Bounded Variation and the Riemann–Stieltjes Integral

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

Bounded variation and total variation on an interval

Definition

Let $a\le b$ and let $f:[a,b]\to\mathbb R$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length). If $a<b$ and $P=(n,t)$ is a partition of $[a,b]$ (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), the variation of $f$ over $P$ is

$V(f,P):=\sum_{i<n}|f(t_{i+1})-f(t_i)|.$

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 $[a,b]$ has the partition with point set $\{a,b\}$ . The function $f$ has bounded variation on $[a,b]$ when this set of sums is bounded above (Lower bound, bounded below, bounded set). In that case its total variation is

$\operatorname{Var}_{[a,b]}(f):=\sup_P V(f,P).$

Completeness of $\mathbb R$ 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, $\operatorname{Var}_{[a,a]}(f):=0$ ; no partition from 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, whose standing hypothesis is $a<b$ , is invoked.

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Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant

Statement

Let $f:[a,b]\to\mathbb R$ have bounded variation. Then

$|f(y)-f(x)|\le\operatorname{Var}_{[a,b]}(f)$ for all $x,y\in[a,b]$ ;
$f$ is bounded on $[a,b]$ ;
$\operatorname{Var}_{[a,b]}(f)=0$ if and only if $f$ is constant.

These claims include the singleton interval $a=b$ .

Facts & Assumptions

Given: Reals $a\le b$ and a bounded-variation function $f:[a,b]\to\mathbb R$ .

[L1]

Total variation is the supremum of the partition sums $V(f,P)$ , with value $0$ on a singleton interval (Bounded variation and total variation on an interval).

[L2]

A point of an interval can be inserted into a partition without deleting its existing points (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 of nonnegative terms dominate every term (Laws of finite sums and finite products).

[L4]

$|u+v|\le |u|+|v|$ in an ordered field (The triangle inequality).

[L5]

A subset of $\mathbb R$ is bounded when the absolute values of its members have a common real bound (Lower bound, bounded below, bounded set).

Proof

technique · direct

1.1

If $a<b$ and $x<y$ lie in $[a,b]$ , insert $x$ and $y$ into the endpoint partition. The resulting partition sum contains $|f(y)-f(x)|$ as a nonnegative term, so $|f(y)-f(x)|\le V(f,P)\le\operatorname{Var}_{[a,b]}(f)$ . The same inequality is $0\le\operatorname{Var}_{[a,b]}(f)$ when $x=y$ , and when $a=b$ only that case occurs.

L1L2L3

2.1

Put $M:=|f(a)|+\operatorname{Var}_{[a,b]}(f)$ . For $x\in[a,b]$ , $|f(x)|\le |f(x)-f(a)|+|f(a)|\le M$ , so $f([a,b])$ is bounded.

step 1.1L4L5

3.1

If the total variation is $0$ , step 1.1 gives $|f(y)-f(x)|=0$ for every $x,y$ , hence $f$ is constant. Conversely, if $f$ is constant then every increment in every partition sum is $0$ , so every sum and its supremum are $0$ ; the singleton convention gives the same conclusion when $a=b$ .

step 1.1L1L3∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Total variation is additive over adjacent subintervals and decreases under restriction

Statement

If $a\le c\le b$ and $f:[a,b]\to\mathbb R$ , then $f$ has bounded variation on $[a,b]$ if and only if its restrictions have bounded variation on $[a,c]$ and $[c,b]$ . In that case

$\operatorname{Var}_{[a,b]}(f)=\operatorname{Var}_{[a,c]}(f)+\operatorname{Var}_{[c,b]}(f).$

Consequently restriction to any subinterval cannot increase total variation.

Facts & Assumptions

Given: Reals $a\le c\le b$ and a function $f:[a,b]\to\mathbb R$ .

[L1]

Total variation is the supremum of finite variation sums, and singleton variation is $0$ (Bounded variation and total variation on an interval).

[L2]

A partition is a finite strictly increasing endpoint list, and a refinement contains every point of the original partition (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 telescope and split at an index (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

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).

[L5]

Closed subintervals are the sets $[u,v]=\{x:u\le x\le v\}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct

1.1

Given partitions $P_1$ of $[a,c]$ and $P_2$ of $[c,b]$ , splice their point lists at $c$ . The resulting partition $P$ of $[a,b]$ satisfies $V(f,P)=V(f,P_1)+V(f,P_2)$ . Thus, whenever $f$ is BV on $[a,b]$ , each restriction is BV and $\operatorname{Var}_{[a,c]}(f)+\operatorname{Var}_{[c,b]}(f)\le\operatorname{Var}_{[a,b]}(f)$ . The assertion is unchanged when $c=a$ or $c=b$ , because the singleton term is $0$ .

L1L2L3L5

1.2

Conversely, insert $c$ into an arbitrary partition $P$ of $[a,b]$ . The refined sum splits into a sum on $[a,c]$ and one on $[c,b]$ , 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 $V(f,P)\le\operatorname{Var}_{[a,c]}(f)+\operatorname{Var}_{[c,b]}(f)$ .

L1L2L3

2.1

Taking the supremum over $P$ in step 1.2 and combining it with step 1.1 proves the equality and the equivalence of boundedness. Applying the equality twice to $[u,v]\subseteq[a,b]$ shows $\operatorname{Var}_{[u,v]}(f)\le\operatorname{Var}_{[a,b]}(f)$ .

step 1.1step 1.2L1L4L5∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

Variation function and positive and negative variations

Definition

Let $f:[a,b]\to\mathbb R$ have bounded variation (Bounded variation and total variation on an interval). Its variation function is

$V_f(x):=\operatorname{Var}_{[a,x]}(f)\qquad(x\in[a,b]),$

so $V_f(a)=0$ . 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 $f$ are

$P_f(x):=\frac{V_f(x)+f(x)-f(a)}2,\qquad N_f(x):=\frac{V_f(x)-f(x)+f(a)}2.$

The names refer to upward and downward accumulated variation; their monotonicity and the resulting decomposition are proved separately.

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The positive and negative variations are nondecreasing and give the Jordan identities

Statement

For a bounded-variation function $f:[a,b]\to\mathbb R$ , the functions $P_f$ and $N_f$ are nondecreasing and

$f(x)=f(a)+P_f(x)-N_f(x),\qquad V_f(x)=P_f(x)+N_f(x).$

Both $P_f(a)$ and $N_f(a)$ are $0$ .

Facts & Assumptions

Given: A bounded-variation function $f:[a,b]\to\mathbb R$ and its functions $V_f,P_f,N_f$ .

[L1]

$V_f,P_f,N_f$ are defined by the displayed formulas in Variation function and positive and negative variations.

[L2]

For $x\le y$ , $V_f(y)-V_f(x)=\operatorname{Var}_{[x,y]}(f)$ (Total variation is additive over adjacent subintervals and decreases under restriction).

[L3]

$|f(y)-f(x)|\le\operatorname{Var}_{[x,y]}(f)$ (Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant).

[L4]

A function is nondecreasing when $x\le y$ implies $g(x)\le g(y)$ (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$ , with the dictionary to monotone sequences).

[L5]

$|u|\ge u$ and $|u|\ge -u$ (Absolute value in an ordered field).

Proof

technique · direct

1.1

For $x\le y$ , [L2] and [L3] give $V_f(y)-V_f(x)\ge|f(y)-f(x)|$ , hence this difference is at least both $f(y)-f(x)$ and $f(x)-f(y)$ . Therefore $P_f(y)-P_f(x)\ge0$ and $N_f(y)-N_f(x)\ge0$ , so both functions are nondecreasing.

L1L2L3L4L5

1.2

Adding and subtracting the defining formulas gives $P_f(x)+N_f(x)=V_f(x)$ and $P_f(x)-N_f(x)=f(x)-f(a)$ . At $x=a$ , $V_f(a)=0$ , so $P_f(a)=N_f(a)=0$ .

L1algebra

2.1

Rearranging the second identity in step 1.2 gives $f=f(a)+P_f-N_f$ , while the first is the asserted variation identity.

step 1.2algebra∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Jordan decomposition for functions of bounded variation

Statement

A real function $f$ on $[a,b]$ has bounded variation if and only if it is a difference of two nondecreasing functions. If $f(a)=0$ , the canonical normalized decomposition is $f=P_f-N_f$ . More generally $f=f(a)+P_f-N_f$ .

It is minimal: if $f=f(a)+g-h$ with nondecreasing $g,h$ and $g(a)=h(a)=0$ , then $P_f(x)\le g(x)$ and $N_f(x)\le h(x)$ for every $x$ .

Facts & Assumptions

Given: A function $f:[a,b]\to\mathbb R$ .

[L1]

For BV $f$ , $P_f,N_f$ are nondecreasing, normalized at $a$ , and $f=f(a)+P_f-N_f$ (The positive and negative variations are nondecreasing and give the Jordan identities).

[L2]

Total variation is the supremum of sums of absolute increments (Bounded variation and total variation on an interval).

[L3]

Nondecreasing means that each forward increment is nonnegative (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$ , with the dictionary to monotone sequences).

[L4]

A partition is a finite increasing point list (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).

[L5]

Finite sums telescope and distribute over addition (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

$|u-v|\le |u|+|v|$ (The triangle inequality).

Proof

technique · direct

1.1

If $f$ is BV, [L1] immediately supplies the stated difference of nondecreasing functions, with the asserted normalization.

L1

1.2

Conversely suppose $f=f(a)+g-h$ with $g,h$ nondecreasing. For a partition $P=(n,t)$ , every forward increment of $g$ and $h$ is nonnegative, so $|f(t_{i+1})-f(t_i)|\le(g(t_{i+1})-g(t_i))+(h(t_{i+1})-h(t_i))$ . Summing and telescoping gives $V(f,P)\le g(b)-g(a)+h(b)-h(a)$ , independent of $P$ ; hence $f$ is BV.

L2L3L4L5L6

2.1

Now assume the decomposition is normalized. On $[a,x]$ , step 1.2 gives $V_f(x)\le g(x)+h(x)$ , while $f(x)-f(a)=g(x)-h(x)$ . Adding these inequalities and dividing by $2$ yields $P_f(x)\le g(x)$ ; subtracting the increment identity from the variation inequality yields $N_f(x)\le h(x)$ .

step 1.2L1L2algebra∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A bounded-variation function has at most countably many discontinuities, all of the first kind

Statement

If $f:[a,b]\to\mathbb R$ has bounded variation, every well-posed one-sided limit of $f$ 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 $f:[a,b]\to\mathbb R$ .

[L1]

Jordan decomposition writes $f=f(a)+P_f-N_f$ with $P_f,N_f$ nondecreasing (Jordan decomposition for functions of bounded variation).

[L2]

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 $c$ is a discontinuity exactly when $\lim_{x \to c^{-}} f(x) < \lim_{x \to c^{+}} f(x)$ , The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ ).

[L3]

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 $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used).

[L4]

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

technique · direct

1.1

Apply [L2] to $P_f$ and $N_f$ . 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 $f=f(a)+P_f-N_f$ . Thus $f$ has no discontinuity of the second kind.

L1L2L4

1.2

If both $P_f$ and $N_f$ are continuous at a point, [L4] makes $f$ continuous there. Hence the discontinuity set of $f$ is contained in the union of the two component discontinuity sets.

L1L4

2.1

Each component discontinuity set is at most countable by [L3]. Given injections of them into $\mathbb N$ , 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 $\mathbb N$ . Step 1.2 therefore makes the discontinuity set of $f$ at most countable, and step 1.1 makes every one of its discontinuities first-kind.

step 1.1step 1.2L3algebra∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

The jumps of a variation function equal the absolute jumps of the original function

Statement

Let $f:[a,b]\to\mathbb R$ have bounded variation and let $V_f(x)=\operatorname{Var}_{[a,x]}(f)$ . At an interior point $c$ ,

$V_f(c+)-V_f(c)=|f(c+)-f(c)|,\qquad V_f(c)-V_f(c-)=|f(c)-f(c-)|.$

The corresponding one-sided formula holds at either endpoint. In particular, $V_f$ is continuous at every point where $f$ is continuous.

Facts & Assumptions

Given: A bounded-variation function $f:[a,b]\to\mathbb R$ , its variation function $V_f$ , and a point $c\in[a,b]$ .

[L1]

$V_f(y)-V_f(x)=\operatorname{Var}_{[x,y]}(f)$ whenever $x\le y$ (Total variation is additive over adjacent subintervals and decreases under restriction).

[L2]

$|f(y)-f(x)|\le\operatorname{Var}_{[x,y]}(f)$ (Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant).

[L3]

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 $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ ).

[L4]

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).

[L5]

Continuity is equality of the relevant limit with the function value (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L6]

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 $s_N$ plus the $N$ -th tail, Every convergent sequence is Cauchy).

[L7]

Geometric sequences with ratio in $(0,1)$ tend to zero (For $|r| < 1$ the sequence $r^k$ is null, and for $|r| > 1$ the sequence $|r|^k$ diverges to $+\infty$ ).

Proof

technique · direct

1.1

Since $V_f$ is nondecreasing and bounded above by $V_f(b)$ , its one-sided limits exist. By [L1] and [L2], $V_f(x)-V_f(c)\ge|f(x)-f(c)|$ for $x>c$ ; passage to the right limit gives $V_f(c+)-V_f(c)\ge|f(c+)-f(c)|$ .

L1L2L3L4

1.2

For the reverse inequality, fix $x_0>c$ and put $x_n=c+(x_0-c)2^{-n}$ . Let $a_n=\operatorname{Var}_{[x_{n+1},x_n]}(f)$ . By repeated additivity, every partial sum of the nonnegative series $\sum_na_n$ is $\operatorname{Var}_{[x_N,x_0]}(f)$ for a suitable $N$ , hence is bounded by $\operatorname{Var}_{[c,x_0]}(f)$ . Its tails therefore tend to zero by [L6], while $x_n\downarrow c$ by [L7].

L1L6L7

2.1

Given $\varepsilon>0$ , take $N$ so large that the series tail from $N$ is below $\varepsilon$ and $|f(y)-f(c+)|<\varepsilon$ whenever $c<y\le x_N$ . For any partition $c=t_0<t_1<\cdots<t_k=x_N$ , choose $m\ge N$ with $x_{m+1}<t_1\le x_m$ . The part after its first increment is at most [step 1.2, L1, L2, L3, L6] $\operatorname{Var}_{[t_1,x_N]}(f)\le\operatorname{Var}_{[x_{m+1},x_N]}(f)=\sum_{n=N}^{m}a_n<\varepsilon,$ while $|f(t_1)-f(c)|\le|f(c+)-f(c)|+\varepsilon$ . Taking the supremum over partitions gives $\operatorname{Var}_{[c,x_N]}(f)\le|f(c+)-f(c)|+2\varepsilon$ . Restriction gives the same bound for $c<x\le x_N$ , and [L2] gives the reverse bound in the limit.

3.1

Thus $\lim_{x\downarrow c}\operatorname{Var}_{[c,x]}(f)=|f(c+)-f(c)|$ , and [L1] proves the right-hand formula. Applying steps 1.2–2.1 to the reversed interval proves the left-hand formula. If $f$ is continuous at $c$ , both absolute jumps vanish by [L5], so $V_f$ is continuous there. Endpoint cases use only the available side.

step 1.1step 2.1L1L3L5∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Homogeneity and subadditivity of total variation

Statement

For bounded-variation functions $f,g:[a,b]\to\mathbb R$ and $c\in\mathbb R$ ,

$\operatorname{Var}(cf)=|c|\operatorname{Var}(f),\qquad \operatorname{Var}(f+g)\le\operatorname{Var}(f)+\operatorname{Var}(g).$

Thus $-f$ , $f+g$ , and every finite linear combination of BV functions are BV; in particular $\operatorname{Var}(-f)=\operatorname{Var}(f)$ .

Facts & Assumptions

Given: BV functions $f,g:[a,b]\to\mathbb R$ and a scalar $c$ .

[L1]

Total variation is the supremum of partition variation sums (Bounded variation and total variation on an interval).

[L2]

A partition is a finite increasing point list (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 distribute over scalar multiplication and addition (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

$|cu|=|c||u|$ (Basic properties of the absolute value).

[L5]

$|u+v|\le|u|+|v|$ (The triangle inequality).

Proof

technique · direct

1.1

For every partition $P$ , [L4] and [L3] give $V(cf,P)=|c|V(f,P)$ . Taking suprema gives $\operatorname{Var}(cf)=|c|\operatorname{Var}(f)$ , including $c=0$ and the singleton interval.

L1L2L3L4

1.2

For every partition, [L5] applied to each increment and then [L3] give $V(f+g,P)\le V(f,P)+V(g,P)\le\operatorname{Var}(f)+\operatorname{Var}(g)$ . Taking the supremum proves subadditivity.

L1L2L3L5

2.1

Step 1.1 with $c=-1$ gives $\operatorname{Var}(-f)=\operatorname{Var}(f)$ . Repeated use of steps 1.1 and 1.2 proves closure under every finite linear combination.

step 1.1step 1.2L3∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Functions of bounded variation form an algebra

Statement

If $f$ and $g$ have bounded variation on $[a,b]$ , so do $f+g$ , $cf$ , and $fg$ . If $|f|\le M_f$ and $|g|\le M_g$ , then

$\operatorname{Var}(fg)\le M_f\operatorname{Var}(g)+M_g\operatorname{Var}(f).$

Facts & Assumptions

Given: BV functions $f,g:[a,b]\to\mathbb R$ .

[L1]

BV functions are closed under sums and scalar multiples (Homogeneity and subadditivity of total variation).

[L2]

Every BV function is bounded (Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant).

[L3]

Total variation is the supremum of partition sums (Bounded variation and total variation on an interval).

[L4]

Partitions and their finite sums are as in 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, Finite sums and finite products, by recursion, and Laws of finite sums and finite products.

[L5]

Absolute value is multiplicative and satisfies the triangle inequality (Basic properties of the absolute value, The triangle inequality).

Proof

technique · direct

1.1

L2L5algebra

2.1

Summing step 1.1 over any partition yields $V(fg,P)\le M_fV(g,P)+M_gV(f,P)\le M_f\operatorname{Var}(g)+M_g\operatorname{Var}(f)$ . Taking the supremum proves the displayed bound and that $fg$ is BV.

step 1.1L3L4

3.1

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.

step 2.1L1∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Every bounded-variation function on a compact interval is Riemann integrable

Statement

Every real-valued function of bounded variation on a compact interval $[a,b]$ is Darboux, equivalently Riemann, integrable.

Facts & Assumptions

Given: A bounded-variation function $f:[a,b]\to\mathbb R$ .

[L1]

Jordan decomposition writes $f=f(a)+P_f-N_f$ with $P_f,N_f$ nondecreasing (Jordan decomposition for functions of bounded variation).

[L2]

A monotone real function on a compact interval is integrable (A monotone function on $[a,b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $|f(b) - f(a)|\,(b-a)/\iota(N)$ ).

[L3]

Linear combinations of integrable functions are integrable and their integrals combine linearly (Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$ ).

[L4]

Darboux integrability is the proper integral notion on $[a,b]$ (The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as $\sup_P L(f,P)$ and $\inf_P U(f,P)$ , Darboux integrability as their equality, and the notation $\int_a^b f$ ).

Proof

technique · direct

1.1

By [L1], $P_f$ and $N_f$ are nondecreasing; by [L2] both are integrable, and the constant function $f(a)$ is integrable.

L1L2L4

2.1

Linearity applied to $f=f(a)+P_f-N_f$ makes $f$ integrable. The singleton interval follows from the zero-integral convention.

step 1.1L1L3L4∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-11Open item page →

Absolute continuity on a compact interval

Definition

Let $a\le b$ and $f:[a,b]\to\mathbb R$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length). The function $f$ is absolutely continuous on $[a,b]$ if for every $\varepsilon>0$ there is $\delta>0$ such that every finite family of subintervals $[u_j,v_j]\subseteq[a,b]$ , indexed by $j<m$ , whose open interiors are pairwise disjoint and which satisfies

$\sum_{j<m}(v_j-u_j)<\delta$

also satisfies

$\sum_{j<m}|f(v_j)-f(u_j)|<\varepsilon.$

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 $m=0$ both sums are $0$ , so the condition is automatic. On $[a,a]$ every permitted interval is a singleton and every endpoint increment is $0$ (Absolute value in an ordered field), so every function on that singleton is absolutely continuous. Absolute continuity implies ordinary continuity (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point); that implication is proved next rather than built into the definition.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

$C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation

Statement

Let $f:[a,b]\to\mathbb R$ .

If $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $f'$ extends continuously to $[a,b]$ , then $f$ is Lipschitz.
Every Lipschitz $f$ is absolutely continuous.
Every absolutely continuous $f$ is continuous and has bounded variation.

Thus, with $C^1$ understood in the endpoint-extension sense of claim 1, $C^1\subseteq\mathrm{Lipschitz}\subseteq AC\subseteq C\cap BV$ on a compact interval.

Facts & Assumptions

Given: A compact interval $[a,b]$ and a function $f:[a,b]\to\mathbb R$ .

[L1]

Absolute continuity is the finite disjoint-interval condition of Absolute continuity on a compact interval.

[L2]

Derivatives and continuity use The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set and Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point.

[L3]

A continuous real function on $[a,b]$ is bounded (A continuous real function on a compact subset of $\mathbb{R}$ is bounded).

[L4]

A continuous function with bounded derivative on an interval is Lipschitz (If $f$ is continuous on an interval $I$ and $|f'| \le M$ at every interior point, then $|f(x) - f(y)| \le M|x-y|$ for all $x,y \in I$ , so $f$ is Lipschitz with constant $M$ and uniformly continuous on $I$ ).

[L5]

The Lipschitz condition is $|f(x)-f(y)|\le L|x-y|$ for one $L\ge0$ (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace).

[L6]

Total variation is the supremum of partition sums (Bounded variation and total variation on an interval, 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).

[L7]

Finite sums split and telescope (Laws of finite sums and finite products).

[L8]

The canonical naturals are cofinal in $\mathbb R$ (Every complete ordered field is Archimedean).

Proof

technique · direct

1.1

Under claim 1, the continuous extension of $f'$ is bounded by some $M\ge0$ on $[a,b]$ by [L3]. The bounded-derivative theorem [L4] then makes $f$ Lipschitz with constant $M$ .

L2L3L4

1.2

If $f$ is Lipschitz with constant $L$ , then for every finite disjoint family, $\sum_{j<m}|f(v_j)-f(u_j)|\le L\sum_{j<m}(v_j-u_j)$ . For $L=0$ any positive $\delta$ works; for $L>0$ choose $\delta=\varepsilon/L$ . This proves absolute continuity, including the empty family.

L1L5L7

1.3

If $f$ is absolutely continuous, apply [L1] to the single interval with endpoints $x,y$ to obtain the usual $\varepsilon$ - $\delta$ continuity condition, so $f$ is continuous.

L1L2

1.4

For bounded variation, take $\delta>0$ from absolute continuity with $\varepsilon=1$ . By [L8] choose a natural $N\ge1$ with $(b-a)/N<\delta$ . Insert the points of the uniform $N$ -partition into an arbitrary partition $P$ . Inside each uniform block, the refined subintervals have disjoint interiors and total length at most $(b-a)/N<\delta$ , so their endpoint oscillations sum to less than $1$ . Summing over the $N$ blocks gives $V(f,P)\le N$ , independent of $P$ . Thus $f$ is BV. If $a=b$ , its variation is $0$ .

L1L6L7L8

2.1

Steps 1.1 through 1.4 prove all three inclusions and the asserted hierarchy.

step 1.1step 1.2step 1.3step 1.4∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-11Open item page →

Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral

Definition

Let $a<b$ , let $f,\alpha:[a,b]\to\mathbb R$ , and let $P=(n,t)$ be a partition (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). A choice of tags $\xi_i\in[t_i,t_{i+1}]$ for $i<n$ makes $(P,\xi)$ a tagged partition as in Tagged partitions of $[a,b]$ , with a tag $\xi_i$ in each subinterval, and the Riemann sum $S(f,P,\xi) = \sum_i f(\xi_i)\,\Delta_i$ . Its Riemann-Stieltjes sum is

$S(f,\alpha;P,\xi):=\sum_{i<n}f(\xi_i)\bigl(\alpha(t_{i+1})-\alpha(t_i)\bigr).$

The function $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ if there is $I\in\mathbb R$ such that for every $\varepsilon>0$ there is $\delta>0$ for which every tagged partition with $\|P\|<\delta$ satisfies $|S(f,\alpha;P,\xi)-I|<\varepsilon$ . Then $I=\int_a^b f\,d\alpha$ .

If $f$ is bounded (Lower bound, bounded below, bounded set) and $\alpha$ is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$ , with the dictionary to monotone sequences), put

$m_i:=\inf_{x\in[t_i,t_{i+1}]}f(x),\qquad M_i:=\sup_{x\in[t_i,t_{i+1}]}f(x),$ $L_\alpha(f,P):=\sum_{i<n}m_i\Delta_i\alpha,\qquad U_\alpha(f,P):=\sum_{i<n}M_i\Delta_i\alpha,$

where $\Delta_i\alpha=\alpha(t_{i+1})-\alpha(t_i)\ge0$ . These are the lower and upper Stieltjes sums. Each subinterval is nonempty and its image under bounded $f$ 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 $[a,a]$ the integral is $0$ ; for $b<a$ set $\int_a^b f\,d\alpha=-\int_b^a f\,d\alpha$ , matching The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ .

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Darboux criterion for Riemann–Stieltjes integrability with a nondecreasing integrator

Statement

Let $a<b$ , let $f:[a,b]\to\mathbb R$ be bounded and let $\alpha:[a,b]\to\mathbb R$ 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 $\alpha$ ; and
for every $\varepsilon>0$ there is a partition $P$ with

$U_\alpha(f,P)-L_\alpha(f,P)<\varepsilon.$

In condition 2, writing $\omega_i(f)$ for the oscillation of $f$ on $[t_i,t_{i+1}]$ , the condition is $\sum_{i<n}\omega_i(f)\Delta_i\alpha<\varepsilon$ .

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 $\alpha$ is continuous, the weighted Darboux condition alone is equivalent to mesh Riemann-Stieltjes integrability.

Facts & Assumptions

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

[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 $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[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 $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ 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 $P$ , $U_\alpha(f,P)-L_\alpha(f,P)=\sum_{i<n}(M_i-m_i)\Delta_i\alpha=\sum_{i<n}\omega_i(f)\Delta_i\alpha$ . 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 $\varepsilon>0$ , choose a partition $P$ fine enough that every tagged sum over $P$ is within $\varepsilon/4$ 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 $U_\alpha(f,P)-L_\alpha(f,P)-\varepsilon/2$ , but by less than $\varepsilon/2$ through $I$ ; hence $U_\alpha(f,P)-L_\alpha(f,P)<\varepsilon$ .

L1L3L4L5

1.3

Mesh integrability also forces continuity of $f$ at every discontinuity $c$ of $\alpha$ . By [L6], the total increment of $\alpha$ 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 $(f(x)-f(c))\Delta\alpha$ . 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 $I_-\le I_+$ . Step 1.1 and condition 2 force $I_-=I_+=:I$ . Given $\varepsilon>0$ , choose $P_0$ with Darboux gap below $\varepsilon/3$ . Around each of its finitely many interior points $c$ , choose a small neighborhood as follows: if $\alpha$ is continuous at $c$ , make the variation of $\alpha$ there so small that twice the bound on $|f|$ times that variation is below the allotted error; if $\alpha$ is discontinuous at $c$ , condition 1 and [L7] make the oscillation of $f$ there so small that its product with $\alpha(b)-\alpha(a)$ 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 $R=P_0\vee Q$ . A tagged sum on $R$ lies between $L_\alpha(f,P_0)$ and $U_\alpha(f,P_0)$ . 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 $\alpha$ , or by the local oscillation of $f$ times the total variation $\alpha(b)-\alpha(a)$ . The choices in step 2.1 make the sum of all crossing errors below $2\varepsilon/3$ . Hence every sufficiently fine tagged sum lies within $\varepsilon$ 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 $\alpha$ is continuous, condition 1 is vacuous.

step 1.1step 1.2step 1.3step 2.1step 3.1∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

The Riemann–Stieltjes integral is unique

Statement

For fixed $f,\alpha:[a,b]\to\mathbb R$ , at most one real number satisfies the mesh-limit condition defining $\int_a^b f\,d\alpha$ .

Facts & Assumptions

Given: Two reals $I,J$ satisfying the defining mesh condition for the same functions $f,\alpha$ .

[L1]

The mesh-limit condition quantifies over every sufficiently fine tagged partition (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L2]

Uniform partitions have mesh $(b-a)/N$ for every natural $N\ge1$ (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, The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L3]

Reciprocal naturals become arbitrarily small (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[L4]

$|u+v|\le|u|+|v|$ and $|u|=0$ exactly when $u=0$ (The triangle inequality, Basic properties of the absolute value).

Proof

technique · direct

1.1

Given $\varepsilon>0$ , choose positive thresholds $\delta_I,\delta_J$ for error $\varepsilon/2$ in the two mesh conditions. By [L3] choose a natural $N\ge1$ whose uniform partition has mesh smaller than both thresholds, and give it arbitrary tags.

L1L2L3choose

2.1

For its sum $S$ , $|I-J|\le|I-S|+|S-J|<\varepsilon$ . Since this holds for every $\varepsilon>0$ , $|I-J|=0$ and $I=J$ . The singleton interval has only the prescribed value $0$ .

step 1.1L1L4∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Refinement and tag-change estimates for Stieltjes sums

Statement

Let $\alpha$ have bounded variation on $[a,b]$ , let $P=(n,t)$ be a partition, and let $Q$ refine $P$ . If the oscillation of $f$ on $[t_i,t_{i+1}]$ is at most $\omega_i$ , then any tagged sum on $P$ and any tagged sum on $Q$ satisfy

$|S_Q-S_P|\le\sum_{i<n}\omega_i\operatorname{Var}_{[t_i,t_{i+1}]}(\alpha).$

In particular, if every $\omega_i\le\omega$ , the bound is $\omega\operatorname{Var}_{[a,b]}(\alpha)$ . Two tagged sums on arbitrary partitions whose intervals all have oscillation at most $\omega$ differ by at most $2\omega\operatorname{Var}_{[a,b]}(\alpha)$ .

Facts & Assumptions

Given: Functions $f,\alpha:[a,b]\to\mathbb R$ , a partition $P$ , a refinement $Q$ , and tags on both.

[L1]

Stieltjes sums are weighted finite sums of integrator increments (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L2]

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).

[L3]

A refinement groups its subintervals inside the coarse ones, and any two partitions have a common refinement (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).

[L4]

Finite sums group and telescope (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

The absolute value of a finite sum is at most the sum of absolute values (The triangle inequality).

Proof

technique · direct

1.1

Inside one coarse interval $[t_i,t_{i+1}]$ , the refined integrator increments telescope to $\alpha(t_{i+1})-\alpha(t_i)$ . Subtract the coarse term by assigning its tag value to every refined increment. Each coefficient difference has absolute value at most $\omega_i$ , so the absolute difference contributed by that block is at most $\omega_i$ times the sum of the absolute refined increments, hence at most $\omega_i\operatorname{Var}_{[t_i,t_{i+1}]}(\alpha)$ .

L1L2L3L4L5

2.1

Summing step 1.1 over the coarse blocks proves the first estimate. If $\omega_i\le\omega$ , additivity of variation gives the uniform bound. The conclusions remain $0$ when the variation or the interval is $0$ .

step 1.1L2L4L5

3.1

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 $2$ .

step 2.1L3L5∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator

Statement

If $f:[a,b]\to\mathbb R$ is continuous and $\alpha:[a,b]\to\mathbb R$ has bounded variation, then $\int_a^b f\,d\alpha$ exists.

Facts & Assumptions

Given: A continuous $f:[a,b]\to\mathbb R$ and a BV function $\alpha:[a,b]\to\mathbb R$ .

[L1]

The refinement estimate bounds differences of tagged sums by oscillation times total variation (Refinement and tag-change estimates for Stieltjes sums).

[L2]

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).

[L3]

Total variation is finite for a BV function (Bounded variation and total variation on an interval).

[L4]

A continuous real function on a compact interval is uniformly continuous (Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L5]

Every Cauchy sequence of reals converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

Uniform partitions of arbitrarily small mesh exist, and common refinements exist (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).

Proof

technique · direct

1.1

Put $V=\operatorname{Var}_{[a,b]}(\alpha)$ . If $V=0$ , every integrator increment is $0$ , so every Stieltjes sum is $0$ and the integral exists. Assume $V>0$ . Given $\varepsilon>0$ , uniform continuity gives $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon/(4V)$ . Any two tagged partitions of mesh below $\delta$ have all local oscillations below $\varepsilon/(4V)$ ; comparing both with their common refinement through [L1] makes their sums differ by less than $\varepsilon$ .

L1L3L4L6

2.1

Choose one tagged uniform partition with mesh tending to $0$ for each natural index and call its sum $s_n$ . Step 1.1 makes $(s_n)$ Cauchy, so [L5] gives a real limit $I$ .

step 1.1L5L6choose

3.1

Given $\varepsilon>0$ , choose the threshold in step 1.1 for error $\varepsilon/2$ and then a uniform sum $s_n$ beyond that threshold with $|s_n-I|<\varepsilon/2$ . Every arbitrary tagged sum with sufficiently small mesh differs from $s_n$ by less than $\varepsilon/2$ , hence is within $\varepsilon$ of $I$ . This is the mesh-limit definition, and [L2] identifies the unique value.

step 1.1step 2.1L1L2L5L6∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

The total-variation bound for a Riemann–Stieltjes integral

Statement

Suppose $\int_a^b f\,d\alpha$ exists, $\alpha$ has bounded variation, and $|f(x)|\le M$ on $[a,b]$ . Then

$\left|\int_a^b f\,d\alpha\right|\le M\operatorname{Var}_{[a,b]}(\alpha).$

Facts & Assumptions

Given: An existing Stieltjes integral, a BV integrator $\alpha$ , and a bound $|f|\le M$ .

[L1]

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).

[L2]

Every sum of absolute integrator increments is bounded by total variation (Bounded variation and total variation on an interval).

[L3]

Finite sums and the triangle inequality give $|\sum u_i|\le\sum|u_i|$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality, Basic properties of the absolute value).

[L4]

Non-strict inequalities pass to limits (Limits preserve non-strict inequalities).

Proof

technique · direct

1.1

Every tagged sum satisfies $|S(f,\alpha;P,\xi)|\le\sum_{i<n}|f(\xi_i)||\Delta_i\alpha|\le M\sum_{i<n}|\Delta_i\alpha|\le M\operatorname{Var}_{[a,b]}(\alpha)$ .

L2L3

2.1

Take a sequence of tagged partitions with mesh tending to $0$ . 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.

step 1.1L1L4∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Linearity and interval additivity of the Riemann–Stieltjes integral

Statement

Whenever the integrals on the right exist,

$\int_a^b(uf+vg)\,d\alpha=u\int_a^b f\,d\alpha+v\int_a^b g\,d\alpha,$ $\int_a^b f\,d(u\alpha+v\beta)=u\int_a^b f\,d\alpha+v\int_a^b f\,d\beta.$

Let $a\le c\le b$ , suppose $\alpha$ has bounded variation, and suppose $f$ is continuous at $c$ . Then integrability on $[a,b]$ is equivalent to integrability on both $[a,c]$ and $[c,b]$ , and

$\int_a^b f\,d\alpha=\int_a^c f\,d\alpha+\int_c^b f\,d\alpha.$

Facts & Assumptions

Given: Functions for which the displayed integrals are defined, scalars $u,v$ , and for additivity a BV integrator $\alpha$ and a cut $c$ where $f$ is continuous.

[L1]

Stieltjes integrability is convergence of all sufficiently fine tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L2]

Such a limit is unique (The Riemann–Stieltjes integral is unique).

[L3]

Partitions can be inserted at $c$ and spliced across $c$ (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).

[L4]

Finite sums distribute and split at an index (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

Continuity at $c$ makes $f(x)$ close to $f(c)$ when $x$ is close to $c$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L6]

Reversal and singleton conventions are those of the oriented integral (The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ ).

[L7]

The sum of the absolute integrator increments over any partition is at most $\operatorname{Var}_{[a,b]}(\alpha)$ (Bounded variation and total variation on an interval).

[L8]

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

technique · direct

1.1

Each tagged sum is exactly linear in $f$ and in $\alpha$ , by distribution in the finite sum. Passing to mesh limits and using uniqueness proves both linearity formulas.

L1L2L4

1.2

For a partition containing $c$ , its Stieltjes sum splits exactly into the sums on the two subintervals. Inserting $c$ into a fine partition changes only the interval containing $c$ . Direct subtraction bounds the difference, for any choices of the old and new tags, by the oscillation of $f$ near $c$ times the sum of the relevant absolute increments of $\alpha$ , hence by that oscillation times $\operatorname{Var}_{[a,b]}(\alpha)$ . Continuity of $f$ at $c$ makes this error tend to zero with the mesh.

L1L3L4L5L7

2.1

If the whole-interval integral exists, take any two sufficiently fine sums on $[a,c]$ and splice each with the same sufficiently fine sum on $[c,b]$ . 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].

step 1.2L1L2L3L6L8∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Riemann–Stieltjes integration by parts

Statement

The integral $\int_a^b f\,d\alpha$ exists if and only if $\int_a^b\alpha\,df$ exists. When either exists,

$\int_a^b f\,d\alpha+\int_a^b\alpha\,df=f(b)\alpha(b)-f(a)\alpha(a).$

Facts & Assumptions

Given: Functions $f,\alpha:[a,b]\to\mathbb R$ .

[L1]

Riemann-Stieltjes integrability is the common mesh limit of tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L2]

A Stieltjes integral is unique (The Riemann–Stieltjes integral is unique).

[L3]

Partitions and endpoint tags are permitted tagged partitions (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).

[L4]

Finite summation by parts exchanges a sequence and its successive increments (Abel summation by parts: with $A_n = \sum_{k<n} a_k$ one has $\sum_{k<n} a_k b_k = A_n b_{n-1} - \sum_{k < n-1} A_{k+1}\,(b_{k+1} - b_k)$ for every $n \ge 1$ , Finite sums and finite products, by recursion, Laws of finite sums and finite products).

Proof

technique · direct

1.1

For a partition $P=(n,t)$ , finite summation by parts gives the exact identity $\sum_{i<n}f(t_i)(\alpha(t_{i+1})-\alpha(t_i))+\sum_{i<n}\alpha(t_{i+1})(f(t_{i+1})-f(t_i))=f(b)\alpha(b)-f(a)\alpha(a)$ .

L3L4

2.1

Suppose $\int f\,d\alpha$ exists and consider an arbitrary tagged sum $S_f(\alpha;P,\eta)$ . Refine each $[t_i,t_{i+1}]$ by inserting its tag $\eta_i$ . On $[t_i,\eta_i]$ tag the complementary $f\,d\alpha$ sum at $t_i$ , and on $[\eta_i,t_{i+1}]$ tag it at $t_{i+1}$ . Direct expansion on the $i$ th interval gives [step 1.1, L1, L2, L3, L4] $\alpha(\eta_i)(f(t_{i+1})-f(t_i))+f(t_i)(\alpha(\eta_i)-\alpha(t_i))+f(t_{i+1})(\alpha(t_{i+1})-\alpha(\eta_i))=f(t_{i+1})\alpha(t_{i+1})-f(t_i)\alpha(t_i).$ The refined mesh does not exceed $\lVert P\rVert$ , so the complementary sums converge to $\int f\,d\alpha$ . Telescoping the displayed identities forces every fine tagged sum for $\int\alpha\,df$ to converge to the endpoint product minus that integral.

3.1

Exchanging $f$ and $\alpha$ proves the converse. Adding the two values yields the displayed formula, including the singleton and reversed-orientation cases.

step 1.1step 2.1L1L2∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A bounded-variation integrand is Riemann–Stieltjes integrable against every continuous integrator

Statement

If $f:[a,b]\to\mathbb R$ has bounded variation and $\alpha:[a,b]\to\mathbb R$ is continuous, then $\int_a^b f\,d\alpha$ exists.

Facts & Assumptions

Given: A BV function $f$ and a continuous function $\alpha$ on $[a,b]$ .

[L1]

A continuous integrand is Stieltjes integrable against a BV integrator (A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator).

[L2]

Existence of $\int\alpha\,df$ is equivalent to existence of $\int f\,d\alpha$ (Riemann–Stieltjes integration by parts).

[L3]

Bounded variation and continuity are those of Bounded variation and total variation on an interval and Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ 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

Since $\alpha$ is continuous and $f$ is BV, [L1] gives the integral $\int_a^b\alpha\,df$ .

L1L3

2.1

Integration by parts [L2] then gives existence of $\int_a^b f\,d\alpha$ and its value $f(b)\alpha(b)-f(a)\alpha(a)-\int_a^b\alpha\,df$ .

step 1.1L2∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A bounded function with finitely many discontinuities is Stieltjes integrable against a continuous bounded-variation integrator

Statement

Let $f:[a,b]\to\mathbb R$ be bounded and have only finitely many discontinuities. If $\alpha:[a,b]\to\mathbb R$ is continuous and has bounded variation, then $\int_a^b f\,d\alpha$ exists.

Facts & Assumptions

Given: A bounded $f$ with finite discontinuity set $D$ , and a continuous BV integrator $\alpha$ on $[a,b]$ .

[L1]

A BV function is the difference of two nondecreasing functions (Jordan decomposition for functions of bounded variation).

[L2]

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).

[L3]

For $a<b$ , 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).

[L4]

Continuous functions on compact intervals are uniformly continuous (Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness).

[L5]

Stieltjes integration is linear in the integrator (Linearity and interval additivity of the Riemann–Stieltjes integral).

[L6]

Continuity makes the increment of $\alpha$ arbitrarily small on sufficiently short intervals around each point (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ 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

If $a=b$ the integral is $0$ by the definition of the Riemann-Stieltjes integral on a singleton interval and there is nothing to prove, so assume $a<b$ , which is the standing hypothesis of [L3].

givenL3

1.2

First suppose that $\alpha$ is continuous and nondecreasing. Write $|f|\le M$ . Given $\varepsilon>0$ , [L6] and finiteness of $D$ allow pairwise disjoint closed intervals $I_x$ about the points $x\in D$ whose total $\alpha$ -increment is less than $\varepsilon/(4M+1)$ .

L6

2.1

On the compact complement of the interiors of the $I_x$ , the function $f$ is continuous and hence uniformly continuous by [L4]. Choose a partition containing all endpoints of the $I_x$ and fine enough that every remaining partition interval has oscillation below $\varepsilon/(1+\alpha(b)-\alpha(a))$ . The intervals meeting $D$ contribute at most $2M$ times their total $\alpha$ -increment, and all other intervals contribute less than $\varepsilon$ . After rescaling the two preliminary bounds, the weighted oscillation sum is arbitrarily small, so [L3] gives $f\in R(\alpha)$ .

step 1.2L3L4

3.1

For a general continuous BV $\alpha$ , [L1] writes $\alpha=\alpha(a)+P_\alpha-N_\alpha$ . Both $P_\alpha$ and $N_\alpha$ are continuous by [L2]. Step 2.1 gives integrability against each, and linearity in the integrator [L5] gives integrability against $\alpha$ .

step 2.1L1L2L5∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A continuous function of a Stieltjes-integrable function is Stieltjes integrable for a nondecreasing integrator

Statement

Suppose $\alpha:[a,b]\to\mathbb R$ is nondecreasing, $f$ is bounded and Riemann–Stieltjes integrable with respect to $\alpha$ , and $\phi$ is continuous on a compact interval containing $f([a,b])$ . Then $\phi\circ f$ is Riemann–Stieltjes integrable with respect to $\alpha$ .

Facts & Assumptions

Given: A nondecreasing $\alpha$ , a bounded $f\in R(\alpha)$ , and a continuous $\phi$ on a compact interval containing the range of $f$ .

[L1]

For $a<b$ , bounded $f$ and nondecreasing $\alpha$ , integrability in the mesh sense is equivalent to the conjunction of two conditions: $f$ is continuous at every discontinuity of $\alpha$ , and for every $\varepsilon>0$ some partition has $\sum_{i<n}\omega_i(f)\Delta_i\alpha<\varepsilon$ (Darboux criterion for Riemann–Stieltjes integrability with a nondecreasing integrator).

[L2]

The function $\phi$ is bounded and uniformly continuous on its compact domain (A continuous real function on a compact subset of $\mathbb{R}$ is bounded, Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness).

[L3]

Finite sums may be split and estimated termwise (Laws of finite sums and finite products).

Proof

technique · direct

1.1

L2

2.1

By [L1], choose a partition $P$ for which $\sum_I\operatorname{osc}_I(f)\,\Delta_I\alpha<\eta\varepsilon/(4K+1)$ . Split its intervals into those with $\operatorname{osc}_I(f)<\eta$ and the rest. The first class contributes less than $\varepsilon/2$ to the weighted oscillation sum of $\phi\circ f$ . In the second class, $\operatorname{osc}_I(\phi\circ f)\le2K$ , while $\eta\sum\Delta_I\alpha\le\sum_I\operatorname{osc}_I(f)\Delta_I\alpha$ ; hence it too contributes less than $\varepsilon/2$ .

step 1.1L1L2L3

3.1

Thus the weighted oscillation condition in [L1] holds for $\phi\circ f$ . The same theorem says that $f$ is continuous at every discontinuity of $\alpha$ ; continuity of $\phi$ makes $\phi\circ f$ continuous there as well. Both clauses of [L1] now give $\phi\circ f\in R(\alpha)$ .

step 2.1L1L2∎

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Every bounded-variation function is uniformly approximable by step functions

Statement

If $f:[a,b]\to\mathbb R$ has bounded variation, then for every $\varepsilon>0$ there is a finite step function $s$ with $\lVert f-s\rVert_\infty<\varepsilon$ . Endpoint values of $s$ may be prescribed to equal those of $f$ .

More precisely, if $E\subseteq[a,b]$ is at most countable and $f$ is continuous at every point of $E$ , the interior breakpoints of $s$ may all be chosen outside $E$ .

Facts & Assumptions

Given: A BV function $f$ , a tolerance $\varepsilon>0$ , and, for the strengthened assertion, an at most countable set $E$ of continuity points of $f$ .

[L1]

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 $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ ).

[L2]

Every nonempty subset of the reals that is bounded above has a supremum (Complete ordered field (least-upper-bound property)).

[L3]

Every nonempty open interval is uncountable (Every nondegenerate interval of $\mathbb{R}$ is uncountable).

Proof

technique · direct

1.1

Fix $\eta>0$ . Let $A$ be the set of $x\in[a,b]$ for which there is a finite chain $a=x_0<\cdots<x_m=x$ such that the oscillation of $f$ on every open interval $(x_{j-1},x_j)$ is below $\eta$ . The set contains $a$ and is bounded above by $b$ , so $c:=\sup A$ exists by [L2].

L2construct

2.1

Suppose $c<b$ . The left limit at $c$ and the right limit at $c$ supplied by [L1] give one-sided intervals on which the oscillation is below $\eta$ . Choose $x\in A$ in the left interval (use $x=a$ if $c=a$ ), append $c$ to its chain if necessary, and then append a point $y>c$ in the right interval. This puts $y$ in $A$ , contradicting that $c$ is an upper bound. Hence $c=b$ . The left limit at $b$ now lets a chain ending sufficiently near $b$ be extended to $b$ . Thus there is a finite partition of $[a,b]$ on each of whose open components the oscillation of $f$ is below $\eta$ . The singleton case is immediate.

step 1.1L1L2

3.1

On each open component choose one value of $f$ , and at every partition point assign the actual value of $f$ . The resulting finite step function differs from $f$ by less than $\eta$ everywhere. Taking $\eta=\varepsilon/4$ leaves room for the strengthened construction.

step 2.1choose

4.1

Only finitely many interior breakpoints lie in $E$ . Around each such breakpoint $c$ , continuity of $f$ gives a small two-sided interval, disjoint from the corresponding intervals for the other breakpoints, on which the oscillation is below $\varepsilon/4$ . By [L3] choose a replacement point outside $E$ 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 $\varepsilon/4$ , and a component can be enlarged at both ends, so its new oscillation is at most $\varepsilon/4+\varepsilon/4+\varepsilon/4<\varepsilon$ , 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 $E$ .

step 3.1L3∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Two bounded-variation functions with no common discontinuity are Riemann–Stieltjes integrable

Statement

Let $f,\alpha:[a,b]\to\mathbb R$ have bounded variation. If no point is a discontinuity of both functions, then $\int_a^b f\,d\alpha$ exists.

Facts & Assumptions

Given: BV functions $f$ and $\alpha$ with disjoint discontinuity sets.

[L1]

The discontinuity set $D_\alpha$ is at most countable (A bounded-variation function has at most countably many discontinuities, all of the first kind).

[L2]

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).

[L3]

Direct subtraction of two Stieltjes sums and the finite-sum triangle inequality give $|S_\alpha(f)-S_\alpha(g)|\le\lVert f-g\rVert_\infty\operatorname{Var}(\alpha)$ (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).

[L4]

Every Cauchy sequence of reals converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).

[L5]

If $\alpha$ 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

technique · direct

1.1

By the no-common-discontinuity hypothesis, $f$ is continuous at every point of $D_\alpha$ . For each $n\ge1$ , [L1] and [L2] provide a finite step function $s_n$ with $\lVert f-s_n\rVert_\infty<1/n$ and all interior breakpoints outside $D_\alpha$ . If an endpoint belongs to $D_\alpha$ , continuity of $f$ there permits the value on the adjacent open component to be changed to $f$ at that endpoint while retaining the same bound after beginning with tolerance $1/(2n)$ . Thus $s_n$ is continuous at every point of $D_\alpha$ .

L1L2

2.1

Each $s_n$ is integrable with respect to $\alpha$ . Its finitely many discontinuities are points where $\alpha$ is continuous by step 1.1. By [L5], choose disjoint neighborhoods of those points whose total local variation is small. Outside them $s_n$ is locally constant, while inside them [L5] bounds differences between fine sums by the small local variation times the finite oscillation of $s_n$ . 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.

step 1.1L4L5

3.1

Given $\varepsilon>0$ , choose $n$ so that $2\operatorname{Var}_{[a,b]}(\alpha)/n<\varepsilon/2$ (the zero-variation case is immediate), and then choose a mesh bound making any two sums of $s_n$ differ by less than $\varepsilon/2$ . By [L3], replacing $s_n$ by $f$ in either sum changes it by at most $\operatorname{Var}(\alpha)/n$ . Hence all sufficiently fine sums of $f$ 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.

step 2.1L3L4∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

The identity integrator recovers the Riemann integral

Statement

Let $a\le b$ . For the identity function $\operatorname{id}(x)=x$ , a bounded $f:[a,b]\to\mathbb R$ is Riemann–Stieltjes integrable with respect to $\operatorname{id}$ exactly when it is Riemann integrable, and then

$\int_a^b f\,d\operatorname{id}=\int_a^b f(x)\,dx.$

For $a=b$ both sides are $0$ by the singleton conventions. For reversed endpoints the statement is about a function defined on the sorted interval: if $a>b$ and $f$ is bounded on $[b,a]$ , then both oriented conventions negate the corresponding sorted integral, so the equality is inherited from the case just proved. The hypothesis $a\le b$ is needed for the displayed clause itself, because $[a,b]$ is empty when $a>b$ and a function typed on it supplies no values to integrate.

Facts & Assumptions

Given: A bounded function $f$ on the compact interval with endpoints $a,b$ .

[L1]

A Stieltjes sum is $\sum_i f(\xi_i)(\alpha(t_{i+1})-\alpha(t_i))$ (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L2]

A Riemann sum is $\sum_i f(\xi_i)(t_{i+1}-t_i)$ (Tagged partitions of $[a,b]$ , with a tag $\xi_i$ in each subinterval, and the Riemann sum $S(f,P,\xi) = \sum_i f(\xi_i)\,\Delta_i$ ).

[L3]

Darboux and tagged-sum Riemann integrability agree (The Darboux and Riemann definitions agree: a bounded $f$ on $[a,b]$ is Darboux integrable with integral $I$ if and only if for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that $|S(f,P,\xi) - I| < \varepsilon$ for every tagged partition of mesh below $\delta$ ).

[L4]

Oriented integrals reverse sign when endpoints are interchanged (The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ ).

Proof

technique · direct

1.1

When $a<b$ and $\alpha=\operatorname{id}$ , every increment $\alpha(t_{i+1})-\alpha(t_i)$ equals $t_{i+1}-t_i$ . Thus [L1] and [L2] are termwise identical for every tagged partition.

L1L2

2.1

Consequently the two mesh limits exist simultaneously and have the same value; [L3] identifies that tagged limit with the Darboux integral. For $a=b$ both conventions give zero, and [L4] handles $a>b$ .

step 1.1L3L4∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A continuously differentiable integrator reduces Stieltjes integration to ordinary integration

Statement

Let $f:[a,b]\to\mathbb R$ be Riemann integrable. Suppose $\alpha$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $\alpha'$ extends continuously to $[a,b]$ . Then $f$ is Riemann–Stieltjes integrable with respect to $\alpha$ and

$\int_a^b f\,d\alpha=\int_a^b f(x)\alpha'(x)\,dx.$

Facts & Assumptions

Given: A Riemann-integrable $f$ and an integrator $\alpha$ with continuous derivative on the compact interval.

[L1]

The mean value theorem represents every increment of $\alpha$ as $\alpha'(\eta_i)(t_{i+1}-t_i)$ (The mean value theorem, as the case $g(x) = x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a < b$ and differentiable on $(a,b)$ there is $c \in (a,b)$ with $f(b) - f(a) = f'(c)(b-a)$ ).

[L2]

The continuous function $\alpha'$ is uniformly continuous (Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness).

[L3]

The continuous function $\alpha'$ is Riemann integrable, so the product $f\alpha'$ is Riemann integrable (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, If $f,g$ are integrable on $[a,b]$ then so are $\lvert f\rvert$ , $f^{2}$ , $fg$ , $\max(f,g)$ and $\min(f,g)$ , and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$ ).

[L4]

Riemann tagged sums converge to the Darboux integral (The Darboux and Riemann definitions agree: a bounded $f$ on $[a,b]$ is Darboux integrable with integral $I$ if and only if for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that $|S(f,P,\xi) - I| < \varepsilon$ for every tagged partition of mesh below $\delta$ ).

Proof

technique · direct

1.1

A Riemann-integrable function is bounded; choose $M$ with $|f|\le M$ . For each partition interval, [L1] gives $\eta_i\in(t_i,t_{i+1})$ such that $\Delta_i\alpha=\alpha'(\eta_i)\Delta_i t$ . Hence [L1] $S_\alpha(f;P,\xi)-\sum_i f(\xi_i)\alpha'(\xi_i)\Delta_i t=\sum_i f(\xi_i)(\alpha'(\eta_i)-\alpha'(\xi_i))\Delta_i t.$

2.1

By [L2], the absolute value of the right side is at most $M(b-a)\omega_{\alpha'}(\lVert P\rVert)$ , which tends to zero with the mesh. By [L3] and [L4], the second sum in step 1.1 tends to $\int_a^b f\alpha'$ . Thus all Stieltjes sums have the same limit, proving both existence and the formula.

step 1.1L2L3L4∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A countable pure-step integrator evaluates a continuous integrand as the absolutely convergent weighted sum of its values at the jumps

Statement

Let $a<b$ . Write $I:\mathbb R\to\mathbb R$ for the unit step $I(t)=0$ for $t\le0$ and $I(t)=1$ for $t>0$ . Let $(s_n)_{n\in\mathbb N}$ be points of the open interval $(a,b)$ , and let $(c_n)_{n\in\mathbb N}$ be reals with $c_n\ge0$ and $\sum c_n$ convergent.

Then for every $x\in[a,b]$ the series $\sum_n c_nI(x-s_n)$ converges, so

$\alpha(x):=\sum_{n=0}^{\infty}c_nI(x-s_n)$

defines a nondecreasing $\alpha:[a,b]\to\mathbb R$ , which therefore has bounded variation.

For every continuous $f:[a,b]\to\mathbb R$ the integral $\int_a^b f\,d\alpha$ exists, the series $\sum_n c_nf(s_n)$ converges absolutely, and

$\int_a^b f\,d\alpha=\sum_{n=0}^{\infty}c_nf(s_n).$

The points $s_n$ are not required to be distinct, and any $c_n$ may be zero.

Facts & Assumptions

Given: Reals $a<b$ , points $s_n\in(a,b)$ , reals $c_n\ge0$ with $\sum c_n$ convergent, and a continuous $f:[a,b]\to\mathbb R$ .

[L1]

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).

[L2]

A real function on $[a,b]$ 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 $\sum_{i<n}|g(t_{i+1})-g(t_i)|$ (Bounded variation and total variation on an interval).

[L3]

If $f:[a,b]\to\mathbb R$ is continuous and $\alpha:[a,b]\to\mathbb R$ has bounded variation, then $\int_a^b f\,d\alpha$ exists (A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator).

[L4]

Whenever the integrals on the right exist, $\int_a^b f\,d(u\alpha+v\beta)=u\int_a^b f\,d\alpha+v\int_a^b f\,d\beta$ (Linearity and interval additivity of the Riemann–Stieltjes integral).

[L5]

If $\int_a^b f\,d\alpha$ exists, $\alpha$ has bounded variation, and $|f(x)|\le M$ on $[a,b]$ , then $\bigl|\int_a^b f\,d\alpha\bigr|\le M\operatorname{Var}_{[a,b]}(\alpha)$ (The total-variation bound for a Riemann–Stieltjes integral).

[L6]

The closed bounded interval $[a,b]$ is compact (Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact), and a continuous real function is bounded on a compact subset of its domain: there is $M\ge0$ with $|f(x)|\le M$ throughout (A continuous real function on a compact subset of $\mathbb{R}$ is bounded, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L7]

The Riemann–Stieltjes sum of a tagged partition $(P,\xi)$ is $S(f,\alpha;P,\xi)=\sum_{i<n}f(\xi_i)(\alpha(t_{i+1})-\alpha(t_i))$ , and $\int_a^b f\,d\alpha=I$ means that for every $\varepsilon>0$ some $\delta>0$ makes $|S(f,\alpha;P,\xi)-I|<\varepsilon$ for every tagged partition of mesh below $\delta$ (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral, 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, Tagged partitions of $[a,b]$ , with a tag $\xi_i$ in each subinterval, and the Riemann sum $S(f,P,\xi) = \sum_i f(\xi_i)\,\Delta_i$ ).

[L8]

If $0\le a_k\le b_k$ for all large $k$ and $\sum b_k$ converges, then $\sum a_k$ converges (If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$ , and divergence of $\sum a_k$ gives divergence of $\sum b_k$ ); a series converging absolutely converges (If $\sum |a_k|$ converges then $\sum a_k$ converges).

[L9]

Continuity of $f$ at $s$ means that for every $\varepsilon>0$ there is $\delta>0$ with $|f(x)-f(s)|<\varepsilon$ whenever $x$ lies in the domain and $|x-s|<\delta$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point); convergence of a real sequence is the usual $\varepsilon$ –threshold condition (Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

Fix $x\in[a,b]$ . Each term $c_nI(x-s_n)$ lies in $[0,c_n]$ , so the partial sums of $\sum_n c_nI(x-s_n)$ are nondecreasing and bounded above by $\sum_n c_n$ . By [L1] the series converges and $\alpha(x)$ is defined, with $0\le\alpha(x)\le\sum_n c_n$ .

givenL1

1.2

Fix $s\in(a,b)$ and put $\sigma(x):=I(x-s)$ on $[a,b]$ . Then $\sigma$ is nondecreasing, hence of bounded variation by [L2]. Let $\varepsilon>0$ and take $\delta>0$ from [L9] for $f$ at $s$ , so that $|f(x)-f(s)|<\varepsilon$ whenever $|x-s|<\delta$ . Let $(P,\xi)$ be a tagged partition of mesh below $\delta$ . The increment $\sigma(t_{i+1})-\sigma(t_i)$ is $1$ when $t_i\le s<t_{i+1}$ and $0$ otherwise, and because $a\le s<b$ exactly one index $i$ satisfies $t_i\le s<t_{i+1}$ . Hence $S(f,\sigma;P,\xi)=f(\xi_i)$ for that index, and $t_i\le\xi_i\le t_{i+1}$ with $t_i\le s<t_{i+1}$ give $|\xi_i-s|\le t_{i+1}-t_i<\delta$ , so $|S(f,\sigma;P,\xi)-f(s)|<\varepsilon$ . By [L7], $\int_a^b f\,d\sigma=f(s)$ .

givenL2L7L9

2.1

If $x\le y$ then $I(x-s_n)\le I(y-s_n)$ for every $n$ , because $I$ is nondecreasing and $x-s_n\le y-s_n$ . Multiplying by $c_n\ge0$ and summing, every partial sum for $x$ is at most the corresponding partial sum for $y$ , so the limits satisfy $\alpha(x)\le\alpha(y)$ by [L1]. Thus $\alpha$ is nondecreasing, and $\alpha=\alpha-0$ exhibits it as a difference of two nondecreasing functions, so [L2] gives $\alpha$ bounded variation.

step 1.1givenL1L2

2.2

For $N\in\mathbb N$ set $\alpha_N(x):=\sum_{n<N}c_nI(x-s_n)$ , 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 $\int_a^b f\,d\alpha_N=\sum_{n<N}c_nf(s_n)$ .

step 1.2L4

3.1

By [L6] there is $M\ge0$ with $|f(x)|\le M$ on $[a,b]$ . Since $0\le|c_nf(s_n)|\le Mc_n$ and $\sum c_n$ converges, [L8] makes $\sum_n c_nf(s_n)$ absolutely convergent, hence convergent. By step 2.1 and [L3] the integral $\int_a^b f\,d\alpha$ exists.

givenstep 2.1L3L6L8

3.2

Set $\beta_N:=\alpha-\alpha_N$ . For each $x$ , $\beta_N(x)=\sum_{n\ge N}c_nI(x-s_n)$ , the tail of the series in step 1.1; the argument of steps 1.1 and 2.1 applies verbatim to it, so $\beta_N$ is nondecreasing with bounded variation. Since $s_n\in(a,b)$ we have $I(a-s_n)=0$ and $I(b-s_n)=1$ , so $\beta_N(a)=0$ and $\beta_N(b)=\sum_{n\ge N}c_n$ . A nondecreasing function has every partition sum equal to $\beta_N(b)-\beta_N(a)$ , because each increment is nonnegative and the sum telescopes, so [L2] gives $\operatorname{Var}_{[a,b]}(\beta_N)=\sum_{n\ge N}c_n$ .

step 1.1step 2.1givenL1L2

4.1

Both $\alpha_N$ and $\beta_N$ are of bounded variation, so [L3] makes $\int_a^b f\,d\alpha_N$ and $\int_a^b f\,d\beta_N$ exist, and $\alpha=\alpha_N+\beta_N$ with [L4] gives $\int_a^b f\,d\alpha=\int_a^b f\,d\alpha_N+\int_a^b f\,d\beta_N$ . Using step 2.2 and then [L5] with the bound $M$ of step 3.1, $\Bigl|\int_a^b f\,d\alpha-\sum_{n<N}c_nf(s_n)\Bigr|=\Bigl|\int_a^b f\,d\beta_N\Bigr|\le M\sum_{n\ge N}c_n.$

step 2.2step 3.1step 3.2L3L4L5

5.1

Convergence of $\sum c_n$ makes its tails $\sum_{n\ge N}c_n$ tend to $0$ as $N$ increases, so given $\varepsilon>0$ the right side of step 4.1 is below $\varepsilon$ for all large $N$ . Hence the partial sums $\sum_{n<N}c_nf(s_n)$ converge to $\int_a^b f\,d\alpha$ , and by step 3.1 that series converges absolutely. By [L1] and [L9] its sum is $\int_a^b f\,d\alpha$ , which is the claimed identity.

step 3.1step 4.1givenL1L9∎

Remark

The two endpoints behave differently, which is why the jumps are confined to the open interval. A jump at $a$ would be harmless: $I(x-a)$ vanishes only at $x=a$ , the increment $\alpha(b)-\alpha(a)$ still records the whole weight, and step 2.1 goes through unchanged because its counting argument needs only $a\le s<b$ . A jump at $b$ genuinely breaks the identity: $I(x-b)=0$ for every $x\in[a,b]$ , so such a term contributes nothing at all to $\alpha$ , yet it would contribute $c_nf(b)$ to the right-hand sum. The hypothesis $s_n\in(a,b)$ excludes that case, and it is the hypothesis Rudin states.

Rudin's Theorem 6.16 additionally requires the $s_n$ to be distinct. Nothing in the proof above uses distinctness, so it is not assumed here.

Continuity of $f$ is not decorative. cex-common-jump-prevents-riemann-stieltjes-integrability exhibits an $f$ and an $\alpha$ sharing a single jump for which no mesh limit exists, and a single step integrator is exactly the $\alpha$ of that counterexample.

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Change of variable for the Riemann–Stieltjes integral

Statement

Let $c<d$ and $a<b$ , and let $\phi:[c,d]\to[a,b]$ be a strictly increasing continuous bijection. For functions $f,\alpha:[a,b]\to\mathbb R$ , one of the two Riemann–Stieltjes integrals below exists if and only if the other does, and in that case

$\int_a^b f\,d\alpha=\int_c^d(f\circ\phi)\,d(\alpha\circ\phi).$

The nondegeneracy hypotheses are not cosmetic. If $c=d$ and $a=b$ both integrals are $0$ 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 $f,\alpha$ 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 $\phi:[c,d]\to[a,b]$ and functions $f,\alpha$ on $[a,b]$ .

[L1]

The inverse $\phi^{-1}$ is continuous (Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f[I]$ , and the inverse $g : f[I] \to I$ is continuous and strictly monotone in the same sense as $f$ ).

[L2]

Both $\phi$ and $\phi^{-1}$ are uniformly continuous on their compact domains (Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness).

[L3]

The Stieltjes integral is the common mesh limit of its tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L4]

A Stieltjes integral, when it exists, is unique (The Riemann–Stieltjes integral is unique).

Proof

technique · direct

1.1

If $Q=(s_j)$ is a partition of $[c,d]$ with tags $\eta_j$ , then $P=(\phi(s_j))$ is a partition of $[a,b]$ with tags $\phi(\eta_j)$ . Direct substitution gives [given] $S_{\alpha\circ\phi}(f\circ\phi;Q,\eta)=S_\alpha(f;P,\phi\circ\eta).$

2.1

By uniform continuity of $\phi$ in [L2], arbitrarily fine $Q$ give arbitrarily fine image partitions $P$ . 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 $\phi^{-1}$ , using [L1] and [L2], proves the converse. Uniqueness [L4] identifies the two limits.

step 1.1L1L2L3L4∎

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Young's partition estimate for rational Hölder exponents

Statement

$|L_{m+1}-L_m|\le K_fK_g(b-a)^{p+q}2^{-m(p+q-1)}.$

Put $r=p+q$ and $C_r:=\frac{2^r}{1-2^{1-r}}.$ If $R$ refines an arbitrary partition $P$ , then their left-endpoint sums satisfy $|L(R)-L(P)|\le C_rK_fK_g(b-a)\lVert P\rVert^{r-1}.$

Facts & Assumptions

Given: Rational Hölder exponents $p,q$ with $p+q>1$ , Hölder constants $K_f,K_g$ , and the stated partitions.

[L1]

Rational powers are monotone and satisfy the exponent laws (Rational powers $a^r$ of a positive base, Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L2]

A geometric series with ratio in $(0,1)$ converges and its tails tend to zero (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges).

[L3]

Finite sums obey the triangle inequality and may be regrouped (Laws of finite sums and finite products, The triangle inequality).

Proof

technique · direct

1.1

Insert a point $v$ between adjacent points $u<w$ . The change from the old left-endpoint term to the two new terms is [given] $(f(v)-f(u))(g(w)-g(v))$ up to sign. Its absolute value is at most $K_fK_g(v-u)^p(w-v)^q$ , hence at most $K_fK_g(w-u)^{p+q}$ by [L1].

2.1

Passing from $D_m$ to $D_{m+1}$ inserts one midpoint in each of $2^m$ intervals of length $(b-a)2^{-m}$ . Summing step 1.1 gives the first displayed bound. More generally, if a partition of an interval $I$ has $k\ge2$ subintervals, some interior point has two adjacent lengths whose sum is at most $2|I|/(k-1)$ : the sum of all such two-interval lengths is at most $2|I|$ . Removing that point therefore changes the left sum by at most $K_fK_g(2|I|/(k-1))^r$ .

step 1.1L1L3

3.1

Remove the extra points of $R$ inside a fixed interval $I$ of $P$ , one at a time, always using step 2.1. The total error is at most $2^rK_fK_g|I|^r\sum_{j\ge1}j^{-r}$ . Grouping the positive integers into $[2^m,2^{m+1})$ bounds this series by $\sum_{m\ge0}2^{-m(r-1)}=(1-2^{1-r})^{-1}$ via [L2]. Thus the error on $I$ is at most $C_rK_fK_g|I|^r$ . Summing over $I\in P$ and using $|I|^r\le\lVert P\rVert^{r-1}|I|$ proves the refinement estimate. If $K_fK_g=0$ or $a=b$ , every error is zero.

step 2.1L1L2L3∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Young's Riemann–Stieltjes existence theorem for rational Hölder exponents

Statement

Let $p,q\in\mathbb Q\cap(0,1]$ with $p+q>1$ . If $f:[a,b]\to\mathbb R$ is $p$ -Hölder and $g:[a,b]\to\mathbb R$ is $q$ -Hölder, then both $\int_a^b f\,dg$ and $\int_a^b g\,df$ exist. They satisfy

$\int_a^b f\,dg+\int_a^b g\,df=f(b)g(b)-f(a)g(a).$

Facts & Assumptions

Given: Hölder functions $f,g$ with rational exponents whose sum exceeds one.

[L1]

The Young partition estimate controls refinement errors by a constant times $\lVert P\rVert^{p+q-1}$ (Young's partition estimate for rational Hölder exponents).

[L2]

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).

[L3]

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).

[L4]

Such a limit is unique (The Riemann–Stieltjes integral is unique).

[L5]

Tails of a geometric series with ratio in $(0,1)$ tend to zero (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges).

Proof

technique · direct

1.1

For the dyadic left sums $L_m$ , the first estimate in [L1] and the geometric-tail fact [L5] make $(L_m)$ a Cauchy sequence. It therefore converges to some $I$ by [L2].

L1L2L5

2.1

Given a partition $P$ , compare it and a sufficiently fine dyadic partition $D_m$ with their common refinement. The second estimate in [L1] bounds the two refinement errors by a constant times $\lVert P\rVert^{r-1}+\lVert D_m\rVert^{r-1}$ . Together with $L_m\to I$ , this shows that every sufficiently fine left-endpoint sum is close to $I$ . Replacing a left endpoint $t_i$ by an arbitrary tag $\xi_i$ changes the $i$ th term by at most $K_fK_g|t_{i+1}-t_i|^r$ ; the total is at most $K_fK_g(b-a)\lVert P\rVert^{r-1}$ . Thus every fine tagged sum tends to $I$ , and [L3] gives $\int f\,dg$ . Interchanging $f$ and $g$ gives $\int g\,df$ .

step 1.1L1L2L3L4

3.1

On every partition, the right-endpoint sum for $f\,dg$ plus the left-endpoint sum for $g\,df$ telescopes exactly to $f(b)g(b)-f(a)g(a)$ . Passing to the two limits established in step 2.1 proves the formula.

step 2.1L3∎

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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 $[a,b]$ with $a<b$ , 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 $a<b$ is part of the statement and not cosmetic: on $[a,a]$ the integral is $0$ 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 $a=b$ 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.

Bounded Variation and the Riemann–Stieltjes Integral

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Bounded variation and total variation on an interval

Definition

Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant

Statement

Facts & Assumptions

Proof

Total variation is additive over adjacent subintervals and decreases under restriction

Statement

Facts & Assumptions

Proof

Variation function and positive and negative variations

Definition

The positive and negative variations are nondecreasing and give the Jordan identities

Statement

Facts & Assumptions

Proof

Jordan decomposition for functions of bounded variation

Statement

Facts & Assumptions

Proof

A bounded-variation function has at most countably many discontinuities, all of the first kind

Statement

Facts & Assumptions

Proof

The jumps of a variation function equal the absolute jumps of the original function

Statement

Facts & Assumptions

Proof

Homogeneity and subadditivity of total variation

Statement

Facts & Assumptions

Proof

Functions of bounded variation form an algebra

Statement

Facts & Assumptions

Proof

Every bounded-variation function on a compact interval is Riemann integrable

Statement

Facts & Assumptions

Proof

Absolute continuity on a compact interval

Definition

C1C^1C1 implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation

Statement

Facts & Assumptions

Proof

Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral

Definition

Darboux criterion for Riemann–Stieltjes integrability with a nondecreasing integrator

Statement

Facts & Assumptions

Proof

The Riemann–Stieltjes integral is unique

Statement

Facts & Assumptions

Proof

Refinement and tag-change estimates for Stieltjes sums

Statement

Facts & Assumptions

Proof

A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator

Statement

Facts & Assumptions

Proof

The total-variation bound for a Riemann–Stieltjes integral

Statement

Facts & Assumptions

Proof

Linearity and interval additivity of the Riemann–Stieltjes integral

Statement

Facts & Assumptions

Proof

Riemann–Stieltjes integration by parts

Statement

Facts & Assumptions

$C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation