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 ( ${A C}_{ω}$ )).

Let $f \in L_{l o c}^{1} (R^{n})$ , let $A \subseteq R^{n}$ , and suppose that for each $x \in A$ there is a family $(E_{r} (x))_{r > 0}$ shrinking nicely to $x$ with constant $α_{x} > 0$ . Then for almost every $x \in A$ , $\lim_{r \to 0^{+}} \frac{1}{λ (E_{r} (x))} \int_{E_{r} (x)} ∣ f (y) - f (x) ∣ d λ (y) = 0,$ hence $\lim_{r \to 0^{+}} \frac{1}{λ (E_{r} (x))} \int_{E_{r} (x)} f (y) d λ (y) = f (x) .$

Facts & Assumptions

Given: The Axiom of Countable Choice, a locally integrable function $f$ , a set $A \subseteq R^{n}$ , and for each $x \in A$ a family $(E_{r} (x))_{r > 0}$ shrinking nicely to $x$ .

[L1]

Shrinking nicely means that for each $x \in A$ there is a constant $α_{x} > 0$ such that $E_{r} (x) \subseteq B (x, r) and λ (E_{r} (x)) \geq α_{x} λ (B (x, r))$ 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

1.1L1L2givenalgebra

Let $x \in A$ be a Lebesgue point of $f$ , as supplied by [L2]. For every [L1, L2, given, algebra] $r > 0$ , [L1] gives $\frac{1}{λ (E_{r} (x))} \int_{E_{r} (x)} ∣ f (y) - f (x) ∣ d λ (y) \leq \frac{1}{α_{x} λ (B (x, r))} \int_{B (x, r)} ∣ f (y) - f (x) ∣ d λ (y) .$

2.1step 1.1

Because $x$ is a Lebesgue point, the right-hand side of step 1.1 tends to [step 1.1] $0$ as $r \to 0^{+}$ . Therefore the left-hand side also tends to $0$ .

3.1step 2.1algebra

Using [step 2.1, algebra] $∣ \frac{1}{λ (E_{r} (x))} \int_{E_{r} (x)} f (y) d λ (y) - f (x) ∣ \leq \frac{1}{λ (E_{r} (x))} \int_{E_{r} (x)} ∣ f (y) - f (x) ∣ d λ (y),$ step 2.1 immediately gives the second limit as well.

4.1L2step 3.1∎

Step 3.1 holds at every Lebesgue point of $f$ , hence for almost every $x \in A$ .

Differentiation holds along families shrinking nicely

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources