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Bounded Variation and the Riemann–Stieltjes Integral
1 · Prerequisites
- Absolute and Conditional Convergence; Rearrangement; Products
- Compactness in Metric Spaces
- Completeness, Completion, and Uniform Continuity
- Construction of the Natural Numbers
- Construction of the Real Numbers via Cauchy Sequences
- Construction of the Real Numbers via Dedekind Cuts
- Continuity, IVT, EVT, and Uniform Continuity
- Countability and Uncountability
- Foundations of the Real Numbers for Analysis
- Limits of Real Functions
- limsup, liminf, and Subsequential Limits
- Metric Spaces
- Monotone Functions, Discontinuities, and Continuity Sets
- Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness
- Order, Zorn's Lemma, and the Axiom of Choice
- Properties of the Integral and the Working FTC
- Relations, Functions, and Quotients
- Roots, Rational Powers, and Classical Inequalities
- Sequences and Limits
- Series: Convergence and the Nonnegative Tests
- Suprema and Infima
- The Derivative and the Mean Value Theorems
- The Riemann Integral: Definition and Integrability
- The ZFC Axioms and the Basic Set Constructions
- Topology of ℝ
2 · Summary
The declared prerequisites supply partitions, Darboux and tagged Riemann sums, algebra and order estimates for proper integrals, the working fundamental theorem, integration by parts, substitution, and the integral test. They also supply monotone functions and their one-sided limits, compactness and uniform continuity, rational powers, and complete-real convergence. This machinery passes from finite variation sums to suprema, decomposes BV functions into monotone parts, and compares Stieltjes sums without importing measure theory.
Total variation, its canonical positive and negative parts, and absolute continuity lead to the Jordan decomposition and the –Lipschitz–AC–BV hierarchy. Riemann–Stieltjes integration then develops through a jump-compatible Darboux criterion, refinement estimates, algebra, integration by parts, and successively broader existence theorems using regulated approximation and the no-common-jump condition. Ordinary integration, differentiable integrators, and monotone reparametrization arise as reductions. A point-removal partition estimate finally yields Young integration for rational Hölder exponents whose sum exceeds one.
3 · Logical flowchart
4 · Definitions, theorems and proofs
Bounded variation and total variation on an interval
Definition
Let and let (Intervals of : the nine order-convex forms, nondegeneracy, and length). If and is a partition of (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions), the variation of over is
The sum is finite (Finite sums and finite products, by recursion, Laws of finite sums and finite products) and nonnegative (Absolute value in an ordered field). The set of all such sums is nonempty, since has the partition with point set . The function has bounded variation on when this set of sums is bounded above (Lower bound, bounded below, bounded set). In that case its total variation is
Completeness of gives this supremum and Suprema and infima are unique makes it unique (Complete ordered field (least-upper-bound property)). On a singleton interval, by convention, ; no partition from Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, whose standing hypothesis is , is invoked.
Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant
Statement
Let have bounded variation. Then
- for all ;
- is bounded on ;
- if and only if is constant.
These claims include the singleton interval .
Facts & Assumptions
Given: Reals and a bounded-variation function .
Total variation is the supremum of the partition sums , with value on a singleton interval (Bounded variation and total variation on an interval).
A point of an interval can be inserted into a partition without deleting its existing points (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite sums of nonnegative terms dominate every term (Laws of finite sums and finite products).
in an ordered field (The triangle inequality).
A subset of is bounded when the absolute values of its members have a common real bound (Lower bound, bounded below, bounded set).
Proof
If and lie in , insert and into the endpoint partition. The resulting partition sum contains as a nonnegative term, so . The same inequality is when , and when only that case occurs.
Put . For , , so is bounded.
If the total variation is , step 1.1 gives for every , hence is constant. Conversely, if is constant then every increment in every partition sum is , so every sum and its supremum are ; the singleton convention gives the same conclusion when .
Total variation is additive over adjacent subintervals and decreases under restriction
Statement
If and , then has bounded variation on if and only if its restrictions have bounded variation on and . In that case
Consequently restriction to any subinterval cannot increase total variation.
Facts & Assumptions
Given: Reals and a function .
Total variation is the supremum of finite variation sums, and singleton variation is (Bounded variation and total variation on an interval).
A partition is a finite strictly increasing endpoint list, and a refinement contains every point of the original partition (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite sums telescope and split at an index (Finite sums and finite products, by recursion, Laws of finite sums and finite products).
A supremum is the least upper bound of a nonempty set bounded above, and is unique (Complete ordered field (least-upper-bound property), Suprema and infima are unique).
Closed subintervals are the sets (Intervals of : the nine order-convex forms, nondegeneracy, and length).
Proof
Given partitions of and of , splice their point lists at . The resulting partition of satisfies . Thus, whenever is BV on , each restriction is BV and . The assertion is unchanged when or , because the singleton term is .
Conversely, insert into an arbitrary partition of . The refined sum splits into a sum on and one on , while inserting a point does not decrease the sum, because the corresponding increment is replaced by two increments whose absolute values dominate it by the triangle inequality. Hence .
Taking the supremum over in step 1.2 and combining it with step 1.1 proves the equality and the equivalence of boundedness. Applying the equality twice to shows .
Variation function and positive and negative variations
Definition
Let have bounded variation (Bounded variation and total variation on an interval). Its variation function is
so . This is defined on every subinterval by Total variation is additive over adjacent subintervals and decreases under restriction and the singleton convention. The positive variation and negative variation of are
The names refer to upward and downward accumulated variation; their monotonicity and the resulting decomposition are proved separately.
The positive and negative variations are nondecreasing and give the Jordan identities
Statement
For a bounded-variation function , the functions and are nondecreasing and
Both and are .
Facts & Assumptions
Given: A bounded-variation function and its functions .
are defined by the displayed formulas in Variation function and positive and negative variations.
Proof
For , [L2] and [L3] give , hence this difference is at least both and . Therefore and , so both functions are nondecreasing.
Adding and subtracting the defining formulas gives and . At , , so .
Rearranging the second identity in step 1.2 gives , while the first is the asserted variation identity.
Jordan decomposition for functions of bounded variation
Statement
A real function on has bounded variation if and only if it is a difference of two nondecreasing functions. If , the canonical normalized decomposition is . More generally .
It is minimal: if with nondecreasing and , then and for every .
Facts & Assumptions
Given: A function .
For BV , are nondecreasing, normalized at , and (The positive and negative variations are nondecreasing and give the Jordan identities).
Total variation is the supremum of sums of absolute increments (Bounded variation and total variation on an interval).
Nondecreasing means that each forward increment is nonnegative (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of , with the dictionary to monotone sequences).
A partition is a finite increasing point list (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite sums telescope and distribute over addition (Finite sums and finite products, by recursion, Laws of finite sums and finite products).
Proof
If is BV, [L1] immediately supplies the stated difference of nondecreasing functions, with the asserted normalization.
Conversely suppose with nondecreasing. For a partition , every forward increment of and is nonnegative, so . Summing and telescoping gives , independent of ; hence is BV.
Now assume the decomposition is normalized. On , step 1.2 gives , while . Adding these inequalities and dividing by yields ; subtracting the increment identity from the variation inequality yields .
A bounded-variation function has at most countably many discontinuities, all of the first kind
Statement
If has bounded variation, every well-posed one-sided limit of exists. Consequently every discontinuity is of the first kind, and the set of discontinuities is at most countable.
Facts & Assumptions
Given: A bounded-variation function .
Jordan decomposition writes with nondecreasing (Jordan decomposition for functions of bounded variation).
Every well-posed one-sided limit of a monotone function exists, so all its discontinuities are of the first kind (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 is a discontinuity exactly when , The left and right limits of at , as limits of the restrictions of to and ).
The discontinuity set of a monotone function on an interval is at most countable (Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into being built from one fixed enumeration of the rationals by least index, so no choice principle is used).
Finite sums and differences preserve existing finite function limits (Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero).
Proof
Apply [L2] to and . At every endpoint or interior point where a one-sided limit is defined, both component limits exist, and [L4] gives the corresponding one-sided limit of . Thus has no discontinuity of the second kind.
If both and are continuous at a point, [L4] makes continuous there. Hence the discontinuity set of is contained in the union of the two component discontinuity sets.
Each component discontinuity set is at most countable by [L3]. Given injections of them into , map the first set to the even naturals and the points belonging only to the second to the odd naturals; this injects their union into . Step 1.2 therefore makes the discontinuity set of at most countable, and step 1.1 makes every one of its discontinuities first-kind.
The jumps of a variation function equal the absolute jumps of the original function
Statement
Let have bounded variation and let . At an interior point ,
The corresponding one-sided formula holds at either endpoint. In particular, is continuous at every point where is continuous.
Facts & Assumptions
Given: A bounded-variation function , its variation function , and a point .
Every relevant one-sided limit of a BV function exists (A bounded-variation function has at most countably many discontinuities, all of the first kind, The left and right limits of at , as limits of the restrictions of to and ).
Finite sums and differences preserve existing one-sided limits (Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero).
Continuity is equality of the relevant limit with the function value (Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point).
A bounded nondecreasing sequence converges; its partial sums are therefore Cauchy, so the sums over all sufficiently remote finite tails are uniformly small (A monotone sequence converges if and only if it is bounded, Series, partial sums, convergence and the sum, divergence, and the tail series, A series converges iff each of its tail series converges, and the sum splits as plus the -th tail, Every convergent sequence is Cauchy).
Geometric sequences with ratio in tend to zero (For the sequence is null, and for the sequence diverges to ).
Proof
Since is nondecreasing and bounded above by , its one-sided limits exist. By [L1] and [L2], for ; passage to the right limit gives .
For the reverse inequality, fix and put . Let . By repeated additivity, every partial sum of the nonnegative series is for a suitable , hence is bounded by . Its tails therefore tend to zero by [L6], while by [L7].
Given , take so large that the series tail from is below and whenever . For any partition , choose with . The part after its first increment is at most [step 1.2, L1, L2, L3, L6] while . Taking the supremum over partitions gives . Restriction gives the same bound for , and [L2] gives the reverse bound in the limit.
Thus , and [L1] proves the right-hand formula. Applying steps 1.2–2.1 to the reversed interval proves the left-hand formula. If is continuous at , both absolute jumps vanish by [L5], so is continuous there. Endpoint cases use only the available side.
Homogeneity and subadditivity of total variation
Statement
For bounded-variation functions and ,
Thus , , and every finite linear combination of BV functions are BV; in particular .
Facts & Assumptions
Given: BV functions and a scalar .
Total variation is the supremum of partition variation sums (Bounded variation and total variation on an interval).
A partition is a finite increasing point list (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite sums distribute over scalar multiplication and addition (Finite sums and finite products, by recursion, Laws of finite sums and finite products).
Proof
For every partition , [L4] and [L3] give . Taking suprema gives , including and the singleton interval.
For every partition, [L5] applied to each increment and then [L3] give . Taking the supremum proves subadditivity.
Step 1.1 with gives . Repeated use of steps 1.1 and 1.2 proves closure under every finite linear combination.
Functions of bounded variation form an algebra
Statement
If and have bounded variation on , so do , , and . If and , then
Facts & Assumptions
Given: BV functions .
BV functions are closed under sums and scalar multiples (Homogeneity and subadditivity of total variation).
Every BV function is bounded (Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant).
Total variation is the supremum of partition sums (Bounded variation and total variation on an interval).
Absolute value is multiplicative and satisfies the triangle inequality (Basic properties of the absolute value, The triangle inequality).
Proof
By [L2] choose with and on . For a partition point pair , the identity gives .
Summing step 1.1 over any partition yields . Taking the supremum proves the displayed bound and that is BV.
Closure under sums and scalar multiples is [L1], and step 2.1 supplies closure under products, so the BV functions form an algebra under pointwise operations.
Every bounded-variation function on a compact interval is Riemann integrable
Statement
Every real-valued function of bounded variation on a compact interval is Darboux, equivalently Riemann, integrable.
Facts & Assumptions
Given: A bounded-variation function .
Jordan decomposition writes with nondecreasing (Jordan decomposition for functions of bounded variation).
A monotone real function on a compact interval is integrable (A monotone function on is Riemann integrable: for the uniform partition into parts the upper minus lower sum telescopes to ).
Linear combinations of integrable functions are integrable and their integrals combine linearly (Integrable functions on form a set closed under sums and scalar multiples, and ).
Darboux integrability is the proper integral notion on (The lower and upper Darboux integrals of a bounded on as and , Darboux integrability as their equality, and the notation ).
Proof
By [L1], and are nondecreasing; by [L2] both are integrable, and the constant function is integrable.
Linearity applied to makes integrable. The singleton interval follows from the zero-integral convention.
Absolute continuity on a compact interval
Definition
Let and (Intervals of : the nine order-convex forms, nondegeneracy, and length). The function is absolutely continuous on if for every there is such that every finite family of subintervals , indexed by , whose open interiors are pairwise disjoint and which satisfies
also satisfies
Finite sums and the empty sum are those of Finite sums and finite products, by recursion and Laws of finite sums and finite products. For both sums are , so the condition is automatic. On every permitted interval is a singleton and every endpoint increment is (Absolute value in an ordered field), so every function on that singleton is absolutely continuous. Absolute continuity implies ordinary continuity (Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point); that implication is proved next rather than built into the definition.
implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation
Statement
Let .
- If is continuous on , differentiable on , and extends continuously to , then is Lipschitz.
- Every Lipschitz is absolutely continuous.
- Every absolutely continuous is continuous and has bounded variation.
Thus, with understood in the endpoint-extension sense of claim 1, on a compact interval.
Facts & Assumptions
Given: A compact interval and a function .
Absolute continuity is the finite disjoint-interval condition of Absolute continuity on a compact interval.
A continuous real function on is bounded (A continuous real function on a compact subset of is bounded).
A continuous function with bounded derivative on an interval is Lipschitz (If is continuous on an interval and at every interior point, then for all , so is Lipschitz with constant and uniformly continuous on ).
The Lipschitz condition is for one (Lipschitz map, -Hölder map for rational , and contraction, Dictionary: for with the metric , continuity and uniform continuity of agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of is compact in the open-cover sense of exactly when it is a compact metric subspace).
Total variation is the supremum of partition sums (Bounded variation and total variation on an interval, Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite sums split and telescope (Laws of finite sums and finite products).
The canonical naturals are cofinal in (Every complete ordered field is Archimedean).
Proof
Under claim 1, the continuous extension of is bounded by some on by [L3]. The bounded-derivative theorem [L4] then makes Lipschitz with constant .
If is Lipschitz with constant , then for every finite disjoint family, . For any positive works; for choose . This proves absolute continuity, including the empty family.
If is absolutely continuous, apply [L1] to the single interval with endpoints to obtain the usual - continuity condition, so is continuous.
For bounded variation, take from absolute continuity with . By [L8] choose a natural with . Insert the points of the uniform -partition into an arbitrary partition . Inside each uniform block, the refined subintervals have disjoint interiors and total length at most , so their endpoint oscillations sum to less than . Summing over the blocks gives , independent of . Thus is BV. If , its variation is .
Steps 1.1 through 1.4 prove all three inclusions and the asserted hierarchy.
Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral
Definition
Let , let , and let be a partition (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions). A choice of tags for makes a tagged partition as in Tagged partitions of , with a tag in each subinterval, and the Riemann sum . Its Riemann-Stieltjes sum is
The function is Riemann-Stieltjes integrable with respect to on if there is such that for every there is for which every tagged partition with satisfies . Then .
If is bounded (Lower bound, bounded below, bounded set) and is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of , with the dictionary to monotone sequences), put
where . These are the lower and upper Stieltjes sums. Each subinterval is nonempty and its image under bounded is bounded above and below, so the suprema exist by Complete ordered field (least-upper-bound property) and the infima by Greatest lower bound (infimum) and Every nonempty set bounded below has an infimum. Finite sums use Finite sums and finite products, by recursion and Laws of finite sums and finite products. On the integral is ; for set , matching The integral with oriented limits: and .
Darboux criterion for Riemann–Stieltjes integrability with a nondecreasing integrator
Statement
Let , let be bounded and let be nondecreasing. Then 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:
- is continuous at every discontinuity of ; and
- for every there is a partition with
In condition 2, writing for the oscillation of on , the condition is .
The hypothesis is required and not cosmetic. On the integral is by Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, so every bounded is integrable, while Partition of as a finite strictly increasing list , 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 reads the value straight off the definition.
In particular, when is continuous, the weighted Darboux condition alone is equivalent to mesh Riemann-Stieltjes integrability.
Facts & Assumptions
Given: A bounded and a nondecreasing .
Tagged, upper, and lower Stieltjes sums are those of Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral.
Common refinements exist and insertion does not increase mesh (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite sums split, telescope, and preserve inequalities termwise (Finite sums and finite products, by recursion, Laws of finite sums and finite products).
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).
Arbitrarily small positive reciprocal naturals exist (For every in a complete ordered field there is a natural with ).
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 is a discontinuity exactly when ).
Continuity controls oscillation in sufficiently small neighborhoods (Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point).
Proof
For every partition , . 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.
Suppose first that the mesh-limit integral is . Given , choose a partition fine enough that every tagged sum over is within of . 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 , but by less than through ; hence .
Mesh integrability also forces continuity of at every discontinuity of . By [L6], the total increment of across every sufficiently small interval straddling 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 and then at an arbitrary point in that interval changes the sum by . Both sums must approach the same mesh limit, so as . The same one-sided argument applies at an endpoint.
Conversely assume both stated conditions. The lower sums have a supremum and the upper sums an infimum , with . Step 1.1 and condition 2 force . Given , choose with Darboux gap below . Around each of its finitely many interior points , choose a small neighborhood as follows: if is continuous at , make the variation of there so small that twice the bound on times that variation is below the allotted error; if is discontinuous at , condition 1 and [L7] make the oscillation of there so small that its product with is below the allotted error. Choose the neighborhoods disjoint and divide the error among their finite number.
Let now have mesh smaller than all those neighborhood radii and let . A tagged sum on lies between and . Comparing a sum on with one on , the intervals of that do not cross a point of contribute at most the Darboux gap. Each crossing interval lies in one chosen neighborhood: its refinement error is bounded either by times the local variation of , or by the local oscillation of times the total variation . The choices in step 2.1 make the sum of all crossing errors below . Hence every sufficiently fine tagged sum lies within of .
Steps 1.2–1.3 prove necessity, steps 2.1–3.1 prove sufficiency, and step 1.1 proves the weighted-oscillation formulation. When is continuous, condition 1 is vacuous.
The Riemann–Stieltjes integral is unique
Statement
For fixed , at most one real number satisfies the mesh-limit condition defining .
Facts & Assumptions
Given: Two reals satisfying the defining mesh condition for the same functions .
The mesh-limit condition quantifies over every sufficiently fine tagged partition (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).
Uniform partitions have mesh for every natural (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, The canonical natural of a field).
Reciprocal naturals become arbitrarily small (For every in a complete ordered field there is a natural with ).
and exactly when (The triangle inequality, Basic properties of the absolute value).
Proof
Given , choose positive thresholds for error in the two mesh conditions. By [L3] choose a natural whose uniform partition has mesh smaller than both thresholds, and give it arbitrary tags.
For its sum , . Since this holds for every , and . The singleton interval has only the prescribed value .
Refinement and tag-change estimates for Stieltjes sums
Statement
Let have bounded variation on , let be a partition, and let refine . If the oscillation of on is at most , then any tagged sum on and any tagged sum on satisfy
In particular, if every , the bound is . Two tagged sums on arbitrary partitions whose intervals all have oscillation at most differ by at most .
Facts & Assumptions
Given: Functions , a partition , a refinement , and tags on both.
Stieltjes sums are weighted finite sums of integrator increments (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).
Total variation bounds every sum of absolute increments and is additive on adjacent subintervals (Bounded variation and total variation on an interval, Total variation is additive over adjacent subintervals and decreases under restriction).
A refinement groups its subintervals inside the coarse ones, and any two partitions have a common refinement (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite sums group and telescope (Finite sums and finite products, by recursion, Laws of finite sums and finite products).
The absolute value of a finite sum is at most the sum of absolute values (The triangle inequality).
Proof
Inside one coarse interval , the refined integrator increments telescope to . Subtract the coarse term by assigning its tag value to every refined increment. Each coefficient difference has absolute value at most , so the absolute difference contributed by that block is at most times the sum of the absolute refined increments, hence at most .
Summing step 1.1 over the coarse blocks proves the first estimate. If , additivity of variation gives the uniform bound. The conclusions remain when the variation or the interval is .
For two arbitrary partitions, pass to their common refinement and apply the uniform estimate once from each original sum to the refined sum. The triangle inequality gives the factor .
A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator
Statement
If is continuous and has bounded variation, then exists.
Facts & Assumptions
Given: A continuous and a BV function .
The refinement estimate bounds differences of tagged sums by oscillation times total variation (Refinement and tag-change estimates for Stieltjes sums).
A Stieltjes mesh limit, when it exists, is unique (The Riemann–Stieltjes integral is unique, Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).
Total variation is finite for a BV function (Bounded variation and total variation on an interval).
A continuous real function on a compact interval is uniformly continuous (Heine-Cantor in : a continuous real function on a compact subset of is uniformly continuous, proved -natively from sequential compactness, Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point).
Uniform partitions of arbitrarily small mesh exist, and common refinements exist (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Proof
Put . If , every integrator increment is , so every Stieltjes sum is and the integral exists. Assume . Given , uniform continuity gives such that implies . Any two tagged partitions of mesh below have all local oscillations below ; comparing both with their common refinement through [L1] makes their sums differ by less than .
Choose one tagged uniform partition with mesh tending to for each natural index and call its sum . Step 1.1 makes Cauchy, so [L5] gives a real limit .
Given , choose the threshold in step 1.1 for error and then a uniform sum beyond that threshold with . Every arbitrary tagged sum with sufficiently small mesh differs from by less than , hence is within of . This is the mesh-limit definition, and [L2] identifies the unique value.
The total-variation bound for a Riemann–Stieltjes integral
Statement
Suppose exists, has bounded variation, and on . Then
Facts & Assumptions
Given: An existing Stieltjes integral, a BV integrator , and a bound .
Stieltjes sums converge to the integral in the mesh sense (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, The Riemann–Stieltjes integral is unique).
Every sum of absolute integrator increments is bounded by total variation (Bounded variation and total variation on an interval).
Finite sums and the triangle inequality give (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality, Basic properties of the absolute value).
Non-strict inequalities pass to limits (Limits preserve non-strict inequalities).
Proof
Every tagged sum satisfies .
Take a sequence of tagged partitions with mesh tending to . Their sums converge to the integral by [L1], and [L4] passes the bound in step 1.1 to the limit. Orientation and the singleton case preserve the same absolute-value inequality.
Linearity and interval additivity of the Riemann–Stieltjes integral
Statement
Whenever the integrals on the right exist,
Let , suppose has bounded variation, and suppose is continuous at . Then integrability on is equivalent to integrability on both and , and
Facts & Assumptions
Given: Functions for which the displayed integrals are defined, scalars , and for additivity a BV integrator and a cut where is continuous.
Stieltjes integrability is convergence of all sufficiently fine tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).
Such a limit is unique (The Riemann–Stieltjes integral is unique).
Partitions can be inserted at and spliced across (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite sums distribute and split at an index (Finite sums and finite products, by recursion, Laws of finite sums and finite products).
Continuity at makes close to when is close to (Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point).
Reversal and singleton conventions are those of the oriented integral (The integral with oriented limits: and ).
The sum of the absolute integrator increments over any partition is at most (Bounded variation and total variation on an interval).
Every Cauchy sequence of real sums has a finite real limit (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).
Proof
Each tagged sum is exactly linear in and in , by distribution in the finite sum. Passing to mesh limits and using uniqueness proves both linearity formulas.
For a partition containing , its Stieltjes sum splits exactly into the sums on the two subintervals. Inserting into a fine partition changes only the interval containing . Direct subtraction bounds the difference, for any choices of the old and new tags, by the oscillation of near times the sum of the relevant absolute increments of , hence by that oscillation times . Continuity of at makes this error tend to zero with the mesh.
If the whole-interval integral exists, take any two sufficiently fine sums on and splice each with the same sufficiently fine sum on . The two whole-interval sums are close, so their common right part cancels and the left sums are Cauchy. Choose uniform left-hand sums with mesh tending to zero; they form a Cauchy sequence and have a limit by [L8]. Every arbitrary sufficiently fine left-hand sum is close to a sufficiently late uniform one, so the entire left-hand mesh family has that limit. The symmetric argument gives the right integral. Conversely, if both restricted integrals exist, splice their fine sums and use step 1.2 to compare with arbitrary whole-interval sums. The exact split gives the displayed value by [L2]. Endpoint cuts and reversed limits follow from [L6].
Riemann–Stieltjes integration by parts
Statement
The integral exists if and only if exists. When either exists,
Facts & Assumptions
Given: Functions .
Riemann-Stieltjes integrability is the common mesh limit of tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).
A Stieltjes integral is unique (The Riemann–Stieltjes integral is unique).
Partitions and endpoint tags are permitted tagged partitions (Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).
Finite summation by parts exchanges a sequence and its successive increments (Abel summation by parts: with one has for every , Finite sums and finite products, by recursion, Laws of finite sums and finite products).
Proof
For a partition , finite summation by parts gives the exact identity .
Suppose exists and consider an arbitrary tagged sum . Refine each by inserting its tag . On tag the complementary sum at , and on tag it at . Direct expansion on the th interval gives [step 1.1, L1, L2, L3, L4] The refined mesh does not exceed , so the complementary sums converge to . Telescoping the displayed identities forces every fine tagged sum for to converge to the endpoint product minus that integral.
Exchanging and proves the converse. Adding the two values yields the displayed formula, including the singleton and reversed-orientation cases.
A bounded-variation integrand is Riemann–Stieltjes integrable against every continuous integrator
Statement
If has bounded variation and is continuous, then exists.
Facts & Assumptions
Given: A BV function and a continuous function on .
A continuous integrand is Stieltjes integrable against a BV integrator (A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator).
Existence of is equivalent to existence of (Riemann–Stieltjes integration by parts).
Bounded variation and continuity are those of Bounded variation and total variation on an interval and Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point.
Proof
Since is continuous and is BV, [L1] gives the integral .
Integration by parts [L2] then gives existence of and its value .
A bounded function with finitely many discontinuities is Stieltjes integrable against a continuous bounded-variation integrator
Statement
Let be bounded and have only finitely many discontinuities. If is continuous and has bounded variation, then exists.
Facts & Assumptions
Given: A bounded with finite discontinuity set , and a continuous BV integrator on .
A BV function is the difference of two nondecreasing functions (Jordan decomposition for functions of bounded variation).
The canonical monotone summands of a continuous BV function are continuous (The jumps of a variation function equal the absolute jumps of the original function).
For , a bounded integrand and a nondecreasing integrator, mesh integrability is equivalent to continuity of the integrand at every discontinuity of the integrator together with the weighted oscillation criterion (Darboux criterion for Riemann–Stieltjes integrability with a nondecreasing integrator).
Continuous functions on compact intervals are uniformly continuous (Heine-Cantor in : a continuous real function on a compact subset of is uniformly continuous, proved -natively from sequential compactness).
Stieltjes integration is linear in the integrator (Linearity and interval additivity of the Riemann–Stieltjes integral).
Continuity makes the increment of arbitrarily small on sufficiently short intervals around each point (Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point).
Proof
If the integral is by the definition of the Riemann-Stieltjes integral on a singleton interval and there is nothing to prove, so assume , which is the standing hypothesis of [L3].
First suppose that is continuous and nondecreasing. Write . Given , [L6] and finiteness of allow pairwise disjoint closed intervals about the points whose total -increment is less than .
On the compact complement of the interiors of the , the function is continuous and hence uniformly continuous by [L4]. Choose a partition containing all endpoints of the and fine enough that every remaining partition interval has oscillation below . The intervals meeting contribute at most times their total -increment, and all other intervals contribute less than . After rescaling the two preliminary bounds, the weighted oscillation sum is arbitrarily small, so [L3] gives .
For a general continuous BV , [L1] writes . Both and are continuous by [L2]. Step 2.1 gives integrability against each, and linearity in the integrator [L5] gives integrability against .
A continuous function of a Stieltjes-integrable function is Stieltjes integrable for a nondecreasing integrator
Statement
Suppose is nondecreasing, is bounded and Riemann–Stieltjes integrable with respect to , and is continuous on a compact interval containing . Then is Riemann–Stieltjes integrable with respect to .
Facts & Assumptions
Given: A nondecreasing , a bounded , and a continuous on a compact interval containing the range of .
For , bounded and nondecreasing , integrability in the mesh sense is equivalent to the conjunction of two conditions: is continuous at every discontinuity of , and for every some partition has (Darboux criterion for Riemann–Stieltjes integrability with a nondecreasing integrator).
The function is bounded and uniformly continuous on its compact domain (A continuous real function on a compact subset of is bounded, Heine-Cantor in : a continuous real function on a compact subset of is uniformly continuous, proved -natively from sequential compactness).
Finite sums may be split and estimated termwise (Laws of finite sums and finite products).
Proof
Choose with . Given , uniform continuity supplies such that implies .
By [L1], choose a partition for which . Split its intervals into those with and the rest. The first class contributes less than to the weighted oscillation sum of . In the second class, , while ; hence it too contributes less than .
Thus the weighted oscillation condition in [L1] holds for . The same theorem says that is continuous at every discontinuity of ; continuity of makes continuous there as well. Both clauses of [L1] now give .
Every bounded-variation function is uniformly approximable by step functions
Statement
If has bounded variation, then for every there is a finite step function with . Endpoint values of may be prescribed to equal those of .
More precisely, if is at most countable and is continuous at every point of , the interior breakpoints of may all be chosen outside .
Facts & Assumptions
Given: A BV function , a tolerance , and, for the strengthened assertion, an at most countable set of continuity points of .
A BV function has finite one-sided limits at every point (A bounded-variation function has at most countably many discontinuities, all of the first kind, The left and right limits of at , as limits of the restrictions of to and ).
Every nonempty subset of the reals that is bounded above has a supremum (Complete ordered field (least-upper-bound property)).
Every nonempty open interval is uncountable (Every nondegenerate interval of is uncountable).
Proof
Fix . Let be the set of for which there is a finite chain such that the oscillation of on every open interval is below . The set contains and is bounded above by , so exists by [L2].
Suppose . The left limit at and the right limit at supplied by [L1] give one-sided intervals on which the oscillation is below . Choose in the left interval (use if ), append to its chain if necessary, and then append a point in the right interval. This puts in , contradicting that is an upper bound. Hence . The left limit at now lets a chain ending sufficiently near be extended to . Thus there is a finite partition of on each of whose open components the oscillation of is below . The singleton case is immediate.
On each open component choose one value of , and at every partition point assign the actual value of . The resulting finite step function differs from by less than everywhere. Taking leaves room for the strengthened construction.
Only finitely many interior breakpoints lie in . Around each such breakpoint , continuity of gives a small two-sided interval, disjoint from the corresponding intervals for the other breakpoints, on which the oscillation is below . By [L3] choose a replacement point outside in that interval and between the neighboring breakpoints. Moving the breakpoint adds only a subinterval from this continuity neighborhood to either adjacent component; the original component oscillation is below , and a component can be enlarged at both ends, so its new oscillation is at most , the three pieces overlapping at the old breakpoints. Sampling again and retaining the actual values at all breakpoints and endpoints gives the required approximation with every interior breakpoint outside .
Two bounded-variation functions with no common discontinuity are Riemann–Stieltjes integrable
Statement
Let have bounded variation. If no point is a discontinuity of both functions, then exists.
Facts & Assumptions
Given: BV functions and with disjoint discontinuity sets.
The discontinuity set is at most countable (A bounded-variation function has at most countably many discontinuities, all of the first kind).
A BV function can be approximated uniformly by step functions whose breakpoints avoid a prescribed countable set of its continuity points (Every bounded-variation function is uniformly approximable by step functions).
Direct subtraction of two Stieltjes sums and the finite-sum triangle inequality give (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, Bounded variation and total variation on an interval, Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality).
Every Cauchy sequence of reals converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).
If is continuous at a point, its variation function is continuous there; refinement errors are bounded by local variation times local oscillation (The jumps of a variation function equal the absolute jumps of the original function, Refinement and tag-change estimates for Stieltjes sums).
Proof
By the no-common-discontinuity hypothesis, is continuous at every point of . For each , [L1] and [L2] provide a finite step function with and all interior breakpoints outside . If an endpoint belongs to , continuity of there permits the value on the adjacent open component to be changed to at that endpoint while retaining the same bound after beginning with tolerance . Thus is continuous at every point of .
Each is integrable with respect to . Its finitely many discontinuities are points where is continuous by step 1.1. By [L5], choose disjoint neighborhoods of those points whose total local variation is small. Outside them is locally constant, while inside them [L5] bounds differences between fine sums by the small local variation times the finite oscillation of . Hence the fine sums are Cauchy. Choose a sequence of uniform tagged sums with mesh tending to zero; [L4] gives its limit, and comparison with a sufficiently late member of this sequence shows that every sufficiently fine tagged sum has the same limit.
Given , choose so that (the zero-variation case is immediate), and then choose a mesh bound making any two sums of differ by less than . By [L3], replacing by in either sum changes it by at most . Hence all sufficiently fine sums of are Cauchy. Choose uniform tagged sums with mesh tending to zero; their sums form a Cauchy sequence and converge by [L4]. Comparing an arbitrary sufficiently fine sum with a late uniform sum proves convergence of the whole mesh family to that sequential limit, which is exactly the defining Stieltjes integral.
The identity integrator recovers the Riemann integral
Statement
Let . For the identity function , a bounded is Riemann–Stieltjes integrable with respect to exactly when it is Riemann integrable, and then
For both sides are by the singleton conventions. For reversed endpoints the statement is about a function defined on the sorted interval: if and is bounded on , then both oriented conventions negate the corresponding sorted integral, so the equality is inherited from the case just proved. The hypothesis is needed for the displayed clause itself, because is empty when and a function typed on it supplies no values to integrate.
Facts & Assumptions
Given: A bounded function on the compact interval with endpoints .
A Stieltjes sum is (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).
A Riemann sum is (Tagged partitions of , with a tag in each subinterval, and the Riemann sum ).
Darboux and tagged-sum Riemann integrability agree (The Darboux and Riemann definitions agree: a bounded on is Darboux integrable with integral if and only if for every real there is a real such that for every tagged partition of mesh below ).
Oriented integrals reverse sign when endpoints are interchanged (The integral with oriented limits: and ).
Proof
When and , every increment equals . Thus [L1] and [L2] are termwise identical for every tagged partition.
Consequently the two mesh limits exist simultaneously and have the same value; [L3] identifies that tagged limit with the Darboux integral. For both conventions give zero, and [L4] handles .
A continuously differentiable integrator reduces Stieltjes integration to ordinary integration
Statement
Let be Riemann integrable. Suppose is continuous on , differentiable on , and extends continuously to . Then is Riemann–Stieltjes integrable with respect to and
Facts & Assumptions
Given: A Riemann-integrable and an integrator with continuous derivative on the compact interval.
The mean value theorem represents every increment of as (The mean value theorem, as the case of Cauchy's: for continuous on with and differentiable on there is with ).
The continuous function is uniformly continuous (Heine-Cantor in : a continuous real function on a compact subset of is uniformly continuous, proved -natively from sequential compactness).
The continuous function is Riemann integrable, so the product is Riemann integrable (A continuous function on is Riemann integrable, by Heine-Cantor and Riemann's criterion, If are integrable on then so are , , , and , and ).
Riemann tagged sums converge to the Darboux integral (The Darboux and Riemann definitions agree: a bounded on is Darboux integrable with integral if and only if for every real there is a real such that for every tagged partition of mesh below ).
Proof
A Riemann-integrable function is bounded; choose with . For each partition interval, [L1] gives such that . Hence [L1]
By [L2], the absolute value of the right side is at most , which tends to zero with the mesh. By [L3] and [L4], the second sum in step 1.1 tends to . Thus all Stieltjes sums have the same limit, proving both existence and the formula.
A countable pure-step integrator evaluates a continuous integrand as the absolutely convergent weighted sum of its values at the jumps
Statement
Let . Write for the unit step for and for . Let be points of the open interval , and let be reals with and convergent.
Then for every the series converges, so
defines a nondecreasing , which therefore has bounded variation.
For every continuous the integral exists, the series converges absolutely, and
The points are not required to be distinct, and any may be zero.
Facts & Assumptions
Given: Reals , points , reals with convergent, and a continuous .
A nondecreasing sequence of reals whose range is bounded above converges, with limit the supremum of its range (A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum); a series converges when its sequence of partial sums converges (Series, partial sums, convergence and the sum, divergence, and the tail series).
A real function on has bounded variation if and only if it is a difference of two nondecreasing functions (Jordan decomposition for functions of bounded variation); the total variation is the supremum of the partition sums (Bounded variation and total variation on an interval).
If is continuous and has bounded variation, then exists (A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator).
Whenever the integrals on the right exist, (Linearity and interval additivity of the Riemann–Stieltjes integral).
If exists, has bounded variation, and on , then (The total-variation bound for a Riemann–Stieltjes integral).
The closed bounded interval is compact (Heine-Borel by bisection: every closed bounded interval is compact), and a continuous real function is bounded on a compact subset of its domain: there is with throughout (A continuous real function on a compact subset of is bounded, Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point).
The Riemann–Stieltjes sum of a tagged partition is , and means that for every some makes for every tagged partition of mesh below (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, Partition of as a finite strictly increasing list , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, Tagged partitions of , with a tag in each subinterval, and the Riemann sum ).
If for all large and converges, then converges (If eventually, convergence of gives convergence of , and divergence of gives divergence of ); a series converging absolutely converges (If converges then converges).
Continuity of at means that for every there is with whenever lies in the domain and (Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point); convergence of a real sequence is the usual –threshold condition (Limits and Cauchy sequences of reals).
Proof
Fix . Each term lies in , so the partial sums of are nondecreasing and bounded above by . By [L1] the series converges and is defined, with .
Fix and put on . Then is nondecreasing, hence of bounded variation by [L2]. Let and take from [L9] for at , so that whenever . Let be a tagged partition of mesh below . The increment is when and otherwise, and because exactly one index satisfies . Hence for that index, and with give , so . By [L7], .
If then for every , because is nondecreasing and . Multiplying by and summing, every partial sum for is at most the corresponding partial sum for , so the limits satisfy by [L1]. Thus is nondecreasing, and exhibits it as a difference of two nondecreasing functions, so [L2] gives bounded variation.
For set , a finite sum. Each summand is a nonnegative multiple of a function of the form treated in step 1.2, so applying [L4] finitely many times, with the integral of each summand supplied by step 1.2, gives .
By [L6] there is with on . Since and converges, [L8] makes absolutely convergent, hence convergent. By step 2.1 and [L3] the integral exists.
Set . For each , , the tail of the series in step 1.1; the argument of steps 1.1 and 2.1 applies verbatim to it, so is nondecreasing with bounded variation. Since we have and , so and . A nondecreasing function has every partition sum equal to , because each increment is nonnegative and the sum telescopes, so [L2] gives .
Both and are of bounded variation, so [L3] makes and exist, and with [L4] gives . Using step 2.2 and then [L5] with the bound of step 3.1,
Convergence of makes its tails tend to as increases, so given the right side of step 4.1 is below for all large . Hence the partial sums converge to , and by step 3.1 that series converges absolutely. By [L1] and [L9] its sum is , which is the claimed identity.
Remark
The two endpoints behave differently, which is why the jumps are confined to the open interval. A jump at would be harmless: vanishes only at , the increment still records the whole weight, and step 2.1 goes through unchanged because its counting argument needs only . A jump at genuinely breaks the identity: for every , so such a term contributes nothing at all to , yet it would contribute to the right-hand sum. The hypothesis excludes that case, and it is the hypothesis Rudin states.
Rudin's Theorem 6.16 additionally requires the to be distinct. Nothing in the proof above uses distinctness, so it is not assumed here.
Continuity of is not decorative. cex-common-jump-prevents-riemann-stieltjes-integrability exhibits an and an sharing a single jump for which no mesh limit exists, and a single step integrator is exactly the of that counterexample.
Change of variable for the Riemann–Stieltjes integral
Statement
Let and , and let be a strictly increasing continuous bijection. For functions , one of the two Riemann–Stieltjes integrals below exists if and only if the other does, and in that case
The nondegeneracy hypotheses are not cosmetic. If and both integrals are by the singleton convention and the identity holds trivially, but no partition exists and the argument below does not apply. If the written endpoints are reversed the intervals are empty, the empty map is vacuously such a bijection, and typed on an empty interval give the displayed integrals no values; that case is excluded rather than asserted.
Facts & Assumptions
Given: A strictly increasing continuous bijection and functions on .
Both and are uniformly continuous on their compact domains (Heine-Cantor in : a continuous real function on a compact subset of is uniformly continuous, proved -natively from sequential compactness).
The Stieltjes integral is the common mesh limit of its tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).
A Stieltjes integral, when it exists, is unique (The Riemann–Stieltjes integral is unique).
Proof
If is a partition of with tags , then is a partition of with tags . Direct substitution gives [given]
By uniform continuity of in [L2], arbitrarily fine give arbitrarily fine image partitions . Thus existence of the left-hand integral in the displayed formula forces the right-hand sums to converge to the same value. Applying the identical argument to , using [L1] and [L2], proves the converse. Uniqueness [L4] identifies the two limits.
Young's partition estimate for rational Hölder exponents
Statement
Let satisfy . Suppose and on . If is the partition into equal intervals and is its left-endpoint Stieltjes sum, then
Put and If refines an arbitrary partition , then their left-endpoint sums satisfy
Facts & Assumptions
Given: Rational Hölder exponents with , Hölder constants , and the stated partitions.
Rational powers are monotone and satisfy the exponent laws (Rational powers of a positive base, Laws of rational exponents, Monotonicity of and of ).
A geometric series with ratio in converges and its tails tend to zero (For , , and for the series diverges).
Finite sums obey the triangle inequality and may be regrouped (Laws of finite sums and finite products, The triangle inequality).
Proof
Insert a point between adjacent points . The change from the old left-endpoint term to the two new terms is [given] up to sign. Its absolute value is at most , hence at most by [L1].
Passing from to inserts one midpoint in each of intervals of length . Summing step 1.1 gives the first displayed bound. More generally, if a partition of an interval has subintervals, some interior point has two adjacent lengths whose sum is at most : the sum of all such two-interval lengths is at most . Removing that point therefore changes the left sum by at most .
Remove the extra points of inside a fixed interval of , one at a time, always using step 2.1. The total error is at most . Grouping the positive integers into bounds this series by via [L2]. Thus the error on is at most . Summing over and using proves the refinement estimate. If or , every error is zero.
Young's Riemann–Stieltjes existence theorem for rational Hölder exponents
Statement
Let with . If is -Hölder and is -Hölder, then both and exist. They satisfy
Facts & Assumptions
Given: Hölder functions with rational exponents whose sum exceeds one.
The Young partition estimate controls refinement errors by a constant times (Young's partition estimate for rational Hölder exponents).
Every Cauchy sequence of real numbers converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).
A Stieltjes integral is the common limit of all sufficiently fine tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).
Such a limit is unique (The Riemann–Stieltjes integral is unique).
Tails of a geometric series with ratio in tend to zero (For , , and for the series diverges).
Proof
For the dyadic left sums , the first estimate in [L1] and the geometric-tail fact [L5] make a Cauchy sequence. It therefore converges to some by [L2].
Given a partition , compare it and a sufficiently fine dyadic partition with their common refinement. The second estimate in [L1] bounds the two refinement errors by a constant times . Together with , this shows that every sufficiently fine left-endpoint sum is close to . Replacing a left endpoint by an arbitrary tag changes the th term by at most ; the total is at most . Thus every fine tagged sum tends to , and [L3] gives . Interchanging and gives .
On every partition, the right-endpoint sum for plus the left-endpoint sum for telescopes exactly to . Passing to the two limits established in step 2.1 proves the formula.
Conventions and proved scope for bounded variation and Stieltjes integration
Statement
Total variation is zero on a singleton, and both ordinary and Riemann–Stieltjes integrals use the oriented convention when endpoints are reversed. Absolute continuity here is formulated with finite disjoint families of intervals.
On a nondegenerate interval with , and for a nondecreasing integrator, the weighted Darboux condition matches the all-fine-mesh definition only together with continuity of the integrand at the integrator's discontinuities; this extra compatibility is vacuous for a continuous integrator. The hypothesis is part of the statement and not cosmetic: on the integral is by convention, so every bounded integrand is integrable there, while a singleton interval admits no partition at all and so the Darboux condition fails; a consumer needing reads the value off the definition instead. A general BV integrator is handled through Jordan decomposition or tagged sums. Finite-step integrators turn the integral of a continuous integrand into a weighted evaluation sum over the jumps, while continuously differentiable integrators reduce the integral of a Riemann-integrable integrand to an ordinary integral against the derivative. The no-common-discontinuity theorem is sharp in view of the companion common-jump counterexample. Young's theorem is proved here only for rational Hölder exponents because arbitrary real exponents are not available at this point in the reading order; the later Real powers for positive bases, with the zero-base positive-exponent convention ↗ is what supplies them. No Lebesgue–Stieltjes measure, almost-everywhere differentiability theorem, or arbitrary-real-exponent Stieltjes theorem is asserted on this page.
5 · Examples, counterexamples and false statements
None yet.
Sources
- William F. Trench, Introduction to Real Analysis, Ch. 3
- Christopher Heil, Absolute Continuity and the Banach-Zaretsky Theorem
- William F. Trench, Introduction to Real Analysis, Definition 3.1.5
- W. Rudin, Principles of Mathematical Analysis, Ch. 6
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.6
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.8
- William F. Trench, Introduction to Real Analysis, Exercise 3.2.9
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.12
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.22
- William F. Trench, Introduction to Real Analysis, Exercise 3.2.8
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.9
- William F. Trench, Introduction to Real Analysis, Exercise 3.2.10
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.10
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.11
- William F. Trench, Introduction to Real Analysis, Section 3.2
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Definition 6.1
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.17
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.16
- W. Rudin, Principles of Mathematical Analysis, Ch. 6, Theorem 6.19
- Nourdin, Nualart, and Peccati, The Breuer–Major theorem in total variation: improved rates under minimal regularity, Section 2.2