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TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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Fubini's theorem on term-by-term differentiation for pointwise sums of nondecreasing functions

Statement

Assume the Axiom of Countable Choice.

Let (Fn)n1 be a sequence (Sequences of reals: bounded, eventually, frequently, tails, subsequences) of nondecreasing functions on [a,b], and suppose that the pointwise sum

F(x):=n1Fn(x)

converges to a finite real number for every x[a,b]. Then F is nondecreasing, and for almost every x(a,b),

F(x)=n1Fn(x).

Facts & Assumptions

Given: Countable choice and a pointwise convergent series of nondecreasing functions F=n1Fn on [a,b].

[A1]

The symbols are those of the statement.

Proof

technique · direct
1.1

Every partial sum GN:=n=1NFn and every tail HN:=n>NFn=FGN is nondecreasing. By A monotone function is differentiable almost everywhere by the rising-sun route, the derivatives of F, all Fn, all GN, and all HN exist on a common full-measure set. On that set, GN=n=1NFn and HN=FGN.

given
2.1

Fix N. Since HN is increasing, the derivative bound theorem For a nondecreasing function, the derivative is measurable and integrable and its integral is bounded by the total increase gives 0ab(Fn=1NFn)dλ=abHN(x)dλ(x)HN(b)HN(a). But HN(b)HN(a)0 because the series defining F(a) and F(b) both converge.

step 1.1
3.1

The nonnegative functions Fn=1NFn decrease pointwise almost everywhere to Fn1Fn. Step 2.1 therefore forces ab(Fn1Fn)dλ=0. Since the integrand is nonnegative, it vanishes almost everywhere. Thus F=n1Fn almost everywhere.

step 2.1
4.1

Steps 1.1 through 3.1 prove the theorem.

step 1.1step 2.1step 3.1

Depends on

Used by

Dependency tree · two levels

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Sources