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Darboux criterion for Riemann–Stieltjes integrability with a nondecreasing integrator

Statement

Let $a < b$ , let $f : [a, b] \to R$ be bounded and let $α : [a, b] \to R$ be nondecreasing. Then $f$ is Riemann-Stieltjes integrable in the mesh sense of Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral if and only if both of the following conditions hold:

$f$ is continuous at every discontinuity of $α$ ; and
for every $ε > 0$ there is a partition $P$ with

$U_{α} (f, P) - L_{α} (f, P) < ε .$

In condition 2, writing $ω_{i} (f)$ for the oscillation of $f$ on $[t_{i}, t_{i + 1}]$ , the condition is $\sum_{i < n} ω_{i} (f) Δ_{i} α < ε$ .

The hypothesis $a < b$ is required and not cosmetic. On $[a, a]$ the integral is $0$ by Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, so every bounded $f$ is integrable, while Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions admits no partition of a singleton interval, so condition 2 asserts the existence of something that does not exist and fails. The equivalence therefore holds only on a nondegenerate interval; a consumer needing $a = b$ reads the value straight off the definition.

In particular, when $α$ is continuous, the weighted Darboux condition alone is equivalent to mesh Riemann-Stieltjes integrability.

Facts & Assumptions

Given: A bounded $f : [a, b] \to R$ and a nondecreasing $α : [a, b] \to R$ .

[L1]

Tagged, upper, and lower Stieltjes sums are those of Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral.

[L2]

Common refinements exist and insertion does not increase mesh (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L3]

Finite sums split, telescope, and preserve inequalities termwise (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

Every nonempty set of reals bounded above has a supremum, and every nonempty set bounded below has an infimum (Complete ordered field (least-upper-bound property), Every nonempty set bounded below has an infimum, Greatest lower bound (infimum), Lower bound, bounded below, bounded set).

[L5]

Arbitrarily small positive reciprocal naturals exist (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

Every discontinuity of a nondecreasing function is witnessed by a positive total one-sided jump (A monotone function on an interval has no discontinuity of the second kind: at every point both relevant one-sided limits exist, and an interior point $c$ is a discontinuity exactly when $\lim_{x \to c^{-}} f (x) < \lim_{x \to c^{+}} f (x)$ ).

[L7]

Continuity controls oscillation in sufficiently small neighborhoods (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

Proof

technique · direct

1.1

For every partition $P$ , $U_{α} (f, P) - L_{α} (f, P) = \sum_{i < n} (M_{i} - m_{i}) Δ_{i} α = \sum_{i < n} ω_{i} (f) Δ_{i} α$ . Refinement can only decrease the upper sum and increase the lower sum, because each refined supremum is no larger and each refined infimum no smaller than its coarse counterpart.

L1L2L3L4

1.2

Suppose first that the mesh-limit integral is $I$ . Given $ε > 0$ , choose a partition $P$ fine enough that every tagged sum over $P$ is within $ε / 4$ of $I$ . In each subinterval choose tags whose values approach its supremum and infimum within a common error small enough, using [L5] and the finite number of intervals. The two resulting tagged sums differ by more than $U_{α} (f, P) - L_{α} (f, P) - ε / 2$ , but by less than $ε / 2$ through $I$ ; hence $U_{α} (f, P) - L_{α} (f, P) < ε$ .

L1L3L4L5

1.3

Mesh integrability also forces continuity of $f$ at every discontinuity $c$ of $α$ . By [L6], the total increment of $α$ across every sufficiently small interval straddling $c$ is bounded below by a fixed positive number. Complete such an interval to an arbitrarily fine partition and keep every other tag fixed. Tagging the straddling interval first at $c$ and then at an arbitrary point $x$ in that interval changes the sum by $(f (x) - f (c)) Δ α$ . Both sums must approach the same mesh limit, so $f (x) \to f (c)$ as $x \to c$ . The same one-sided argument applies at an endpoint.

L1L2L6L7

2.1

Conversely assume both stated conditions. The lower sums have a supremum $I_{-}$ and the upper sums an infimum $I_{+}$ , with $I_{-} \leq I_{+}$ . Step 1.1 and condition 2 force $I_{-} = I_{+} = : I$ . Given $ε > 0$ , choose $P_{0}$ with Darboux gap below $ε / 3$ . Around each of its finitely many interior points $c$ , choose a small neighborhood as follows: if $α$ is continuous at $c$ , make the variation of $α$ there so small that twice the bound on $∣ f ∣$ times that variation is below the allotted error; if $α$ is discontinuous at $c$ , condition 1 and [L7] make the oscillation of $f$ there so small that its product with $α (b) - α (a)$ is below the allotted error. Choose the neighborhoods disjoint and divide the error among their finite number.

step 1.1L4L6L7

3.1

Let $Q$ now have mesh smaller than all those neighborhood radii and let $R = P_{0} \lor Q$ . A tagged sum on $R$ lies between $L_{α} (f, P_{0})$ and $U_{α} (f, P_{0})$ . Comparing a sum on $Q$ with one on $R$ , the intervals of $Q$ that do not cross a point of $P_{0}$ contribute at most the Darboux gap. Each crossing interval lies in one chosen neighborhood: its refinement error is bounded either by $2 \sup ∣ f ∣$ times the local variation of $α$ , or by the local oscillation of $f$ times the total variation $α (b) - α (a)$ . The choices in step 2.1 make the sum of all crossing errors below $2 ε / 3$ . Hence every sufficiently fine tagged sum lies within $ε$ of $I$ .

step 2.1L1L2L3L4L7

4.1

Steps 1.2–1.3 prove necessity, steps 2.1–3.1 prove sufficiency, and step 1.1 proves the weighted-oscillation formulation. When $α$ is continuous, condition 1 is vacuous.

step 1.1step 1.2step 1.3step 2.1step 3.1∎

Depends on

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Sources

W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.6 (standard reference, not scraped)
William F. Trench, Introduction to Real Analysis, Ch. 3 (standard reference, not scraped)