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The centered Hardy-Littlewood maximal function is Borel measurable

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $f \in L_{l o c}^{1} (R^{n})$ . Then the centered maximal function $M f : R^{n} \to [0, \infty]$ is Borel measurable.

Facts & Assumptions

Given: The Axiom of Countable Choice and a locally integrable function $f$ on $R^{n}$ .

[L1]

The centered maximal function is $M f (x) = \sup_{r > 0} A_{r} ∣ f ∣ (x) .$ (The centered and uncentered Hardy-Littlewood maximal functions)

[L2]

For every locally integrable function $g$ , the map $(x, r) \mapsto A_{r} g (x)$ on $R^{n} \times (0, \infty)$ is continuous. (Ball averages vary continuously with the centre and radius)

Proof

technique · direct

1.1given

Fix a real $t$ . If $t < 0$ , then ${M f > t} = R^{n}$ , which is open. [given] Assume from now on that $t \geq 0$ .

1.2L1L2givenchoose

Let $x \in {M f > t}$ . By [L1], there is $r > 0$ with $A_{r} ∣ f ∣ (x) > t$ . Apply [L1, L2, given, choose] to the locally integrable function $∣ f ∣$ : continuity of $y \mapsto A_{r} ∣ f ∣ (y)$ at $x$ gives $δ > 0$ such that $∥ y - x ∥_{2} < δ ⟹ A_{r} ∣ f ∣ (y) > t .$ Hence $B (x, δ) \subseteq {M f > t}$ .

2.1step 1.2

Step 1.2 shows that every point of ${M f > t}$ is interior, so this [step 1.2] superlevel set is open.

3.1step 1.1step 2.1∎

Every strict superlevel set of $M f$ is open, so $M f$ is Borel measurable. [step 1.1, step 2.1]

Depends on

Used by

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., sentence after Lemma 3.16 (standard reference, not scraped)