How Refinement Tightens Darboux Bounds

Why Partition Refinement Matters

A lower Darboux sum gives a rectangular estimate from below, while an upper Darboux sum gives one from above. Their accuracy depends on the partition: dividing an interval into smaller pieces can make the rectangles fit the function more closely.

The key question is how the bounds respond to a refinement. Adding partition points cannot push the lower sum down or the upper sum up. The change may be strict, or one of the sums may stay the same. To see why, it is useful to examine exactly what happens when a single subinterval is split.

Darboux Sums and Refinements

Let \(f:[a,b]\to\mathbb{R}\) be bounded, and let \(P=\{x_0,x_1,\ldots,x_n\}\) be a partition of \([a,b]\), where \(a=x_0<x_1<\cdots<x_n=b\). On each subinterval \([x_{i-1},x_i]\), let \(m_i\) be the infimum of \(f\) and \(M_i\) its supremum. Boundedness ensures these numbers are finite; the infimum or supremum need not be attained.

Each term is the height of a rectangle, given by the infimum or supremum on that subinterval, multiplied by its width. The terms add to give the corresponding sum for the whole partition.

A partition \(Q\) is a refinement of \(P\) if it contains every point of \(P\), possibly together with additional points. Thus, a refinement divides some of the original subintervals into smaller ones and leaves the others unchanged.

What One Split Changes

Focus on one subinterval \([u,v]\) of \(P\), and choose a point \(c\) with \(u<c<v\). Write \(m\) and \(M\) for the infimum and supremum of \(f\) on \([u,v]\). After splitting at \(c\), write \(m_1,M_1\) for the infimum and supremum on \([u,c]\), and \(m_2,M_2\) for those on \([c,v]\).

Every value of \(f\) on either smaller interval is also a value of \(f\) on the original interval. Consequently, the infimum on the original interval is no greater than either smaller-interval infimum, and the original supremum is no less than either smaller-interval supremum:

Before the split, this subinterval contributes \(m(v-u)\) to the lower sum and \(M(v-u)\) to the upper sum. After the split, its contributions are \(m_1(c-u)+m_2(v-c)\) and \(M_1(c-u)+M_2(v-c)\). Subtracting the old contributions gives the exact changes:

The widths \(c-u\) and \(v-c\) are positive. The inequalities above therefore show that \(\Delta L\geq 0\) and \(\Delta U\leq 0\). All other subintervals keep their original contributions, so the total lower sum cannot decrease and the total upper sum cannot increase.

Worked Example: Splitting an Interval for an Increasing Function

Worked Example: Splitting Only Part of a Partition

Strict Improvement Is Not Guaranteed

The formulas also explain why adding a point may leave one or both sums unchanged. If the infimum on \([u,v]\) is still the infimum on both pieces, then \(m_1=m_2=m\), so \(\Delta L=0\). Likewise, if the supremum on both pieces remains \(M\), then \(\Delta U=0\). A partition point is useful for tightening a bound only when it separates values in a way that changes the relevant infima or suprema.

From One Split to a Refinement

A refinement may add several points, but the same comparison applies at each split. One can add the new points one at a time. At every step, a single existing subinterval is divided, so its lower contribution cannot decrease and its upper contribution cannot increase; the other contributions are unchanged.

In particular, if \(Q\) refines \(P\), the resulting sums satisfy

This conclusion concerns the sums associated with the two partitions. It does not say that the refined lower sum must equal the refined upper sum, or that either one must equal the integral. It says that refinement moves the two Darboux bounds inward: the lower estimate stays put or rises, and the upper estimate stays put or falls.

A Common Mistake: Assuming Both Sums Change Strictly

It is tempting to treat refinement as an automatic improvement in the strict sense. For example, someone might reason that two smaller intervals must each have a larger infimum and a smaller supremum than the original interval, and conclude that both sums always change strictly. That inference is false: the smaller intervals can have the same infimum or supremum as the original one.

For a concrete case, take the constant function \(f(x)=5\) on \([0,2]\). With \(P=\{0,2\}\), both sums equal \(5\cdot2=10\). After splitting at \(1\), both contributions are \(5\cdot1\), so each sum is still \(5+5=10\). Refinement has not made either sum worse, but neither has changed. The correct statements use \(\leq\) and \(\geq\), not strict inequalities.

A related calculation error is to subtract the original contribution from the refined one in the wrong order. With the convention used here, \(\Delta L=L_{\text{new}}-L_{\text{old}}\) is nonnegative, while \(\Delta U=U_{\text{new}}-U_{\text{old}}\) is nonpositive. Keeping those signs straight makes it easier to check a computation.

Why the Bounds Tighten

For an interval of fixed width, a smaller infimum lowers the rectangle used in a lower sum, while a larger supremum raises the rectangle used in an upper sum. Splitting reverses the relevant comparison: the infimum on each piece cannot be below the infimum on the whole interval, and the supremum on each piece cannot be above the original supremum. The refined lower rectangles therefore cannot have less total height-times-width, and the refined upper rectangles cannot have more.

The separate bounds also move closer together or remain equally far apart. Indeed, when the lower sum rises or stays fixed and the upper sum falls or stays fixed, the difference between upper and lower sums cannot increase. This is the sense in which refinement tightens the Darboux bounds. How to use that shrinking gap to plan estimates is a separate question; here, the essential calculation is the local effect of each split.