Upper and Lower Sums

From Partitions to Upper and Lower Sums

In “Partitions of an Interval,” a partition was introduced as a finite subdivision of $[a,b]$. To use a partition to approximate the area associated with a bounded function, we need a way to assign a height to each subinterval. A single sample value is not enough for a definition that captures the full range of the function on that subinterval. Instead, we use its infimum and supremum there. Multiplying these heights by the subinterval’s length produces lower and upper sums.

Let $f:[a,b]\to\mathbb{R}$ be bounded, and let $P=\{x_0,x_1,\ldots,x_n\}$ be a partition, where $a=x_0<x_1<\cdots<x_n=b$. For each subinterval $[x_{i-1},x_i]$, the set of values of $f$ is nonempty and bounded. It therefore has a real infimum and supremum. Denote them by $m_i$ and $M_i$, respectively.

m_i=\inf\{f(x):x\in[x_{i-1},x_i]\}, \qquad M_i=\sup\{f(x):x\in[x_{i-1},x_i]\}.

The infimum and supremum are bounds for all values of $f$ on that subinterval, whether or not the function actually attains them. This matters for functions that are discontinuous or whose extreme values are approached but not reached. Since $f$ is bounded on the whole interval, all the $m_i$ and $M_i$ are finite.

Definition: For a bounded function $f:[a,b]\to\mathbb{R}$ and a partition $P=\{x_0,\ldots,x_n\}$, the lower sum $L(f,P)$ and upper sum $U(f,P)$ are $$ L(f,P)=\sum_{i=1}^{n}m_i(x_i-x_{i-1}), \qquad U(f,P)=\sum_{i=1}^{n}M_i(x_i-x_{i-1}), $$ where $m_i$ and $M_i$ are the infimum and supremum of $f$ on $[x_{i-1},x_i]$.

Each term is a height multiplied by a width. If the function takes negative values, a term can be negative; these are sums of signed quantities, not necessarily areas of rectangles above the horizontal axis. Also, adjacent subintervals share an endpoint. That causes no difficulty: the shared point belongs to both closed intervals, and each subinterval’s infimum and supremum are computed using its own full set of function values.

Calculating the Sums

Worked Example: The Square Function on Two Equal Pieces

Let $f(x)=x^2$ on $[0,2]$, with partition $P=\{0,1,2\}$. On $[0,1]$, the values range from $0$ to $1$, so $m_1=0$ and $M_1=1$. On $[1,2]$, they range from $1$ to $4$, so $m_2=1$ and $M_2=4$. Both subintervals have length $1$. Therefore

L(f,P)=0\cdot1+1\cdot1=1, \qquad U(f,P)=1\cdot1+4\cdot1=5.

The calculation uses the least and greatest values on each whole subinterval, not a value chosen at a midpoint. On the first interval the lower and upper heights are $0$ and $1$; on the second they are $1$ and $4$.

Worked Example: An Absolute-Value Function with a Minimum Inside a Piece

Consider $f(x)=|x-1|$ on $[0,2]$ and the partition

P=\left\{0,\frac12,\frac32,2\right\}.

On $[0,\frac12]$, the function decreases from $1$ to $\frac12$, so $m_1=\frac12$ and $M_1=1$. On $[\frac12,\frac32]$, it reaches its minimum $0$ at $x=1$, and its largest value is $\frac12$ at either endpoint. Thus $m_2=0$ and $M_2=\frac12$. On $[\frac32,2]$, its values range from $\frac12$ to $1$, so $m_3=\frac12$ and $M_3=1$. The widths are $\frac12,1,\frac12$. Hence

L(f,P)=\frac12\cdot\frac12+0\cdot1+\frac12\cdot\frac12=\frac12,

and

U(f,P)=1\cdot\frac12+\frac12\cdot1+1\cdot\frac12=\frac32.

The middle interval illustrates why the height must be determined from the full subinterval: its minimum occurs in the interior, not at an endpoint.

Worked Example: A Function Whose Values Are Dense in Two Levels

On $[0,1]$, define $f(x)=1$ when $x$ is rational and $f(x)=0$ when $x$ is irrational. Every subinterval of positive length contains both rational and irrational numbers. Therefore on every subinterval of any partition, the infimum is $0$ and the supremum is $1$. For any partition $P=\{x_0,\ldots,x_n\}$,

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

The last sum telescopes to $x_n-x_0=1-0=1$. Neither the lower nor the upper sum changes when the partition changes. The example also shows why infimum and supremum, rather than minimum and maximum, are the general definitions: they describe bounds even when no continuity is assumed.

Lower Sums Are Below Upper Sums

For each subinterval, every function value lies between its infimum and supremum. Since subinterval lengths are positive, multiplying these bounds by the lengths and adding preserves the inequalities.

Theorem (Lower Sum Does Not Exceed Upper Sum): If $f:[a,b]\to\mathbb{R}$ is bounded and $P$ is a partition of $[a,b]$, then $L(f,P)\leq U(f,P)$.

Proof. For each $i$, the definitions of infimum and supremum give $m_i\leq M_i$. Also, $x_i-x_{i-1}>0$, because the partition nodes are strictly increasing. Thus

m_i(x_i-x_{i-1})\leq M_i(x_i-x_{i-1}).

Adding these inequalities for $i=1,\ldots,n$ gives

\sum_{i=1}^{n}m_i(x_i-x_{i-1}) \leq \sum_{i=1}^{n}M_i(x_i-x_{i-1}),

which is exactly $L(f,P)\leq U(f,P)$. $\square$

The inequality is not generally strict. For a constant function, each subinterval has the same infimum and supremum, so the lower and upper sums are equal. The theorem also applies when some or all of the function values are negative, because multiplying by a positive subinterval length still preserves the inequality.

What Refinement Does to the Sums

The Mesh Does Not Increase Under Refinement theorem from “Partitions of an Interval” describes how subinterval widths behave. For upper and lower sums, refinement has a complementary effect: it can only decrease the upper sum and increase the lower sum. Splitting an interval can give tighter bounds on the function’s values on its smaller pieces.

Theorem (Upper and Lower Sums Under Refinement): Let $f:[a,b]\to\mathbb{R}$ be bounded. If $Q$ is a refinement of a partition $P$, then $$ L(f,P)\leq L(f,Q) \qquad\text{and}\qquad U(f,Q)\leq U(f,P). $$

Proof. Consider one subinterval $[x_{i-1},x_i]$ of $P$. The nodes of $Q$ divide it into finitely many subintervals, say $[y_{j-1},y_j]$ for $j=r,\ldots,s$. Each smaller subinterval is contained in $[x_{i-1},x_i]$. Let $m_i,M_i$ be the infimum and supremum on the original subinterval, and let $m'_j,M'_j$ be those on the smaller ones. Containment gives

m_i\leq m'_j\leq M'_j\leq M_i \qquad (r\leq j\leq s).

The smaller intervals fill the original interval, so their lengths sum to its length:

\sum_{j=r}^{s}(y_j-y_{j-1})=x_i-x_{i-1}.

Multiplying the first and last bounds by the corresponding positive lengths and summing over the smaller intervals yields

m_i(x_i-x_{i-1}) \leq \sum_{j=r}^{s}m'_j(y_j-y_{j-1}),

and

\sum_{j=r}^{s}M'_j(y_j-y_{j-1}) \leq M_i(x_i-x_{i-1}).

Now sum these inequalities over all subintervals of $P$. Every subinterval of $Q$ occurs in exactly one of the groups of pieces lying inside a subinterval of $P$. The resulting left and right sums are $L(f,P)$, $L(f,Q)$, $U(f,Q)$, and $U(f,P)$, respectively. Therefore $L(f,P)\leq L(f,Q)$ and $U(f,Q)\leq U(f,P)$. $\square$

The inequalities may be equalities. For example, refining a partition does not change the sums of a constant function. More generally, the theorem guarantees only that refinement moves the two sums in the stated directions; it does not guarantee a strict change.

Comparing Sums from Different Partitions

Two partitions need not refine one another, so the refinement theorem alone does not directly compare their sums. The Existence of a Common Refinement theorem from “Partitions of an Interval” resolves this: combine the nodes of the two partitions to obtain a partition refining both.

Theorem (Comparison Across Partitions): Let $f:[a,b]\to\mathbb{R}$ be bounded, and let $P$ and $Q$ be any two partitions of $[a,b]$. Then $$ L(f,P)\leq U(f,Q). $$

Proof. Let $R$ be a common refinement of $P$ and $Q$, whose existence was established in “Partitions of an Interval.” Since $R$ refines $P$, the Upper and Lower Sums Under Refinement theorem gives $L(f,P)\leq L(f,R)$. The Lower Sum Does Not Exceed Upper Sum theorem gives $L(f,R)\leq U(f,R)$. Finally, since $R$ refines $Q$, the refinement theorem gives $U(f,R)\leq U(f,Q)$. Combining these inequalities,

L(f,P)\leq L(f,R)\leq U(f,R)\leq U(f,Q).

Hence $L(f,P)\leq U(f,Q)$, as claimed. $\square$

This comparison is stronger than comparing lower and upper sums for the same partition: every lower sum is at most every upper sum, even when the partitions have unrelated nodes. It does not assert that one lower sum is always below another lower sum, or that one upper sum is always above another upper sum. Those comparisons need a refinement relationship or additional information.

Why Upper and Lower Sums Matter

For a fixed partition, each subinterval contributes a term based on the full range of $f$ there. The lower sum uses the least possible height and the upper sum the greatest possible height. Refinement makes the bounds more localized: the lower sum cannot go down, and the upper sum cannot go up. This is the basic mechanism by which partitions can tighten the gap between the two sums.

A common pitfall is to replace the infimum and supremum with a function value at a convenient point, such as a midpoint. Such a value can be useful for other approximations, but it does not by itself give the lower or upper sum defined here. Another pitfall is to assume that a supremum or infimum must be attained. Boundedness ensures finite infima and suprema, but continuity or attainment is not part of these definitions. The rational-irrational example demonstrates that upper and lower sums can be defined even for a highly discontinuous function.

At this stage, upper and lower sums are attached to individual partitions. The next tutorial develops the Darboux upper and lower integrals by considering how these sums behave across all partitions, rather than selecting just one subdivision.

Check Your Understanding

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

For a bounded function on a subinterval, what do $m_i$ and $M_i$ represent, and must either value be attained by the function?
Why does multiplying $m_i\leq M_i$ by the subinterval length preserve the inequality?
If $Q$ refines $P$, which direction do the lower sums move, and which direction do the upper sums move?
How does a common refinement help compare $L(f,P)$ with $U(f,Q)$ when neither partition refines the other?
For the function that is $1$ on rational inputs and $0$ on irrational inputs in $[0,1]$, what are its lower and upper sums for any partition, and why?

What You'll Learn