12\cdot2+\frac12\cdot1=\frac32. $$

But $g$ is decreasing, so the suprema occur at the left endpoints and the infima at the right endpoints. Thus the upper sum is $(1/2)\cdot3+(1/2)\cdot2=5/2$, while the lower sum is $3/2$. The right-endpoint sum is the lower sum, not the upper sum. The integral is $2$, as the antiderivative $3x-x^2$ gives $(3-1)-0=2$; indeed, this value lies between $3/2$ and $5/2$. The correction is to determine the local supremum and infimum from the function on each entire subinterval, rather than assume that a chosen endpoint has a particular role.

Partitions are useful because their sums provide finite bounds without requiring an antiderivative or an explicit description of the integral. For step functions, intervals around jumps can be made short; for affine functions, the endpoint bounds can be computed exactly. The definitions also make clear why a single tagged sum is only an approximation, while equality of the lower and upper integrals certifies a Riemann integral.

Check Your Understanding

Use the definitions of the tagged sum and Darboux sums to answer the following.

For $f(x)=4$ on $[1,5]$, find the lower sum, upper sum, and tagged sum for an arbitrary partition.
For the step function in the example, compute the lower and upper sums when $\delta=1/3$, and verify that their average is $4$.
For $f(x)=x$ on $[0,1]$, determine whether right endpoints give the upper or lower sum on an equal partition.
Explain why any tagged sum lies between the lower and upper sums for its partition, including when the function takes negative values.
For $g(x)=3-2x$ on $[0,1]$, compute the upper and lower sums for the partition $\{0,1/2,1\}$.

What You'll Learn