Partitions of an Interval

Partitions as the Basic Structure of a Subdivision

In “Why Integration Requires a Definition,” a partition of \([a,b]\) was introduced as a finite increasing list of points beginning at \(a\) and ending at \(b\). Those points cut the interval into adjacent subintervals. This tutorial develops the structure of partitions themselves: how to compare two subdivisions, how to combine them, and how to control the width of their pieces. These ideas will be needed when sums associated with different partitions are compared.

Write a partition as \(P=\{x_0,x_1,\ldots,x_n\}\), where \(a=x_0<x_1<\cdots<x_n=b\). Its nodes are the listed points, and its subintervals are \([x_{i-1},x_i]\) for \(1\leq i\leq n\). The length of the \(i\)th subinterval is \(\Delta x_i=x_i-x_{i-1}\). The mesh, denoted by \(\|P\|\), is the largest of these lengths:

Because a partition has finitely many subintervals, this maximum exists. Every subinterval has positive length, so \(\|P\|>0\). The mesh records the width of the largest piece, not the average width or the number of nodes. Consequently, a small mesh guarantees that every piece is small.

Thus, refining a partition means adding nodes without removing any old ones. Each subinterval of the original partition may be split into one or more smaller subintervals. The order of the nodes remains increasing, and the endpoints \(a\) and \(b\) remain in the partition.

Refinement Controls the Mesh

A central benefit of refinement is that it never creates a wider subinterval than those already present. This fact follows from the way the nodes of the original partition are retained.

Proof. Let \([u,v]\) be any subinterval of \(Q\). Since every node of \(P\) is also a node of \(Q\), there cannot be a node of \(P\) strictly between \(u\) and \(v\): such a node would also be a \(Q\)-node between consecutive nodes \(u\) and \(v\), which is impossible. Therefore \([u,v]\) lies within one subinterval \([x_{i-1},x_i]\) of \(P\). It follows that

This holds for every subinterval \([u,v]\) of \(Q\). Taking the largest of their lengths gives \(\|Q\|\leq\|P\|\). \(\square\)

The theorem gives a useful guarantee, but not an equivalence: a smaller mesh alone does not mean that one partition refines another. Refinement is a condition on nodes, whereas mesh is a single numerical measurement. For instance, on \([0,1]\), the partitions \(\{0,\frac12,1\}\) and \(\{0,\frac13,\frac23,1\}\) both have mesh at most \(\frac12\), but neither refines the other because neither partition contains all the nodes of the other.

Combining Two Partitions

In later arguments, two different partitions may need to be compared using one subdivision that respects both. The natural construction is to collect all their nodes, remove duplicates, and put the resulting points in increasing order. The result is a common refinement.

Proof. Take the union of the finite sets of nodes of \(P\) and \(Q\), remove any repeated points, and list the remaining points in increasing order:

Call this partition \(R\). Every node of \(P\) and every node of \(Q\) occurs in the union, so \(R\) refines both partitions. The Mesh Does Not Increase Under Refinement theorem gives \(\|R\|\leq\|P\|\) and \(\|R\|\leq\|Q\|\). Hence \(R\) is a common refinement with the stated mesh bounds. \(\square\)

The construction does not require the partitions to have the same number of subintervals or to place their nodes in matching locations. If a node is already present in both partitions, it appears just once in the common refinement. Each interval between consecutive nodes of \(R\) lies within a subinterval of \(P\) and within a subinterval of \(Q\).

Making the Mesh as Small as Needed

A common refinement is useful for comparing subdivisions, and a separate elementary construction guarantees that partitions with arbitrarily small mesh are available. Divide the interval into \(n\) equal pieces. Choosing \(n\) sufficiently large makes each piece shorter than any prescribed positive bound.

Proof. Choose a positive integer \(n\) such that \(n>(b-a)/\eta\), which is possible because the positive integers are unbounded. Set

Then \(x_0=a\), \(x_n=b\), and \(x_i-x_{i-1}=(b-a)/n>0\) for every \(i\). Thus these points form a partition, and every subinterval has the same length. Its mesh is

where the strict inequality follows from \(n>(b-a)/\eta\). \(\square\)

Why These Partition Facts Matter

The definitions and results here provide a way to coordinate subdivisions without assuming that different partitions line up. Refinement preserves all old nodes and introduces only additional cuts. The mesh theorem then ensures that these additional cuts cannot make any piece wider. A common refinement puts two partitions into one framework, while uniform partitions ensure that an arbitrarily small mesh can always be achieved.

There are two pitfalls to avoid. First, a partition with smaller mesh need not refine another partition: mesh records only the largest subinterval length, not the locations of the nodes. Second, refining a partition does not necessarily make its mesh strictly smaller. If the longest subinterval is left intact, its length still determines the mesh. What is guaranteed is the non-strict inequality \(\|Q\|\leq\|P\|\).

In the Riemann definition, estimates are required to hold for every sufficiently fine tagged partition. The facts established here help organize such comparisons: a partition can be refined to include nodes from another partition, and uniform subdivisions can make all subintervals small. The next tutorial uses these partition structures to define and compare upper and lower sums.