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TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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The two FTC forms for a Riemann–Stieltjes integral with a C1 integrator

Statement

Let a<b. Suppose α:[a,b]R is continuous, differentiable on (a,b), and its derivative extends to a continuous function q:[a,b]R.

  1. If f:[a,b]R is Riemann integrable and continuous at c, then A(x):=axfdα is differentiable at c and A(c)=f(c)q(c).
  2. If f is Riemann integrable, G:[a,b]R is continuous and differentiable on (a,b), and G(x)=f(x)q(x) there, then abfdα=G(b)G(a).

Endpoint derivatives in clause 1 are relative. Clause 2 does not divide by q and remains valid where q vanishes.

Facts & Assumptions

Given: The functions in the statement.

[L1]

For a continuous integrator whose interior derivative extends continuously as q, every Riemann-integrable f is Riemann--Stieltjes integrable and uvfdα=uvfq on each closed subinterval (A continuously differentiable integrator reduces Stieltjes integration to ordinary integration).

[L3]

The integral function of an integrable function is differentiable at each continuity point, with derivative equal to the integrand (The first fundamental theorem: if f is integrable on [a,b] and continuous at c, then F(c)=f(c); in particular a continuous f has F as a primitive).

[L4]

A continuous function with an interior derivative admitting an integrable extension satisfies Newton--Leibniz (Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative).

Proof

technique · reduction
1.1

By [L1], A(x)=axf(t)q(t)dt. The product fq is integrable by [L2] and is continuous at c.

givenL1L2
1.2

Under the hypotheses of clause 2, [L1] and [L2] give abfdα=abfq, while fq is an integrable extension of the interior derivative of G.

givenL1L2
2.1

Applying [L3] to the ordinary integral function in step 1.1 gives A(c)=f(c)q(c), including either relative endpoint case.

step 1.1L3
3.1

Applying [L4] to G gives abfq=G(b)G(a), and step 1.2 proves the second clause without any division by q.

step 1.2L4

Depends on

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