Proof Monotone Functions Are Integrable

From Uniform Partitions to Arbitrary Partitions

In “Integrability of Monotone Functions,” the exact Darboux gap on a uniform partition was found by telescoping the increases or decreases across its cells. That proof gives an especially simple formula. The same mechanism works on an arbitrary partition, even though the cells need not have equal lengths. The resulting estimate is useful when a partition is chosen for some other reason, or when we want to control the error of a particular sum.

Let \(P=\{x_0,\ldots,x_n\}\) be a partition of \([a,b]\), so \(a=x_0<x_1<\cdots<x_n=b\). The length of its \(i\)th cell is \(x_i-x_{i-1}\), and its mesh is the length of its largest cell. For a monotone function, the oscillation on each cell is the difference between its endpoint values, with the order depending on whether the function is nondecreasing or nonincreasing. The key observation is that the cell lengths may vary, but every one of them is at most the mesh.

Proof. Suppose first that \(f\) is nondecreasing. On the cell \([x_{i-1},x_i]\), its infimum is \(f(x_{i-1})\) and its supremum is \(f(x_i)\). Consequently, the difference between the upper and lower sums is

Every increment \(f(x_i)-f(x_{i-1})\) is nonnegative. Each cell length is at most \(\|P\|\), so replacing each length by \(\|P\|\) can only increase the sum. The increments telescope:

In this case \(f(b)-f(a)\geq0\), so it equals \(|f(b)-f(a)|\). The gap is nonnegative because each upper sum is at least its corresponding lower sum.

If \(f\) is nonincreasing, the oscillation on the \(i\)th cell is \(f(x_{i-1})-f(x_i)\), which is nonnegative. Thus

This also covers a constant function, for which every cell increment and the endpoint difference are zero. The estimate is proved in both directions of monotonicity. \(\square\)

Why the Mesh Bound Gives a Proof of Integrability

The Uniform-Partition Gap for a Monotone Function gives an exact formula for equal-length cells. The theorem just proved applies to every partition and gives an inequality: the actual gap can be smaller than the mesh times the total endpoint change. For equal-length cells, the mesh is \((b-a)/n\), and the earlier exact formula shows that this upper bound is attained. On unequal cells, some of the larger cell lengths may be paired with smaller increments, so equality need not hold.

We now use the general estimate to give a quantitative certificate for integrability. The Darboux Criterion says that a bounded function is Riemann integrable if, for every positive tolerance, a partition can be found whose upper–lower sum gap is below that tolerance. Boundedness for monotone functions on a closed interval follows from the Boundedness of a Monotone Function established earlier. The new estimate tells us how fine a partition suffices.

Proof. Suppose \(V>0\), and let \(\varepsilon>0\). By the Mesh Bound for the Darboux Gap of a Monotone Function, a partition with \(\|P\|<\varepsilon/V\) satisfies

Partitions with arbitrarily small mesh exist; for example, the Uniform Partitions Have Arbitrarily Small Mesh proposition guarantees one with mesh below \(\varepsilon/V\). Thus the Darboux Criterion applies. If \(V=0\), monotonicity and the endpoint bounds imply that every value of \(f\) equals \(f(a)=f(b)\). Hence \(f\) is constant and has zero gap on every partition. In either case \(f\) is Riemann integrable. \(\square\)

This is a proof in the sense that it supplies the needed boundedness and arbitrarily small Darboux gaps, then applies the criterion. It also tells us how to select a partition from a desired tolerance. The Integrability of Monotone Functions is the established conclusion; the mesh estimate is a sharper tool for obtaining explicit certificates.

Tagged Sums and an Explicit Error Bound

The same partition gap controls more than integrability. Choose a tag \(\xi_i\in[x_{i-1},x_i]\) in each cell and form the tagged sum \(S(f,P)=\sum_{i=1}^{n}f(\xi_i)(x_i-x_{i-1})\). The value of each tag lies between the infimum and supremum on its cell, so the whole sum lies between the lower and upper sums. For an integrable function, the integral also lies between those two sums, by the Integral Lies Between the Partition Sums proposition. Thus a small Darboux gap controls the error of every choice of tags at once.

Proof. On each cell, \(\inf f\leq f(\xi_i)\leq\sup f\). Multiplying these inequalities by the nonnegative cell length and summing over all cells gives \(L(f,P)\leq S(f,P)\leq U(f,P)\). Since \(f\) is integrable, the Integral Lies Between the Partition Sums proposition gives \(L(f,P)\leq\int_a^b f(x)\,dx\leq U(f,P)\). Both the tagged sum and the integral therefore belong to the interval \([L(f,P),U(f,P)]\), so their distance is at most the length of that interval:

Applying the Mesh Bound for the Darboux Gap of a Monotone Function proves the second inequality. \(\square\)

What to Keep in View

The proof depends on two separate controls. Monotonicity makes all cell increments have the same sign, so their sum telescopes to the endpoint change. The mesh bounds the length multiplying each increment. Without monotonicity, increases and decreases can alternate, and the sum of absolute cell increments need not be controlled by the endpoint difference. Thus the estimate is not a general bound for arbitrary bounded functions.

A second point is that small mesh alone is not the whole estimate: the endpoint change also appears. For a fixed monotone function on a fixed interval, that change is finite, so choosing a sufficiently small mesh makes the gap small. If the function is constant, the endpoint change is zero and every Darboux gap vanishes, regardless of mesh. At the other extreme, a large endpoint change requires a finer partition for the same tolerance.

To use this proof efficiently, first identify the direction of monotonicity, then compute the endpoint change, and finally choose the mesh in relation to the desired tolerance. For a concrete partition, the estimate also bounds every tagged-sum error, though the actual gap can be smaller than the bound.