Proving Darboux Refinement Monotonicity

Why refinement has a predictable effect

A partition divides an interval into finitely many pieces. On each piece, the function’s infimum gives the height of a lower rectangle, while its supremum gives the height of an upper rectangle. Refining a partition means adding points, so each new piece lies inside one of the old pieces.

The key fact is local: on a smaller interval, the infimum cannot be below the infimum on the containing interval, and the supremum cannot be above the supremum on the containing interval. The Darboux refinement inequalities follow by applying that fact to every new piece and adding the resulting comparisons.

Partitions and Darboux sums

Let \(f:[a,b]\to\mathbb{R}\) be bounded, where \(a<b\). A partition \(P\) of \([a,b]\) is a finite ordered set of points \(a=x_0<x_1<\cdots<x_n=b\). Its subintervals are \([x_{i-1},x_i]\), and their widths are \(\Delta x_i=x_i-x_{i-1}>0\).

Boundedness ensures that every infimum and supremum in these definitions is a finite real number. The function need not attain either value: “infimum” and “supremum” apply whether or not the smallest or largest function value occurs on the subinterval.

A partition \(Q\) is a refinement of \(P\) if it contains every point of \(P\), possibly along with additional points. Thus every subinterval of \(Q\) is contained in a subinterval of \(P\). The claim to prove is that for every bounded \(f\) and every such pair of partitions,

These inequalities compare sums for the same function on the same interval. They do not require continuity, integrability, or attainment of extrema. The essential requirements are that the function is bounded and that the new partition refines the old one.

The local comparison behind the proof

First consider a single old subinterval \([u,v]\), and split it at a point \(c\) with \(u<c<v\). Let \(m\) and \(M\) be the infimum and supremum of \(f\) on \([u,v]\). Let \(m_1,M_1\) be the infimum and supremum on \([u,c]\), and \(m_2,M_2\) those on \([c,v]\).

Both smaller intervals are subsets of \([u,v]\). Therefore every function value on either smaller interval is also a function value on the old interval. The infimum over the old interval is consequently no greater than either smaller-interval infimum; likewise, the old supremum is no less than either smaller-interval supremum:

Multiplying these comparisons by the positive widths preserves their directions. Since \(v-u=(c-u)+(v-c)\), the new lower contribution is at least the old one, and the new upper contribution is at most the old one:

This is the whole local mechanism. A lower sum uses the smallest available height on each interval, and restricting an interval can only raise that infimum. An upper sum uses the largest available height, and restricting an interval can only lower that supremum.

Worked Example: Splitting a single interval

Complete proof for inserting one point

Suppose \(Q\) is obtained from a partition \(P\) by inserting just one point \(c\) into one of its subintervals \([u,v]\). All other subintervals, and therefore their contributions to both sums, stay unchanged. On the affected interval, use the infima and suprema \(m,M,m_1,M_1,m_2,M_2\) defined above.

The change in the lower sum is exactly the new contribution minus the old contribution. Rearranging it makes the nonnegative terms visible:

Both widths are positive, and both differences \(m_1-m\) and \(m_2-m\) are nonnegative because each smaller interval is contained in the old one. This proves \(L(f,P)\leq L(f,Q)\).

For the upper sum, subtract the new contribution from the old contribution:

Here \(M-M_1\) and \(M-M_2\) are nonnegative, again because each smaller interval is contained in the old one. Thus \(U(f,Q)\leq U(f,P)\). Unchanged intervals contribute zero to the differences, so the argument proves both inequalities for the entire partition. It also shows the precise changes in the sums, not merely their directions.

From one inserted point to any refinement

An arbitrary refinement may insert several points into each old subinterval. The same comparison applies to all of its new pieces. If \([x_{i-1},x_i]\) is an old subinterval, let its infimum and supremum be \(m_i\) and \(M_i\). Write its refined pieces as \([y_{j-1},y_j]\), with infimum \(m'_j\) and supremum \(M'_j\) on each piece. Every such piece lies within the old interval, so \(m'_j\geq m_i\) and \(M'_j\leq M_i\).

Adding the lower contributions over the pieces of this old interval gives

because the widths of the refined pieces add to the old interval’s width. The upper contributions satisfy the reverse comparison:

Now add these inequalities over all old subintervals. Every refined piece belongs to exactly one old subinterval, so the sums on the left combine to the full sums for the refinement, and the sums on the right combine to the sums for the old partition.

This proof also gives a useful route to the general result: a finite refinement can be formed by inserting its additional points one at a time. Each insertion has the proved effect, so after all insertions the lower sum has not decreased and the upper sum has not increased.

Worked Example: Several inserted points

Worked Example: Refinement without continuity

A common mistake: counting intervals instead of comparing extrema

A tempting but incorrect argument is: “A refinement adds subintervals, so there are more rectangles; therefore the upper sum must increase.” The number of terms does not determine the sum’s direction. Both the widths and the extrema attached to those widths change.

For \(f(x)=x\) on \([0,2]\), the old upper sum is \(2\cdot2=4\). After splitting at \(1\), the actual suprema are \(1\) on \([0,1]\) and \(2\) on \([1,2]\), giving the new upper sum \(1\cdot1+2\cdot1=3\), not a larger value. The old maximum \(2\) cannot be used as the supremum on both smaller intervals: it is not the supremum on \([0,1]\). The correction is to compare each new interval’s own supremum with that of the old interval containing it. For lower sums, compare infima in the same way.

Why the inequalities matter

Refinement gives a systematic way to tighten Darboux bounds. Every lower sum from a coarser partition is no greater than every lower sum from its refinement; every upper sum from the refinement is no greater than the corresponding coarser upper sum. Thus adding partition points narrows the gap between the lower and upper estimates for that partition, or leaves one or both bounds unchanged.

The proof relies on containment, not on a graph’s shape or on an assumption that the function behaves smoothly. Its reusable steps are: identify which old interval contains each new interval, compare their infima and suprema, multiply by the positive widths, and add. Keeping those steps explicit prevents errors about which direction each Darboux sum moves.