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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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Differentiation holds along families shrinking nicely

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)).

Let fLloc1(Rn), let ARn, and suppose that for each xA there is a family (Er(x))r>0 shrinking nicely to x with constant αx>0. Then for almost every xA, limr0+1λ(Er(x))Er(x)f(y)f(x)dλ(y)=0, hence limr0+1λ(Er(x))Er(x)f(y)dλ(y)=f(x).

Facts & Assumptions

Given: The Axiom of Countable Choice, a locally integrable function f, a set ARn, and for each xA a family (Er(x))r>0 shrinking nicely to x.

[L1]

Shrinking nicely means that for each xA there is a constant αx>0 such that Er(x)B(x,r)andλ(Er(x))αxλ(B(x,r)) for every r>0. (A family shrinking nicely to a point)

[L2]

Almost every point of f is a Lebesgue point. (Almost every point is a Lebesgue point of a locally integrable function)

Proof

technique · direct
1.1

Let xA be a Lebesgue point of f, as supplied by [L2]. For every [L1, L2, given, algebra] r>0, [L1] gives 1λ(Er(x))Er(x)f(y)f(x)dλ(y)1αxλ(B(x,r))B(x,r)f(y)f(x)dλ(y).

L1L2givenalgebra
2.1

Because x is a Lebesgue point, the right-hand side of step 1.1 tends to [step 1.1] 0 as r0+. Therefore the left-hand side also tends to 0.

step 1.1
3.1

Using [step 2.1, algebra] 1λ(Er(x))Er(x)f(y)dλ(y)f(x)1λ(Er(x))Er(x)f(y)f(x)dλ(y), step 2.1 immediately gives the second limit as well.

step 2.1algebra
4.1

Step 3.1 holds at every Lebesgue point of f, hence for almost every xA.

L2step 3.1

Depends on

Used by

Dependency tree · two levels

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