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✓ 18 results · all verified · 6 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 12 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Maximal Function and Lebesgue Differentiation

1 · Prerequisites

2 · Summary

This page builds the Euclidean maximal-function and differentiation package in the route fixed by the MT-17 design. Ball averages and the centered and uncentered maximal functions are defined separately, measurability of the centered maximal function is proved as a theorem rather than treated as obvious, and the weak (1,1) estimate keeps the honest 5^n constant from the Vitali selection argument.

The second half of the page turns the maximal inequality into differentiation: continuous compactly supported functions differentiate first, then the full Lebesgue differentiation theorem follows by density and error-set control, and the page closes with Lebesgue points, density points, nicely shrinking families, differentiation of measures, and the L^1 first fundamental theorem of calculus. In the current library route, the measure-theoretic parts of this package inherit the Axiom of Countable Choice from the published Lebesgue measure regularity and density inputs they cite.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-04Open item page →

A locally integrable function on $R^{n}$

Definition

Let $f : R^{n} \to C$ be a measurable function (Borel measurable and Lebesgue measurable functions on $R^{n}$ ). We say that $f$ is locally integrable on $R^{n}$ if for every Euclidean ball $B (x, r)$ with $r > 0$ (Open ball, closed ball and sphere in a metric space) one has $\int_{B (x, r)} ∣ f (y) ∣ d λ (y) < \infty,$ so the restriction of $f$ to each ball is integrable in the sense of Integrable real and complex functions, and their integrals.

The set of such functions is denoted by $L_{l o c}^{1} (R^{n})$ . When convenient, the same notation is also used for the corresponding almost-everywhere equivalence classes.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04Open item page →

Euclidean balls have positive finite Lebesgue measure

Statement

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

Let $x \in R^{n}$ and let $r > 0$ . Then the Euclidean ball $B (x, r)$ is Lebesgue measurable and satisfies $0 < λ (B (x, r)) < \infty .$

Facts & Assumptions

Given: The Axiom of Countable Choice, a point $x \in R^{n}$ , and a real radius $r > 0$ .

[L1]

The Euclidean ball is $B (x, r) = {y \in R^{n} : ∥ y - x ∥_{2} < r}$ . (Open ball, closed ball and sphere in a metric space)

[L2]

Every box between its open and closed forms is Lebesgue measurable with its usual volume. (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included)

Proof

technique · direct

1.1

If $∥ y - x ∥_{\infty} < r / \sqrt{n}$ , then [L1, algebra] $∥ y - x ∥_{2} \leq \sqrt{n} ∥ y - x ∥_{\infty} < r,$ so the open cube $x + (- r / \sqrt{n}, r / \sqrt{n})^{n}$ is contained in $B (x, r)$ . On the other hand, $∥ y - x ∥_{2} < r$ implies $∣ y_{i} - x_{i} ∣ < r$ for every coordinate, so $B (x, r) \subseteq x + (- r, r)^{n}$ .

L1algebra

2.1

The two cubes from step 1.1 are measurable by [L2], with measures [step 1.1, L2, algebra] $(2 r / \sqrt{n})^{n} > 0 and (2 r)^{n} < \infty .$ Since $B (x, r)$ lies between them, monotonicity gives $0 < (2 r / \sqrt{n})^{n} \leq λ (B (x, r)) \leq (2 r)^{n} < \infty .$

step 1.1L2algebra

3.1

Therefore $B (x, r)$ is Lebesgue measurable of positive finite measure. [step 2.1]

step 2.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-04Open item page →

The average of a locally integrable function over a Euclidean ball

Definition

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

Let $f \in L_{l o c}^{1} (R^{n})$ (A locally integrable function on $R^{n}$ ), let $x \in R^{n}$ , and let $r > 0$ . The ball average of $f$ over $B (x, r)$ is $A_{r} f (x) : = \frac{1}{λ (B (x, r))} \int_{B (x, r)} f (y) d λ (y) .$

The denominator is well defined because Euclidean balls have positive finite Lebesgue measure shows that every Euclidean ball has positive finite Lebesgue measure.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-04Open item page →

The centered and uncentered Hardy-Littlewood maximal functions

Definition

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

Let $f \in L_{l o c}^{1} (R^{n})$ . The centered Hardy-Littlewood maximal function of $f$ is $M f (x) : = \sup_{r > 0} A_{r} ∣ f ∣ (x) = \sup_{r > 0} \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) ∣ d λ (y) .$

The uncentered Hardy-Littlewood maximal function of $f$ is $M^{*} f (x) : = \sup_{x \in B (y, r)} \frac{1}{λ (B (y, r))} \int_{B (y, r)} ∣ f (z) ∣ d λ (z),$ where the supremum is over all Euclidean balls containing $x$ .

Both functions take values in $[0, \infty]$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-04Open item page →

Sublinear operators and weak or strong type $(p, q)$ bounds

Definition

Let $(X, A, μ)$ and $(Y, B, ν)$ be measure spaces, and let $T$ assign to each measurable function on $X$ a measurable function on $Y$ .

The operator $T$ is sublinear if for all scalars $a, b$ and measurable functions $f, g$ one has $∣ T (a f + b g) ∣ \leq ∣ a ∣ ∣ T f ∣ + ∣ b ∣ ∣ T g ∣ .$

For $1 \leq p \leq \infty$ and $1 \leq q \leq \infty$ , we say that $T$ is of strong type $(p, q)$ if there is a constant $C \geq 0$ such that $∥ T f ∥_{L^{q} (ν)} \leq C ∥ f ∥_{L^{p} (μ)}$ for every $f \in L^{p} (μ)$ (The space $L^{p} (μ)$ as the quotient by null functions, The space $L^{\infty} (μ)$ of essentially bounded measurable functions).

For $1 \leq p < \infty$ and $1 \leq q < \infty$ , we say that $T$ is of weak type $(p, q)$ if there is a constant $C \geq 0$ such that for every $f \in L^{p} (μ)$ and every $t > 0$ , $ν ({∣ T f ∣ > t}) \leq {(\frac{C ∥ f ∥_{L^{p} (μ)}}{t})}^{q} .$ Equivalently, the distribution function of $T f$ (The distribution function of absolute value) obeys $A_{T f} (t) \leq {(\frac{C ∥ f ∥_{L^{p} (μ)}}{t})}^{q} (t > 0) .$

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-04Open item page →

Lebesgue points and the Lebesgue set of an $L_{l o c}^{1}$ class

Definition

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

Let $[f] \in L_{l o c}^{1} (R^{n})$ (A locally integrable function on $R^{n}$ ).

For a representative $f$ of that class, a point $x \in R^{n}$ is a Lebesgue point of $f$ if $\lim_{r \to 0^{+}} \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) - f (x) ∣ d λ (y) = 0.$

The Lebesgue set of the class $[f]$ is the set $Leb ([f]) : = {x \in R^{n} : some representative of [f] has x as a Lebesgue point} .$

The later theorem on Lebesgue points shows that different representatives of the same $L_{l o c}^{1}$ class change this set only by a null set.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-04Open item page →

Density of a measurable set at a point

Definition

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

Let $E \subseteq R^{n}$ be Lebesgue measurable (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ), and let $x \in R^{n}$ . If the limit exists, the density of $E$ at $x$ is $Θ (E, x) : = \lim_{r \to 0^{+}} \frac{λ (E \cap B (x, r))}{λ (B (x, r))},$ where the denominator is positive and finite by Euclidean balls have positive finite Lebesgue measure.

When $Θ (E, x) = 1$ , we call $x$ a density-one point of $E$ ; when $Θ (E, x) = 0$ , we call $x$ a density-zero point of $E$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-04Open item page →

A family shrinking nicely to a point

Definition

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

Let $x \in R^{n}$ . A family $(E_{r})_{r > 0}$ of Lebesgue measurable subsets of $R^{n}$ shrinks nicely to $x$ if there is a constant $α > 0$ such that for every $r > 0$ , $E_{r} \subseteq B (x, r) and λ (E_{r}) \geq α λ (B (x, r)) .$

The constant $α$ is part of the data: later comparison estimates depend on it explicitly.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04Open item page →

Vitali covering lemma for Euclidean balls with fivefold dilates

Statement

For a ball $B (x, r)$ , write $5 B (x, r) : = B (x, 5 r)$ .

Let ${B_{1}, \dots, B_{m}}$ be a finite family of Euclidean balls in $R^{n}$ . Then there is a pairwise disjoint subfamily ${B_{i_{1}}, \dots, B_{i_{ℓ}}}$ such that $⋃_{j = 1}^{m} B_{j} \subseteq ⋃_{k = 1}^{ℓ} 5 B_{i_{k}} .$ Consequently, $λ ⁣ (⋃_{j = 1}^{m} B_{j}) \leq 5^{n} \sum_{k = 1}^{ℓ} λ (B_{i_{k}}) .$
Let $(B_{j})_{j \geq 1}$ be a countable family of Euclidean balls whose radii are bounded above. Then there is a finite or countably infinite index set $I \subseteq N_{\geq 1}$ such that $(B_{i})_{i \in I}$ is pairwise disjoint and $⋃_{j \geq 1} B_{j} \subseteq ⋃_{i \in I} 5 B_{i} .$

Facts & Assumptions

Given: A family of Euclidean balls in $R^{n}$ .

[L1]

The Euclidean balls are the sets $B (x, r) = {y : ∥ y - x ∥_{2} < r}$ . (Open ball, closed ball and sphere in a metric space)

[L2]

Lebesgue measure scales by $c^{n}$ under dilation by $c > 0$ . In particular, for every ball $B$ , $λ (5 B) = 5^{n} λ (B) .$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it)

Proof

technique · direct

1.1

For the finite family, choose $B_{i_{1}}$ with maximal radius among [given, choose] $B_{1}, \dots, B_{m}$ . Having chosen disjoint balls $B_{i_{1}}, \dots, B_{i_{s - 1}}$ , choose $B_{i_{s}}$ with maximal radius among the remaining balls disjoint from all earlier choices, and stop when none remain. The chosen subfamily is pairwise disjoint by construction.

givenchoose

1.2

Now let $(B_{j})_{j \geq 1}$ be countable with radii bounded above, and put $R : = \sup_{j \geq 1} r_{j} .$ For each integer $m \geq 0$ , let $I_{m} : = {j \geq 1 : 2^{- m - 1} R < r_{j} \leq 2^{- m} R} .$ Process the classes $I_{0}, I_{1}, \dots$ in this order, and within each class inspect the indices in increasing order. Retain $B_{j}$ exactly when it is disjoint from every ball already retained. The retained subfamily is pairwise disjoint by construction.

givenconstructalgebra

2.1

Let $B_{j}$ be one of the original balls. If it was chosen, then [step 1.1, L1, choose, algebra] $B_{j} \subseteq 5 B_{j}$ . If it was not chosen, let $B_{i_{s}}$ be the first chosen ball that meets it. Since the choice at stage $s$ had maximal radius among the remaining disjoint balls, the radius of $B_{j}$ is at most that of $B_{i_{s}}$ . Pick $z \in B_{j}$ and choose $w \in B_{j} \cap B_{i_{s}}$ . Then $∥ z - c_{i_{s}} ∥_{2} \leq ∥ z - w ∥_{2} + ∥ w - c_{j} ∥_{2} + ∥ c_{j} - c_{i_{s}} ∥_{2} < r_{j} + r_{j} + r_{i_{s}} \leq 5 r_{i_{s}},$ so $z \in 5 B_{i_{s}}$ . Therefore every original ball lies in the union of the fivefold dilates of the chosen balls.

step 1.1L1choosealgebra

2.2

Let $B_{j}$ be any original ball that was not chosen in the countable construction, and let $j \in I_{m}$ . When the algorithm inspected $j$ , some previously chosen ball $B_{i_{s}}$ already met $B_{j}$ ; otherwise $B_{j}$ would have been retained. If $i_{s} \in I_{ℓ}$ with $ℓ < m$ , then $r_{j} \leq 2^{- m} R \leq 2^{- ℓ - 1} R < r_{i_{s}} .$ If instead $ℓ = m$ , then both balls lie in the same dyadic class, so $r_{j} \leq 2^{- m} R < 2 r_{i_{s}} .$ In either case, $r_{j} < 2 r_{i_{s}} .$ Choose $z \in B_{j}$ and $w \in B_{j} \cap B_{i_{s}}$ . Then $∥ z - c_{i_{s}} ∥_{2} < r_{j} + r_{j} + r_{i_{s}} < 2 (2 r_{i_{s}}) + r_{i_{s}} = 5 r_{i_{s}},$ so again $B_{j} \subseteq 5 B_{i_{s}}$ . Let $I \subseteq N_{\geq 1}$ be the set of retained indices. This set is finite or countably infinite, and chosen balls are also contained in their own fivefold dilates; hence $⋃_{j \geq 1} B_{j} \subseteq ⋃_{i \in I} 5 B_{i} .$

step 1.2L1choosealgebra

3.1

Since the chosen balls are pairwise disjoint, [step 2.1, L2, algebra] $λ ⁣ (⋃_{j = 1}^{m} B_{j}) \leq λ ⁣ (⋃_{k = 1}^{ℓ} 5 B_{i_{k}}) \leq \sum_{k = 1}^{ℓ} λ (5 B_{i_{k}}) = 5^{n} \sum_{k = 1}^{ℓ} λ (B_{i_{k}}) .$ This proves part 1.

step 2.1L2algebra

4.1

Steps 3.1 and 2.2 prove the finite and countable forms.

step 3.1step 2.2∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04Open item page →

Ball averages vary continuously with the centre and radius

Statement

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

Let $f \in L_{l o c}^{1} (R^{n})$ . The map $Φ : R^{n} \times (0, \infty) \to C, Φ (x, r) : = A_{r} f (x),$ is continuous.

Facts & Assumptions

Given: The Axiom of Countable Choice, a locally integrable function $f$ , a point $(x, r) \in R^{n} \times (0, \infty)$ , and a sequence $(x_{k}, r_{k}) \to (x, r)$ with $r_{k} > 0$ .

[L1]

The ball average is $A_{r} f (x) = \frac{1}{λ (B (x, r))} \int_{B (x, r)} f (y) d λ (y) .$ (The average of a locally integrable function over a Euclidean ball)

[L2]

Euclidean balls have positive finite Lebesgue measure. (Euclidean balls have positive finite Lebesgue measure)

[L3]

Lebesgue measure scales by $c^{n}$ under dilation by $c > 0$ . (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it)

[L4]

Dominated convergence passes pointwise almost-everywhere limits through an integrable majorant. (Dominated convergence)

Proof

technique · direct

1.1

Choose $ρ > 0$ with $0 < ρ < r / 2$ . For all sufficiently large $k$ , [given, choose, algebra] $∣ r_{k} - r ∣ < ρ$ and $∥ x_{k} - x ∥_{2} < ρ$ . Then $B (x, r - ρ) \subseteq B (x_{k}, r_{k}) \subseteq B (x, r + 2 ρ) .$ Since $f$ is locally integrable, the function $g : = ∣ f ∣ 1_{B (x, r + 2 ρ)}$ is integrable.

givenchoosealgebra

1.2

Put $χ_{k} : = 1_{B (x_{k}, r_{k})}$ and $χ : = 1_{B (x, r)}$ . [algebra] If $y \notin \partial B (x, r)$ , then $∥ y - x ∥_{2} \neq r$ , so for all sufficiently large $k$ the membership of $y$ in $B (x_{k}, r_{k})$ agrees with its membership in $B (x, r)$ . Thus $χ_{k} (y) \to χ (y)$ for every $y \notin \partial B (x, r)$ .

algebra

1.3

The boundary sphere satisfies [L2, L3, algebra] $\partial B (x, r) \subseteq B (x, r + \frac{1}{m}) ∖ B (x, r - \frac{1}{m}) (m > 1 / r) .$ By [L2] and [L3], $λ ⁣ (B (x, r + \frac{1}{m}) ∖ B (x, r - \frac{1}{m})) = λ (B (0, 1)) ((r + \frac{1}{m})^{n} - (r - \frac{1}{m})^{n}) \to 0,$ so $λ (\partial B (x, r)) = 0$ .

L2L3algebra

1.4

By [L2] and [L3], [L2, L3, algebra] $λ (B (x_{k}, r_{k})) = λ (B (0, 1)) r_{k}^{n} \to λ (B (0, 1)) r^{n} = λ (B (x, r)) .$ The limit denominator is positive by [L2].

L2L3algebra

2.1

Steps 1.1, 1.2, and 1.3 let us apply [L4] to [step 1.1, step 1.2, step 1.3, L4] $f χ_{k}$ , dominated by $g$ , and obtain $\int f (y) χ_{k} (y) d λ (y) \to \int f (y) χ (y) d λ (y) .$ In other words, $\int_{B (x_{k}, r_{k})} f d λ \to \int_{B (x, r)} f d λ .$

step 1.1step 1.2step 1.3L4

3.1

Combining steps 2.1 and 1.4 yields [L1, step 2.1, step 1.4, algebra] $A_{r_{k}} f (x_{k}) \to A_{r} f (x) .$ Since the approximating sequence was arbitrary, $Φ$ is continuous.

L1step 2.1step 1.4algebra∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04Open item page →

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

1.1

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

given

1.2

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 $B (x, δ) \subseteq {M f > t}$ .

L1L2givenchoose

2.1

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

step 1.2

3.1

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

step 1.1step 2.1∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04Open item page →

The centered and uncentered maximal functions are pointwise comparable

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 for every $x \in R^{n}$ , $M f (x) \leq M^{*} f (x) \leq 2^{n} M f (x) .$

Facts & Assumptions

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

[L1]

The centered maximal function takes the supremum over balls centered at $x$ , while the uncentered maximal function takes the supremum over all balls that contain $x$ . (The centered and uncentered Hardy-Littlewood maximal functions)

[L2]

Lebesgue measure scales by $∣ c ∣^{n}$ under dilation by $c \neq 0$ . (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it)

Proof

technique · direct

1.1

Every ball centered at $x$ is in particular a ball containing $x$ , so the [L1] supremum defining $M f (x)$ is taken over a smaller family than the one defining $M^{*} f (x)$ . Therefore $M f (x) \leq M^{*} f (x)$ .

1.2

Let $B (y, r)$ be any ball containing $x$ . If $z \in B (y, r)$ , then [L1, algebra] $∥ z - x ∥_{2} \leq ∥ z - y ∥_{2} + ∥ y - x ∥_{2} < r + r = 2 r,$ so $B (y, r) \subseteq B (x, 2 r)$ . Hence $\int_{B (y, r)} ∣ f ∣ \leq \int_{B (x, 2 r)} ∣ f ∣ d λ .$

L1algebra

2.1

Step 1.2 and [L2] give [step 1.2, L1, L2, algebra] $\frac{1}{λ (B (y, r))} \int_{B (y, r)} ∣ f ∣ \leq \frac{λ (B (x, 2 r))}{λ (B (y, r))} \cdot \frac{1}{λ (B (x, 2 r))} \int_{B (x, 2 r)} ∣ f ∣ = 2^{n} A_{2 r} ∣ f ∣ (x) \leq 2^{n} M f (x) .$ Taking the supremum over all balls $B (y, r)$ containing $x$ yields $M^{*} f (x) \leq 2^{n} M f (x)$ .

step 1.2L1L2algebra

3.1

Combining steps 1.1 and 2.1 gives the claimed comparison. [step 1.1, step 2.1]

step 1.1step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04Open item page →

The centered Hardy-Littlewood maximal operator is weak type $(1, 1)$

Statement

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

Let $f \in L^{1} (R^{n})$ and let $t > 0$ . Then $λ ({x \in R^{n} : M f (x) > t}) \leq \frac{5^{n}}{t} ∥ f ∥_{1} .$ In particular, the centered maximal operator is of weak type $(1, 1)$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, a function $f \in L^{1} (R^{n})$ , and a real number $t > 0$ .

[L1]

The centered maximal function is $M f (x) = \sup_{r > 0} \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) ∣ d λ (y) .$ (The centered and uncentered Hardy-Littlewood maximal functions)

[L2]

For a locally integrable function, the ball-average map $(x, r) \mapsto \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) ∣ d λ$ is continuous on $R^{n} \times (0, \infty)$ . (Ball averages vary continuously with the centre and radius)

[L3]

Lebesgue measure on $R^{n}$ is inner regular by compact sets on open sets. (Assuming countable choice, the Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets)

[L4]

A finite family of balls admits a disjoint subfamily whose fivefold dilates cover the original union, with the measure estimate $λ ⁣ (⋃_{j} B_{j}) \leq 5^{n} \sum_{k} λ (B_{i_{k}}) .$ (Vitali covering lemma for Euclidean balls with fivefold dilates)

[L5]

The $L^{1}$ norm is $∥ f ∥_{1} = \int_{R^{n}} ∣ f ∣ d λ .$ (The class $L^{1} (μ)$ of integrable functions)

Proof

technique · direct

1.1

Put $E_{t} : = {x \in R^{n} : M f (x) > t}$ . If $x \in E_{t}$ , then [L1] gives a radius $r_{x} > 0$ such that $\int_{B (x, r_{x})} ∣ f (y) ∣ d λ (y) > t λ (B (x, r_{x})) .$ By [L2] applied to $∣ f ∣$ at $(x, r_{x})$ , there is $δ_{x} > 0$ such that the same strict inequality holds with $x$ replaced by every $y \in B (x, δ_{x})$ and the radius kept equal to $r_{x}$ . Hence $B (x, δ_{x}) \subseteq E_{t}$ , so $E_{t}$ is open. Now let [L1, L2, given, choose] $K \subseteq E_{t}$ be compact. The balls $B (x, r_{x})$ with $x \in K$ cover $K$ , so compactness yields a finite subcover $B (x_{1}, r_{1}), \dots, B (x_{m}, r_{m})$ .

L1L2givenchoose

2.1

Apply [L4] to that finite subcover. There are pairwise disjoint balls [step 1.1, L4, L5, algebra] $B_{i_{1}}, \dots, B_{i_{ℓ}}$ among $B (x_{1}, r_{1}), \dots, B (x_{m}, r_{m})$ such that $K \subseteq ⋃_{j = 1}^{m} B (x_{j}, r_{j}) \subseteq ⋃_{k = 1}^{ℓ} 5 B_{i_{k}}$ and therefore $λ (K) \leq 5^{n} \sum_{k = 1}^{ℓ} λ (B_{i_{k}}) .$ Each chosen ball still satisfies the witness inequality from step 1.1, so $t \sum_{k = 1}^{ℓ} λ (B_{i_{k}}) < \sum_{k = 1}^{ℓ} \int_{B_{i_{k}}} ∣ f ∣ d λ \leq \int_{R^{n}} ∣ f ∣ d λ = ∥ f ∥_{1},$ because the chosen balls are pairwise disjoint. Hence $λ (K) \leq \frac{5^{n}}{t} ∥ f ∥_{1} .$

step 1.1L4L5algebra

3.1

By [L3], the open set $E_{t}$ is the supremum of the measures of its compact [step 2.1, L3, algebra] subsets. Step 2.1 gives the same upper bound for every compact $K \subseteq E_{t}$ , so $λ (E_{t}) \leq \frac{5^{n}}{t} ∥ f ∥_{1} .$

step 2.1L3algebra

4.1

This is exactly the weak type $(1, 1)$ estimate for the centered maximal [step 3.1] operator.

step 3.1∎

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04Open item page →

The centered maximal operator is bounded on $L^{\infty}$

Statement

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

Let $f \in L^{\infty} (R^{n})$ . Then $M f (x) \leq ∥ f ∥_{\infty} (x \in R^{n}) .$ In particular, $M$ is of strong type $(\infty, \infty)$ with operator norm at most $1$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a function $f \in L^{\infty} (R^{n})$ .

[L1]

The centered maximal function is $M f (x) = \sup_{r > 0} \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) ∣ d λ (y) .$ (The centered and uncentered Hardy-Littlewood maximal functions)

[L2]

The $L^{\infty}$ norm is the essential supremum. In particular, $∣ f ∣ \leq ∥ f ∥_{\infty}$ almost everywhere. (The space $L^{\infty} (μ)$ of essentially bounded measurable functions)

Proof

technique · direct

1.1

Fix $x \in R^{n}$ and $r > 0$ . By [L2], [L1, L2, given, algebra] $\int_{B (x, r)} ∣ f (y) ∣ d λ (y) \leq ∥ f ∥_{\infty} λ (B (x, r)) .$ Dividing by $λ (B (x, r))$ gives $A_{r} ∣ f ∣ (x) \leq ∥ f ∥_{\infty} .$

L1L2givenalgebra

2.1

Taking the supremum of the inequality from step 1.1 over all $r > 0$ yields [step 1.1, L1, algebra] $M f (x) \leq ∥ f ∥_{\infty}$ . Since $x$ was arbitrary, the centered maximal operator is bounded on $L^{\infty}$ with norm at most $1$ .

step 1.1L1algebra∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04Open item page →

Marcinkiewicz interpolation from weak $(1, 1)$ and strong $(\infty, \infty)$

Statement

Let $(X, A, μ)$ be a measure space, let $T$ be a sublinear operator on measurable functions, and suppose:

$T$ is of weak type $(1, 1)$ with constant $A$ ;
$T$ is of strong type $(\infty, \infty)$ with constant $B$ .

Then for every $1 < p < \infty$ and every $f \in L^{p} (μ)$ , $∥ T f ∥_{p} \leq 2 {(\frac{A p}{p - 1})}^{1 / p} B^{1 - 1 / p} ∥ f ∥_{p} .$ In particular, $T$ is of strong type $(p, p)$ for every $1 < p < \infty$ .

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , a sublinear operator $T$ , constants $A, B \geq 0$ , an exponent $1 < p < \infty$ , and a function $f \in L^{p} (μ)$ .

[L1]

Sublinearity, weak type $(1, 1)$ , and strong type $(\infty, \infty)$ are as defined in Sublinear operators and weak or strong type $(p, q)$ bounds.

[L2]

The distribution function of a measurable function $g$ is $A_{g} (t) = μ ({∣ g ∣ > t}) .$ (The distribution function of absolute value)

[L3]

For $0 < p < \infty$ , $\int ∣ g ∣^{p} d μ = p \int_{0}^{\infty} t^{p - 1} μ ({∣ g ∣ > t}) d t$ for every measurable $g$ . (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function)

Proof

technique · direct

1.1

Fix $t > 0$ and $η > 0$ , and put $C : = B + η > 0$ . Since the strong [L1, given, construct, algebra] $(\infty, \infty)$ bound with constant $B$ also holds with the larger constant $C$ , split $f = f_{t}^{>} + f_{t}^{\leq}, f_{t}^{>} : = f 1_{{∣ f ∣ > t / (2 C)}}, f_{t}^{\leq} : = f 1_{{∣ f ∣ \leq t / (2 C)}} .$ Sublinearity gives $∣ T f ∣ \leq ∣ T f_{t}^{>} ∣ + ∣ T f_{t}^{\leq} ∣ .$ Since $∥ f_{t}^{\leq} ∥_{\infty} \leq t / (2 C)$ , the strong $(\infty, \infty)$ bound with constant $C$ yields $∣ T f_{t}^{\leq} ∣ \leq C ∥ f_{t}^{\leq} ∥_{\infty} \leq t / 2.$ Therefore ${∣ T f ∣ > t} \subseteq {∣ T f_{t}^{>} ∣ > t / 2} .$

L1givenconstructalgebra

2.1

Apply the weak $(1, 1)$ bound to $f_{t}^{>}$ : [L1, step 1.1, algebra] $μ ({∣ T f ∣ > t}) \leq μ ({∣ T f_{t}^{>} ∣ > t / 2}) \leq \frac{2 A}{t} ∥ f_{t}^{>} ∥_{1} = \frac{2 A}{t} \int_{{∣ f ∣ > t / (2 C)}} ∣ f ∣ d μ .$

L1step 1.1algebra

3.1

Using [L3] with $g = T f$ and then step 2.1, [L2, L3, step 2.1, algebra] $∥ T f ∥_{p}^{p} = p \int_{0}^{\infty} t^{p - 1} μ ({∣ T f ∣ > t}) d t \leq 2 A p \int_{0}^{\infty} t^{p - 2} (\int_{{∣ f ∣ > t / (2 C)}} ∣ f ∣ d μ) d t .$

L2L3step 2.1algebra

4.1

The integrand in step 3.1 is nonnegative, so Tonelli's theorem for [step 3.1, algebra] nonnegative integrals lets us swap the order: $∥ T f ∥_{p}^{p} \leq 2 A p \int_{X} ∣ f (x) ∣ (\int_{0}^{2 C ∣ f (x) ∣} t^{p - 2} d t) d μ (x) .$ Because $p > 1$ , $\int_{0}^{2 C ∣ f (x) ∣} t^{p - 2} d t = \frac{(2 C ∣ f (x) ∣)^{p - 1}}{p - 1},$ so $∥ T f ∥_{p}^{p} \leq \frac{2^{p} A p}{p - 1} C^{p - 1} \int_{X} ∣ f ∣^{p} d μ .$

step 3.1algebra

5.1

Taking $p$ th roots in step 4.1 gives [step 4.1, algebra] $∥ T f ∥_{p} \leq 2 {(\frac{A p}{p - 1})}^{1 / p} C^{1 - 1 / p} ∥ f ∥_{p} .$ Because $η > 0$ was arbitrary, letting $η ↓ 0$ yields $∥ T f ∥_{p} \leq 2 {(\frac{A p}{p - 1})}^{1 / p} B^{1 - 1 / p} ∥ f ∥_{p} .$ Thus $T$ is of strong type $(p, p)$ for every $1 < p < \infty$ .

step 4.1algebra∎

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The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$

Statement

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

Let $1 < p < \infty$ . Then there is a constant $C_{n, p}$ such that every $f \in L^{p} (R^{n})$ satisfies $∥ M f ∥_{p} \leq C_{n, p} ∥ f ∥_{p} .$ By comparison, the same is true for the uncentered maximal function.

Facts & Assumptions

Given: The Axiom of Countable Choice, an exponent $1 < p < \infty$ , and a function $f \in L^{p} (R^{n})$ .

[L1]

The centered maximal operator is of weak type $(1, 1)$ with constant $5^{n}$ . (The centered Hardy-Littlewood maximal operator is weak type $(1, 1)$ )

[L2]

The centered maximal operator is of strong type $(\infty, \infty)$ with operator norm at most $1$ . (The centered maximal operator is bounded on $L^{\infty}$ )

[L3]

A sublinear operator of weak type $(1, 1)$ and strong type $(\infty, \infty)$ is of strong type $(p, p)$ for every $1 < p < \infty$ . (Marcinkiewicz interpolation from weak $(1, 1)$ and strong $(\infty, \infty)$ )

[L4]

One has $M f \leq M^{*} f \leq 2^{n} M f$ pointwise. (The centered and uncentered maximal functions are pointwise comparable)

Proof

technique · direct

1.1

The maximal operator is sublinear by definition of supremum and absolute [L1, L2, L3, given, algebra] values. Apply [L3] with the constants from [L1] and [L2]. This yields a constant $C_{n, p}$ such that $∥ M f ∥_{p} \leq C_{n, p} ∥ f ∥_{p} .$

L1L2L3givenalgebra

2.1

Step 1.1 proves the centered estimate. Then [L4] gives [step 1.1, L4, algebra] $∥ M^{*} f ∥_{p} \leq 2^{n} ∥ M f ∥_{p} \leq 2^{n} C_{n, p} ∥ f ∥_{p},$ so the uncentered estimate follows as well.

step 1.1L4algebra∎

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Riesz-Thorin interpolation theorem

Statement

Let $(X, A, μ)$ and $(Y, B, ν)$ be measure spaces. Let $1 \leq p_{0}, p_{1} < \infty$ , let $1 < q_{0}, q_{1} < \infty$ , let $0 < θ < 1$ , and define $\frac{1}{p_{θ}} : = \frac{1 - θ}{p_{0}} + \frac{θ}{p_{1}}, \frac{1}{q_{θ}} : = \frac{1 - θ}{q_{0}} + \frac{θ}{q_{1}},$ with the convention $1 / \infty : = 0$ .

Suppose $T$ is a linear operator on the finite simple functions of finite measure support on $X$ , and suppose $∥ T f ∥_{q_{0}} \leq M_{0} ∥ f ∥_{p_{0}}, ∥ T f ∥_{q_{1}} \leq M_{1} ∥ f ∥_{p_{1}}$ for every such $f$ . Then $T$ extends uniquely to a bounded linear operator $\tilde{T} : L^{p_{θ}} (μ) \to L^{q_{θ}} (ν)$ satisfying $∥ \tilde{T} f ∥_{q_{θ}} \leq M_{0}^{1 - θ} M_{1}^{θ} ∥ f ∥_{p_{θ}} (f \in L^{p_{θ}} (μ)) .$

Facts & Assumptions

Given: The operator $T$ on finite simple functions of finite measure support, endpoint bounds with constants $M_{0}, M_{1}$ , and a parameter $0 < θ < 1$ .

[L1]

For finite $p$ , simple functions with finite-measure support are dense in $L^{p}$ . (Simple functions with finite-measure support are dense in $L^{p} (μ)$ for $1 \leq p < \infty$ )

[L2]

For $1 \leq q < \infty$ , the $L^{q}$ norm is the supremum of pairings against unit $L^{q^{'}}$ functions. (The $L^{p}$ norm is the supremum of pairings against unit $L^{q}$ functions)

[L3]

Each $L^{q}$ with $1 \leq q \leq \infty$ is complete. (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ )

Proof

technique · direct

1.1

First assume that $f$ and $g$ are finite simple functions of finite support [L1, L2, given, choose] on $X$ and $Y$ , respectively, with $g$ chosen from $L^{q_{θ}^{'}} (ν)$ and $∥ g ∥_{q_{θ}^{'}} = 1$ . Write $f = \sum_{j = 1}^{m} a_{j} 1_{E_{j}}, g = \sum_{k = 1}^{ℓ} b_{k} 1_{F_{k}},$ with the sets $E_{j}, F_{k}$ pairwise disjoint and of finite measure.

L1L2givenchoose

2.1

Define the analytic families [step 1.1, construct, algebra] $f_{z} : = \sum_{j = 1}^{m} α_{j} ∣ a_{j} ∣^{p_{θ} ((1 - z) / p_{0} + z / p_{1})} 1_{E_{j}},$ $g_{z} : = \sum_{k = 1}^{ℓ} β_{k} ∣ b_{k} ∣^{q_{θ}^{'} ((1 - z) / q_{0}^{'} + z / q_{1}^{'})} 1_{F_{k}},$ where $α_{j} = a_{j} / ∣ a_{j} ∣$ and $β_{k} = b_{k} / ∣ b_{k} ∣$ when the coefficient is nonzero and $0$ otherwise. Then $f_{θ} = f$ and $g_{θ} = g$ . For real $t$ , direct calculation on each simple coefficient gives $∥ f_{i t} ∥_{p_{0}} = ∥ f ∥_{p_{θ}}^{p_{θ} / p_{0}}, ∥ f_{1 + i t} ∥_{p_{1}} = ∥ f ∥_{p_{θ}}^{p_{θ} / p_{1}},$ and likewise $∥ g_{i t} ∥_{q_{0}^{'}} = ∥ g ∥_{q_{θ}^{'}}^{q_{θ}^{'} / q_{0}^{'}} = 1, ∥ g_{1 + i t} ∥_{q_{1}^{'}} = ∥ g ∥_{q_{θ}^{'}}^{q_{θ}^{'} / q_{1}^{'}} = 1.$

step 1.1constructalgebra

3.1

Put [step 2.1, given, algebra] $Φ (z) : = \int_{Y} (T f_{z}) (y) g_{z} (y) d ν (y) .$ Because $f_{z}$ and $g_{z}$ are finite linear combinations of exponentials in $z$ , $Φ$ is continuous on the closed strip $S = {z \in C : 0 \leq Re z \leq 1}$ and holomorphic on its interior. For real $t$ , the endpoint bounds and Holder give $∣ Φ (i t) ∣ \leq M_{0} ∥ f_{i t} ∥_{p_{0}} ∥ g_{i t} ∥_{q_{0}^{'}} \leq M_{0} ∥ f ∥_{p_{θ}},$ $∣ Φ (1 + i t) ∣ \leq M_{1} ∥ f_{1 + i t} ∥_{p_{1}} ∥ g_{1 + i t} ∥_{q_{1}^{'}} \leq M_{1} ∥ f ∥_{p_{θ}} .$

step 2.1givenalgebra

4.1

Fix $δ > 0$ and define [step 3.1, construct, algebra] $Ψ_{δ} (z) : = Φ (z) (M_{0} + δ)^{z - 1} (M_{1} + δ)^{- z} .$ By step 3.1, $∣ Ψ_{δ} ∣$ is at most $1$ on the two boundary lines of the strip. Multiplying once more by $\exp (ε (z^{2} - 1))$ and applying the maximum-modulus principle on large rectangles inside the strip shows that $∣ Ψ_{δ} (z) ∣ \leq 1$ throughout $S$ . Evaluating at $z = θ$ and letting first $ε ↓ 0$ and then $δ ↓ 0$ yields $∣ Φ (θ) ∣ \leq M_{0}^{1 - θ} M_{1}^{θ} ∥ f ∥_{p_{θ}} .$

step 3.1constructalgebra

5.1

Since $Φ (θ) = \int (T f) g d ν$ , step 4.1 gives [L1, L2, step 4.1, algebra] $∣ \int (T f) g d ν ∣ \leq M_{0}^{1 - θ} M_{1}^{θ} ∥ f ∥_{p_{θ}}$ for every unit $g \in L^{q_{θ}^{'}} (ν)$ that is finite simple with finite support. By density [L1] and norm recovery [L2], it follows that $∥ T f ∥_{q_{θ}} \leq M_{0}^{1 - θ} M_{1}^{θ} ∥ f ∥_{p_{θ}}$ for every finite simple $f$ of finite support.

L1L2step 4.1algebra

6.1

Now let $f \in L^{p_{θ}} (μ)$ . By [L1], choose finite simple functions [L1, L3, step 5.1, algebra] $f_{n}$ of finite support with $f_{n} \to f$ in $L^{p_{θ}} (μ)$ . Step 5.1 makes $(T f_{n})$ Cauchy in $L^{q_{θ}} (ν)$ , so [L3] gives a limit $h \in L^{q_{θ}} (ν)$ with $∥ h - T f_{n} ∥_{q_{θ}} \to 0.$ If $(g_{n})$ is another such approximating sequence for $f$ , then step 5.1 applied to $f_{n} - g_{n}$ shows $∥ T f_{n} - T g_{n} ∥_{q_{θ}} \leq M_{0}^{1 - θ} M_{1}^{θ} ∥ f_{n} - g_{n} ∥_{p_{θ}} \to 0,$ so the limit $h$ is independent of the chosen approximation. Define $\tilde{T} f : = h .$ Passing to the limit in step 5.1 yields $∥ \tilde{T} f ∥_{q_{θ}} \leq M_{0}^{1 - θ} M_{1}^{θ} ∥ f ∥_{p_{θ}} .$

L1L3step 5.1algebra

7.1

The definition in step 6.1 extends $T$ , because a constant approximating [step 6.1, algebra] sequence may be used when $f$ is already finite simple of finite support. Applying step 6.1 to $f + g$ and to $c f$ shows that $\tilde{T}$ is linear, since linearity holds termwise on every approximating sequence. If $S$ is any other bounded linear extension of $T$ to $L^{p_{θ}} (μ)$ , then for every $f \in L^{p_{θ}} (μ)$ and every approximating sequence $(f_{n})$ from step 6.1, $∥ S f - \tilde{T} f ∥_{q_{θ}} \leq ∥ S (f - f_{n}) ∥_{q_{θ}} + ∥ \tilde{T} (f_{n} - f) ∥_{q_{θ}} \to 0,$ so $S = \tilde{T}$ .

step 6.1algebra

8.1

Steps 5.1, 6.1, and 7.1 prove the interpolated bounded extension theorem.

step 5.1step 6.1step 7.1∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04Open item page →

Continuous compactly supported functions are recovered by small ball averages

Statement

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

Let $g \in C_{c} (R^{n})$ and let $x \in R^{n}$ . Then $\lim_{r \to 0^{+}} A_{r} g (x) = g (x) .$

Facts & Assumptions

Given: The Axiom of Countable Choice, a function $g \in C_{c} (R^{n})$ , and a point $x \in R^{n}$ .

[L1]

The class $C_{c} (R^{n})$ consists of continuous functions on $R^{n}$ with compact support. (The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$ )

[L2]

The ball average is $A_{r} g (x) = \frac{1}{λ (B (x, r))} \int_{B (x, r)} g (y) d λ (y) .$ (The average of a locally integrable function over a Euclidean ball)

Proof

technique · direct

1.1

Let $ε > 0$ . Since $g$ is continuous at $x$ by [L1], there is [L1, given] $δ > 0$ such that $∥ y - x ∥_{2} < δ ⟹ ∣ g (y) - g (x) ∣ < ε .$

L1given

2.1

For $0 < r < δ$ , every $y \in B (x, r)$ satisfies the hypothesis of step 1.1, [step 1.1, L2, algebra] so $∣ A_{r} g (x) - g (x) ∣ \leq \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ g (y) - g (x) ∣ d λ (y) \leq ε .$

step 1.1L2algebra

3.1

Because step 2.1 holds for every $ε > 0$ , one has [step 2.1] $A_{r} g (x) \to g (x)$ as $r \to 0^{+}$ .

step 2.1∎

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Lebesgue differentiation theorem on $R^{n}$

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 $\lim_{r \to 0^{+}} A_{r} f (x) = f (x)$ for Lebesgue-almost every $x \in R^{n}$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a function $f \in L_{l o c}^{1} (R^{n})$ .

[L1]

Continuous compactly supported functions are recovered by small ball averages at every point. (Continuous compactly supported functions are recovered by small ball averages)

[L2]

For $1 \leq p < \infty$ , $C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ . ( $C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ )

[L3]

Chebyshev-Markov controls superlevel sets by the integral. (Chebyshev-Markov inequality for the integral)

[L4]

The centered maximal operator is weak type $(1, 1)$ . (The centered Hardy-Littlewood maximal operator is weak type $(1, 1)$ )

[L5]

A countable union of null sets is null. (A countable union of measure-zero sets has measure zero, by countable choice)

Proof

technique · direct

1.1

For each integer $m \geq 1$ and each integer $j \geq 1$ , apply [L2] to the [L2, given, choose, construct] $L^{1}$ function $f 1_{B (0, m + 1)}$ and choose $g_{m, j} \in C_{c} (R^{n})$ such that $\int_{B (0, m + 1)} ∣ f - g_{m, j} ∣ d λ < \frac{2^{- 2 j}}{5^{n}} .$ Set $h_{m, j} : = (f - g_{m, j}) 1_{B (0, m + 1)} .$

L2givenchooseconstruct

2.1

Let [L3, L4, step 1.1, algebra] $E_{m, j} : = {x \in B (0, m) : M h_{m, j} (x) > 2^{- j}} \cup {x \in B (0, m) : ∣ h_{m, j} (x) ∣ > 2^{- j}} .$ By [L4] and [L3], $λ (E_{m, j}) \leq \frac{5^{n}}{2^{- j}} ∥ h_{m, j} ∥_{1} + \frac{1}{2^{- j}} ∥ h_{m, j} ∥_{1} < 2^{- j} + 5^{- n} 2^{- j} \leq 2^{1 - j} .$

L3L4step 1.1algebra

3.1

For fixed $m$ , put [step 2.1, algebra] $N_{m} : = ⋂_{N = 1}^{\infty} ⋃_{j \geq N} E_{m, j} .$ The sets $⋃_{j \geq N} E_{m, j}$ decrease with $N$ , and by step 2.1 $λ ⁣ (⋃_{j \geq N} E_{m, j}) \leq \sum_{j \geq N} λ (E_{m, j}) \leq \sum_{j \geq N} 2^{1 - j} = 2^{2 - N} .$ Hence $λ (N_{m}) = 0$ .

step 2.1algebra

4.1

Let $x \in B (0, m) ∖ N_{m}$ . Then there is $N_{x}$ such that [L1, step 1.1, step 3.1, algebra] $x \notin E_{m, j}$ for every $j \geq N_{x}$ . Fix such a $j$ and take $0 < r < 1$ . Since $x \in B (0, m)$ , one has $B (x, r) \subseteq B (0, m + 1)$ , so $A_{r} f (x) - f (x) = (A_{r} g_{m, j} (x) - g_{m, j} (x)) + A_{r} h_{m, j} (x) - h_{m, j} (x) .$ Therefore $∣ A_{r} f (x) - f (x) ∣ \leq ∣ A_{r} g_{m, j} (x) - g_{m, j} (x) ∣ + M h_{m, j} (x) + ∣ h_{m, j} (x) ∣ \leq ∣ A_{r} g_{m, j} (x) - g_{m, j} (x) ∣ + 2^{1 - j} .$ Now [L1] gives $A_{r} g_{m, j} (x) \to g_{m, j} (x)$ , so ${lim sup}_{r \to 0^{+}} ∣ A_{r} f (x) - f (x) ∣ \leq 2^{1 - j}$ for every $j \geq N_{x}$ . Letting $j \to \infty$ yields $A_{r} f (x) \to f (x)$ .

L1step 1.1step 3.1algebra

5.1

The bad set for radius differentiation is contained in [step 3.1, step 4.1, L5] $⋃_{m \geq 1} N_{m}$ , which is null by [L5]. Thus the convergence holds for almost every $x \in R^{n}$ .

step 3.1step 4.1L5∎

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Almost every point is a Lebesgue point of a locally integrable function

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 Lebesgue set of the $L_{l o c}^{1}$ class of $f$ has full Lebesgue measure.

Equivalently, for almost every $x \in R^{n}$ , $\lim_{r \to 0^{+}} \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) - f (x) ∣ d λ (y) = 0.$

Facts & Assumptions

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

[L1]

A point belongs to the Lebesgue set exactly when the averaged oscillation above tends to $0$ . (Lebesgue points and the Lebesgue set of an $L_{l o c}^{1}$ class)

[L2]

A property holds almost everywhere when its exceptional set is contained in a measurable null set. (Measure-null sets and almost-everywhere statements relative to a measure)

[L3]

The rationals are countably infinite, and the product of two at most countable sets is at most countable. ( $Q$ is countably infinite, A product of two at most countable sets is at most countable)

[L4]

Rationals are dense in the reals. (The rationals embed densely in the reals)

[L5]

A countable union of null sets is null. (A countable union of measure-zero sets has measure zero, by countable choice)

[L6]

If $u \in L_{l o c}^{1} (R^{n})$ , then $A_{r} u (x) \to u (x)$ for almost every $x$ . (Lebesgue differentiation theorem on $R^{n}$ )

Proof

technique · direct

1.1

Let [L3, L5, L6, given, construct] $D : = {a + i b : a, b \in Q} .$ By [L3], $D$ is countable. For each $c \in D$ , the function $u_{c} : = ∣ f - c ∣$ is locally integrable, so [L6] gives a null set $N_{c}$ such that $\lim_{r \to 0^{+}} A_{r} u_{c} (x) = u_{c} (x) = ∣ f (x) - c ∣$ for every $x \notin N_{c}$ . Put $N : = ⋃_{c \in D} N_{c} .$ By [L5], $N$ is null.

L3L5L6givenconstruct

2.1

Fix $x \notin N$ and let $ε > 0$ . By density [L4], choose [step 1.1, L4, algebra] $c \in D$ with $∣ f (x) - c ∣ < ε$ . Then for every $r > 0$ , $\frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) - f (x) ∣ d λ (y) \leq A_{r} ∣ f - c ∣ (x) + ∣ c - f (x) ∣ .$ Taking $r \to 0^{+}$ and using step 1.1 gives ${lim sup}_{r \to 0^{+}} \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) - f (x) ∣ d λ (y) \leq 2 ε .$ Since $ε$ is arbitrary, the limit is $0$ .

step 1.1L4algebra

3.1

Step 2.1 holds for every $x \notin N$ , and $N$ is null. By [L1] and [L2], [L1, L2, step 1.1, step 2.1] the Lebesgue set of the class of $f$ has full measure.

L1L2step 1.1step 2.1∎

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Lebesgue density theorem

Statement

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

Let $E \subseteq R^{n}$ be Lebesgue measurable. Then

$Θ (E, x) = 1$ for almost every $x \in E$ ;
$Θ (E, x) = 0$ for almost every $x \in E^{c}$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a Lebesgue measurable set $E \subseteq R^{n}$ .

[L1]

The density of $E$ at $x$ is $Θ (E, x) = \lim_{r \to 0^{+}} \frac{λ (E \cap B (x, r))}{λ (B (x, r))}$ when the limit exists. (Density of a measurable set at a point)

[L2]

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

[L3]

Every Euclidean ball has positive finite measure. (Euclidean balls have positive finite Lebesgue measure)

Proof

technique · direct

1.1

Because $0 \leq 1_{E} \leq 1$ and balls have finite measure by [L3], the [L2, L3, given, algebra] indicator $1_{E}$ is locally integrable. Apply [L2] to $1_{E}$ . There is a null set $N$ such that every $x \notin N$ is a Lebesgue point of $1_{E}$ .

L2L3givenalgebra

2.1

Let $x \in E ∖ N$ . Then $1_{E} (x) = 1$ , and for every $r > 0$ , [L1, step 1.1, algebra] $\frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ 1_{E} (y) - 1 ∣ d λ (y) = 1 - \frac{λ (E \cap B (x, r))}{λ (B (x, r))} .$ Since the left-hand side tends to $0$ , [L1] gives $Θ (E, x) = 1$ .

L1step 1.1algebra

2.2

Let $x \in E^{c} ∖ N$ . Then $1_{E} (x) = 0$ , and for every $r > 0$ , [L1, step 1.1, algebra] $\frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ 1_{E} (y) - 0 ∣ d λ (y) = \frac{λ (E \cap B (x, r))}{λ (B (x, r))} .$ Again the left-hand side tends to $0$ , so [L1] gives $Θ (E, x) = 0$ .

L1step 1.1algebra

3.1

Steps 2.1 and 2.2 hold outside the null set $N$ , so they prove the two [step 1.1, step 2.1, step 2.2] density conclusions.

step 1.1step 2.1step 2.2∎

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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

1.1

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

L1L2givenalgebra

2.1

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 $0$ .

step 1.1

3.1

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.

step 2.1algebra

4.1

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

L2step 3.1∎

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Differentiation of sigma-finite Borel measures finite on compact sets

Statement

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

Let $ν$ be a sigma-finite Borel measure on $R^{n}$ that is finite on compact sets. Write $ν = ν_{a} + ν_{s}, ν_{a} ≪ λ, ν_{s} ⊥ λ,$ for its Lebesgue decomposition relative to Lebesgue measure, and choose a measurable representative $f$ of the Radon-Nikodym class $d ν_{a} / d λ$ . Then for Lebesgue-almost every $x \in R^{n}$ , $\lim_{r \to 0^{+}} \frac{ν (B (x, r))}{λ (B (x, r))} = f (x) .$ More generally, let $A \subseteq R^{n}$ , and suppose that for each $x \in A$ a family $(E_{r} (x))_{r > 0}$ of Borel sets shrinking nicely to $x$ is specified. Then for Lebesgue-almost every $x \in A$ , $\lim_{r \to 0^{+}} \frac{ν (E_{r} (x))}{λ (E_{r} (x))} = f (x) .$

Facts & Assumptions

Given: The Axiom of Countable Choice and a sigma-finite Borel measure $ν$ on $R^{n}$ that is finite on compact sets.

[L1]

Such a measure admits a Lebesgue decomposition $ν = ν_{a} + ν_{s}, ν_{a} ≪ λ, ν_{s} ⊥ λ,$ and the Radon-Nikodym class $d ν_{a} / d λ$ has a measurable representative $f$ . (Every sigma-finite signed measure admits a Lebesgue decomposition relative to a sigma-finite positive measure, A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density, The Radon-Nikodym derivative as an almost-everywhere equivalence class)

[L2]

If $(E_{r})$ shrinks nicely to $x$ , then $\frac{1}{λ (E_{r})} \int_{E_{r}} f d λ \to f (x)$ for almost every $x$ . (Differentiation holds along families shrinking nicely)

[L3]

A finite family of balls admits a disjoint subfamily whose fivefold dilates cover the union. (Vitali covering lemma for Euclidean balls with fivefold dilates)

[L4]

Assuming the Axiom of Countable Choice, Lebesgue measure is inner regular by compact subsets on measurable sets in $R^{n}$ . (Assuming countable choice, the Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets)

[L5]

Increasing measurable unions pass through positive measures. (Continuity from below for measures)

[F1]

Every Borel measure on $R^{n}$ that is finite on compact sets is regular on its Borel sets: for each Borel set $E$ , $ν (E) = \inf {ν (U) : E \subseteq U and U is open} .$ (Rudin, Theorem 2.18)

Proof

technique · direct

1.1

By [L1], write $ν = ν_{a} + ν_{s}$ with $ν_{a} = f d λ$ , $ν_{a} ≪ λ$ , and $ν_{s} ⊥ λ$ . Choose a Borel set $N$ with $λ (N) = 0$ on which $ν_{s}$ is concentrated. For every Borel set $E$ , absolute continuity and concentration give $ν_{s} (E) = ν_{s} (E \cap N) = ν (E \cap N) \geq 0, ν_{a} (E) = ν_{a} (E ∖ N) = ν (E ∖ N) \geq 0.$ Thus both components are positive and $ν_{s} \leq ν$ . Changing $f$ on a null set does not affect the claim, so take $f \geq 0$ . Since every closed Euclidean ball is compact, for every $x$ and $R > 0$ , $\int_{B (x, R)} ∣ f ∣ d λ = ν_{a} (B (x, R)) \leq ν (\overline{B (x, R)}) < \infty .$ Hence $f \in L_{l o c}^{1} (R^{n})$ .

L1givenchoosealgebra

2.1

Let $(E_{r} (x))_{r > 0}$ be a family of Borel sets shrinking nicely to $x$ with constant $α_{x} > 0$ . Then $\frac{ν (E_{r} (x))}{λ (E_{r} (x))} = \frac{1}{λ (E_{r} (x))} \int_{E_{r} (x)} f d λ + \frac{ν_{s} (E_{r} (x))}{λ (E_{r} (x))},$ while positivity and the defining comparison give $0 \leq \frac{ν_{s} (E_{r} (x))}{λ (E_{r} (x))} \leq \frac{1}{α_{x}} \frac{ν_{s} (B (x, r))}{λ (B (x, r))} .$ Consequently [L2] and step 1.1 reduce both conclusions to proving $\frac{ν_{s} (B (x, r))}{λ (B (x, r))} ⟶ 0$ for almost every $x$ .

L1L2step 1.1givenalgebra

2.2

Put $G : = R^{n} ∖ N$ , so $G$ is Borel, $λ (R^{n} ∖ G) = 0$ , and $ν_{s} (G) = 0$ . For each $m \geq 1$ , define $D_{m} (x) : = \sup_{0 < r < 1 / m} \frac{ν_{s} (B (x, r))}{λ (B (x, r))} .$ For fixed $r > 0$ , if $0 < δ < r$ and $∥ x - y ∥_{2} < δ$ , then $B (x, r - δ) \subseteq B (y, r) .$ By [L5], $ν_{s} (B (x, r - δ)) ↑ ν_{s} (B (x, r))$ as $δ ↓ 0$ , so $x \mapsto ν_{s} (B (x, r))$ is lower semicontinuous. Translation invariance gives the fixed positive denominator $λ (B (x, r)) = λ (B (0, r))$ , so each $D_{m}$ is lower semicontinuous and each set ${x \in R^{n} : D_{m} (x) > \frac{1}{k}}$ is open. For $k \geq 1$ , put $F_{k} : = G \cap ⋂_{m \geq 1} {x \in R^{n} : D_{m} (x) > \frac{1}{k}} .$ Then $F_{k}$ is Borel. If $x \in G$ and $L (x) : = {lim sup}_{r \to 0^{+}} \frac{ν_{s} (B (x, r))}{λ (B (x, r))} > 0,$ choose $k$ with $1 / k < L (x)$ . Since $D_{m} (x) \geq L (x) > 1 / k$ for every $m$ , one has $x \in F_{k}$ . Thus ${x \in G : L (x) > 0} \subseteq ⋃_{k \geq 1} F_{k},$ and it is enough to prove $λ (F_{k}) = 0$ for every $k$ .

step 1.1L5givenconstructalgebra

3.1

Fix $k \geq 1$ and $ε > 0$ . Because $ν_{s} (G) = 0$ and $ν_{s}$ is a Borel measure finite on compact sets, [F1] gives an open set $U_{ε} \supseteq G$ with $ν_{s} (U_{ε}) < ε$ . Let $K \subseteq F_{k}$ be compact, and let $B$ be the family of all balls $B (x, r)$ such that $x \in K, B (x, r) \subseteq U_{ε}, ν_{s} (B (x, r)) > \frac{1}{k} λ (B (x, r)) .$ This family covers $K$ : for any $x \in K$ , openness gives an $m \geq 1$ with $B (x, 1 / m) \subseteq U_{ε}$ , and $x \in F_{k}$ gives an $r < 1 / m$ satisfying the displayed strict inequality. Compactness supplies a finite subcover of $K$ from $B$ . Apply [L3] to that finite family. There are pairwise disjoint chosen balls $B_{1}, \dots, B_{q}$ among it such that $K \subseteq ⋃_{j = 1}^{q} 5 B_{j} .$ Hence $λ (K) \leq 5^{n} \sum_{j = 1}^{q} λ (B_{j}) < 5^{n} k \sum_{j = 1}^{q} ν_{s} (B_{j}) \leq 5^{n} k ν_{s} (U_{ε}) < 5^{n} k ε .$

F1L3step 2.2givenconstructalgebra

4.1

Since $ε > 0$ was arbitrary, step 3.1 gives $λ (K) = 0$ for every compact $K \subseteq F_{k}$ . Because $F_{k}$ is Borel by step 2.2, [L4] implies $λ (F_{k}) = \sup {λ (K) : K \subseteq F_{k}, K compact} = 0.$ Step 2.2 now shows that the set where $L (x) > 0$ is contained in the null set $(R^{n} ∖ G) \cup ⋃_{k \geq 1} F_{k} .$ The ratios defining $L$ are nonnegative, so $\lim_{r \to 0^{+}} \frac{ν_{s} (B (x, r))}{λ (B (x, r))} = 0$ for almost every $x$ .

step 2.2step 3.1L4algebra

5.1

Combine step 4.1 with the comparison in step 2.1 and the differentiation theorem [L2] for the locally integrable representative $f$ . For any specified Borel families $(E_{r} (x))_{r > 0}$ shrinking nicely to the points of $A$ , this gives $\frac{ν (E_{r} (x))}{λ (E_{r} (x))} \to f (x)$ for almost every $x \in A$ . Taking $A = R^{n}$ and $E_{r} (x) = B (x, r)$ gives the ball conclusion.

L2step 1.1step 2.1step 4.1∎

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The indefinite integral of an $L^{1}$ function is differentiable almost everywhere

Statement

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

Let $a < b$ and let $f \in L^{1} ([a, b])$ . Define $F (x) : = \int_{a}^{x} f (t) d t (a \leq x \leq b) .$ Then $F$ is differentiable for almost every $x \in (a, b)$ and $F^{'} (x) = f (x)$ at every such point.

Facts & Assumptions

Given: The Axiom of Countable Choice, reals $a < b$ , and a function $f \in L^{1} ([a, b])$ .

[L1]

The $L^{1}$ norm is the integral of the absolute value. (The class $L^{1} (μ)$ of integrable functions)

[L2]

Differentiation along families shrinking nicely recovers the point value at almost every point. (Differentiation holds along families shrinking nicely)

Proof

technique · direct

1.1

Extend $f$ by $0$ outside $[a, b]$ , obtaining a function $g$ on [L1, given, construct, algebra] $R$ . Because $g$ is bounded by $∣ f ∣$ on $[a, b]$ and vanishes elsewhere, $\int_{R} ∣ g (t) ∣ d t = \int_{a}^{b} ∣ f (t) ∣ d t < \infty,$ so $g \in L_{l o c}^{1} (R)$ . For $x \in (a, b)$ and every $r > 0$ , define $E_{r}^{+} (x) : = [x, x + r), E_{r}^{-} (x) : = (x - r, x] .$ Each family shrinks nicely to $x$ with constant $α = \frac{1}{2}$ , because $E_{r}^{\pm} (x) \subseteq B (x, r)$ and $λ (E_{r}^{\pm} (x)) = r = \frac{1}{2} λ (B (x, r))$ .

L1givenconstructalgebra

2.1

Apply [L2] to the set $A : = (a, b)$ and the family $E_{r}^{+} (x)$ from step [L2, step 1.1, algebra] 1.1. This gives a full-measure subset $A_{+} \subseteq (a, b)$ such that $\lim_{r \to 0^{+}} \frac{1}{r} \int_{x}^{x + r} g (t) d t = g (x) (x \in A_{+}) .$ Applying [L2] again to the family $E_{r}^{-} (x)$ gives another full-measure subset $A_{-} \subseteq (a, b)$ such that $\lim_{r \to 0^{+}} \frac{1}{r} \int_{x - r}^{x} g (t) d t = g (x) (x \in A_{-}) .$ Hence both one-sided limits hold for every $x \in A_{+} \cap A_{-}$ , which still has full measure in $(a, b)$ . At such an $x$ , if $h > 0$ is small then $\frac{F (x + h) - F (x)}{h} = \frac{1}{h} \int_{x}^{x + h} f (t) d t = \frac{1}{h} \int_{x}^{x + h} g (t) d t,$ while for $h < 0$ , $\frac{F (x + h) - F (x)}{h} = \frac{1}{∣ h ∣} \int_{x + h}^{x} g (t) d t .$ Both one-sided limits therefore equal $g (x) = f (x)$ .

L2step 1.1algebra

3.1

Hence $F^{'} (x) = f (x)$ for almost every $x \in (a, b)$ .

step 2.1∎

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The Hardy-Littlewood maximal operator is not strong type $(1, 1)$

Statement refuted

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

The centered Hardy-Littlewood maximal operator maps $L^{1} (R^{n})$ to $L^{1} (R^{n})$ .

More strongly, if $f \in L^{1} (R^{n})$ satisfies $∥ f ∥_{1} > 0$ , then $M f \notin L^{1} (R^{n})$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a function $f \in L^{1} (R^{n})$ with $∥ f ∥_{1} > 0$ .

[L1]

The centered maximal function is $M f (x) = \sup_{r > 0} \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) ∣ d λ (y) .$ (The centered and uncentered Hardy-Littlewood maximal functions)

[L2]

The $L^{1}$ norm is $∥ f ∥_{1} = \int_{R^{n}} ∣ f ∣ d λ .$ (The class $L^{1} (μ)$ of integrable functions)

[L3]

Lebesgue measure is sigma-finite, and every bounded measurable set has finite measure. (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure)

[L4]

Lebesgue measure on balls scales like $r^{n}$ under dilation. (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it)

Counterexample

technique · direct

1.1

Since $∥ f ∥_{1} > 0$ , the integral in [L2] is positive. Hence there is [L2, L3, given, choose, algebra] $ε > 0$ such that the measurable set $E : = {y \in R^{n} : ∣ f (y) ∣ > ε}$ has positive measure. Because $R^{n} = ⋃_{m \geq 1} B (0, m)$ and measure is countably subadditive, some $R \geq 1$ satisfies $0 < λ (F) < \infty, F : = E \cap B (0, R),$ the finiteness coming from [L3].

L2L3givenchoosealgebra

2.1

Let $x \in R^{n}$ with $∥ x ∥_{2} \geq 2 R$ . If $y \in F$ , then [step 1.1, L1, L4, algebra] $∥ y - x ∥_{2} \leq ∥ y ∥_{2} + ∥ x ∥_{2} \leq R + ∥ x ∥_{2} \leq \frac{3}{2} ∥ x ∥_{2},$ so $F \subseteq B (x, \frac{3}{2} ∥ x ∥_{2})$ . Since $∣ f ∣ \geq ε$ on $F$ , [L1] gives $M f (x) \geq \frac{1}{λ (B (x, \frac{3}{2} ∥ x ∥_{2}))} \int_{F} ∣ f (y) ∣ d λ (y) \geq \frac{ε λ (F)}{λ (B (0, 1)) (\frac{3}{2} ∥ x ∥_{2})^{n}} = \frac{C}{∥ x ∥_{2}^{n}}$ for a positive constant $C$ .

step 1.1L1L4algebra

3.1

For each integer $k \geq 1$ , set [step 2.1, L4, algebra] $A_{k} : = B (0, 2^{k + 1} R) ∖ B (0, 2^{k} R) .$ On $A_{k}$ one has $∥ x ∥_{2} \leq 2^{k + 1} R$ , so step 2.1 yields $M f (x) \geq \frac{C}{(2^{k + 1} R)^{n}} (x \in A_{k}) .$ Therefore $\int_{A_{k}} M f d λ \geq \frac{C}{(2^{k + 1} R)^{n}} λ (A_{k}) .$ Using [L4] again, $λ (A_{k}) = λ (B (0, 1)) ((2^{k + 1} R)^{n} - (2^{k} R)^{n}),$ so the right-hand side is a positive constant independent of $k$ . Since the annuli $A_{k}$ are pairwise disjoint, the integral of $M f$ over $⋃_{k \geq 1} A_{k}$ diverges.

step 2.1L4algebra

4.1

Thus $M f \notin L^{1} (R^{n})$ whenever $∥ f ∥_{1} > 0$ , so the [step 3.1] strong type $(1, 1)$ claim is false.

step 3.1∎

5 · Examples, counterexamples and false statements

None yet.

The Maximal Function and Lebesgue Differentiation

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

A locally integrable function on RnRn

Definition

Euclidean balls have positive finite Lebesgue measure

Statement

Facts & Assumptions

Proof

The average of a locally integrable function over a Euclidean ball

Definition

The centered and uncentered Hardy-Littlewood maximal functions

Definition

Sublinear operators and weak or strong type (p,q)(p,q) bounds

Definition

Lebesgue points and the Lebesgue set of an Lloc1Lloc1​ class

Definition

Density of a measurable set at a point

Definition

A family shrinking nicely to a point

Definition

Vitali covering lemma for Euclidean balls with fivefold dilates

Statement

Facts & Assumptions

Proof

Ball averages vary continuously with the centre and radius

Statement

Facts & Assumptions

Proof

The centered Hardy-Littlewood maximal function is Borel measurable

Statement

Facts & Assumptions

Proof

The centered and uncentered maximal functions are pointwise comparable

Statement

Facts & Assumptions

Proof

The centered Hardy-Littlewood maximal operator is weak type (1,1)(1,1)

Statement

Facts & Assumptions

Proof

The centered maximal operator is bounded on L∞L∞

Statement

Facts & Assumptions

Proof

Marcinkiewicz interpolation from weak (1,1)(1,1) and strong (∞,∞)(∞,∞)

Statement

Facts & Assumptions

Proof

The centered maximal operator is bounded on Lp(Rn)Lp(Rn) for 1<p<∞1<p<∞

Statement

Facts & Assumptions

Proof

Riesz-Thorin interpolation theorem

Statement

Facts & Assumptions

Proof

Continuous compactly supported functions are recovered by small ball averages

Statement

Facts & Assumptions

Proof

Lebesgue differentiation theorem on RnRn

Statement

Facts & Assumptions

Proof

Almost every point is a Lebesgue point of a locally integrable function

Statement

Facts & Assumptions

Proof

Lebesgue density theorem

Statement

Facts & Assumptions

Proof

Differentiation holds along families shrinking nicely

Statement

Facts & Assumptions

Proof

Differentiation of sigma-finite Borel measures finite on compact sets

A locally integrable function on $R^{n}$

Sublinear operators and weak or strong type $(p, q)$ bounds

Lebesgue points and the Lebesgue set of an $L_{l o c}^{1}$ class

The centered Hardy-Littlewood maximal operator is weak type $(1, 1)$

The centered maximal operator is bounded on $L^{\infty}$

Marcinkiewicz interpolation from weak $(1, 1)$ and strong $(\infty, \infty)$

The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$

Lebesgue differentiation theorem on $R^{n}$

The indefinite integral of an $L^{1}$ function is differentiable almost everywhere

The Hardy-Littlewood maximal operator is not strong type $(1, 1)$