The Maximal Function and Lebesgue Differentiation

If $\mathrm{\parallel }y-x{\mathrm{\parallel }}_{\mathrm{\infty }}<r\mathrm{/}\sqrt{n}$, then [L1, algebra] $\mathrm{\parallel }y-x{\mathrm{\parallel }}_2\le \sqrt{n}\mathrm{\parallel }y-x{\mathrm{\parallel }}_{\mathrm{\infty }}<r,$ so the open cube $x+(-r\mathrm{/}\sqrt{n},r\mathrm{/}\sqrt{n}{)}^n$ is contained in $B(x,r)$. On the other hand, $\mathrm{\parallel }y-x{\mathrm{\parallel }}_2<r$ implies $\mathrm{\mid }{y}_i-x_i\mathrm{\mid }<r$ for every coordinate, so $B(x,r)\subseteq x+(-r,r{)}^n$.

The ball $B(x,r)$ is open in the Euclidean metric by [L3], hence Borel and Lebesgue measurable by [L4] under the stated Countable Choice. The two cubes from step 1.1 are measurable by [L2], with measures [step 1.1, L2, L3, L4, algebra] $(2r\mathrm{/}\sqrt{n}{)}^n>0\text{and}(2r{)}^n<\mathrm{\infty }\mathrm{.}$ Since $B(x,r)$ lies between them, monotonicity [L5] gives $0<(2r\mathrm{/}\sqrt{n}{)}^n\le \lambda (B(x,r))\le (2r{)}^n<\mathrm{\infty }\mathrm{.}$

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

If the finite family is empty, use the empty subfamily. Otherwise retain a ball of maximal radius, breaking ties by least index. Repeat among the remaining balls disjoint from every retained ball, and stop when none remain. There are at most $m$ selections. The retained balls are pairwise disjoint by construction, and the tie rule makes every selection determined.

For the countable family, fix a positive upper bound $R$ for its radii. Let $I_m=\{j\ge 1:2^{-m-1}R<r_j\le 2^{-m}R\}$ for $m\ge 0$; these partition the positive radii. Given the retained indices in earlier classes, inspect $j=1,2,\dots$ and retain $j$ exactly when $j\in I_m$ and its ball is disjoint from every earlier retained ball. Inner recursion on $j$ and outer recursion on $m$ define this rule by [L3], taking the union of each inner sequence before the next class. Every decision is determined, so there is no countable choice of balls. The retained subfamily is pairwise disjoint.

In the finite family, a chosen ball lies in its own fivefold dilate. For an unchosen $B_j$, take the first chosen ball $B_{i_s}$ meeting it. At that stage $B_j$ was eligible, so $r_j\le r_{i_s}$. For $z\in B_j$ and $w\in B_j\cap B_{i_s}$, the centre distance satisfies $\mathrm{\parallel }{c}_j-c_{i_s}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{c}_j-w{\mathrm{\parallel }}_2+\mathrm{\parallel }w-c_{i_s}{\mathrm{\parallel }}_2<r_j+r_{i_s}$. Hence $\mathrm{\parallel }z-c_{i_s}{\mathrm{\parallel }}_2\le \mathrm{\parallel }z-c_j{\mathrm{\parallel }}_2+\mathrm{\parallel }{c}_j-c_{i_s}{\mathrm{\parallel }}_2<2r_j+r_{i_s}\le 3r_{i_s}<5r_{i_s}$. Thus every original ball lies in a chosen fivefold dilate.

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 $i_s\in I_{\mathrm{\ell }}$ with $\mathrm{\ell }<m$, then $r_j\le 2^{-m}R\le 2^{-\mathrm{\ell }-1}R<r_{i_s}\mathrm{.}$ If instead $\mathrm{\ell }=m$, then both balls lie in the same dyadic class, so $r_j\le 2^{-m}R<2r_{i_s}\mathrm{.}$ In either case, $r_j<2r_{i_s}\mathrm{.}$ Choose $z\in B_j$ and $w\in B_j\cap B_{i_s}$. Then $\mathrm{\parallel }z-c_{i_s}{\mathrm{\parallel }}_2<r_j+r_j+r_{i_s}<2(2r_{i_s})+r_{i_s}=5r_{i_s},$ so again $B_j\subseteq 5B_{i_s}$. Let $I\subseteq {\mathbb{N}}_{\ge 1}$ 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 ${\bigcup }_{j\ge 1}{B}_j\subseteq {\bigcup }_{i\in I}5B_i\mathrm{.}$

For the measure consequence, assume countable choice as required by [L2]. Finite subadditivity and the established inclusion give $\lambda ({\bigcup }_{j=1}^m{B}_j)\le \lambda ({\bigcup }_{k=1}^{\mathrm{\ell }}5B_{i_k})\le {\sum }_{k=1}^{\mathrm{\ell }}\lambda (5B_{i_k})=5^n{\sum }_{k=1}^{\mathrm{\ell }}\lambda (B_{i_k})$. This is the only use of countable choice in this proof.

Steps 3.1 and 2.2 prove the finite and countable forms.

Choose $\rho >0$ with $0<\rho <r\mathrm{/}2$. For all sufficiently large $k$, [given, choose, algebra] $\mathrm{\mid }{r}_k-r\mathrm{\mid }<\rho$ and $\mathrm{\parallel }{x}_k-x{\mathrm{\parallel }}_2<\rho$. Then $B(x_k,r_k)\subseteq B(x,r+2\rho )\mathrm{.}$ Since $f$ is locally integrable, the function $g:=\mathrm{\mid }f\mathrm{\mid }{\mathbf{1}}_{B(x,r+2\rho )}$ is integrable.

Put ${\chi }_k:={\mathbf{1}}_{B(x_k,r_k)}$ and $\chi :={\mathbf{1}}_{B(x,r)}$. [algebra] If $y\mathrm{\notin }\mathrm{\partial }B(x,r)$, then $\mathrm{\parallel }y-x{\mathrm{\parallel }}_2\mathrm{\ne }r$, 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 ${\chi }_k(y)\to \chi (y)$ for every $y\mathrm{\notin }\mathrm{\partial }B(x,r)$.

The boundary sphere satisfies [L2, L3, algebra] $\mathrm{\partial }B(x,r)\subseteq B(x,r+\frac{1}{m})\setminus B(x,r-\frac{1}{m})(m>1\mathrm{/}r)\mathrm{.}$ By [L2] and [L3], $\lambda \text{\hspace{0.17em}⁣}(B(x,r+\frac{1}{m})\setminus B(x,r-\frac{1}{m}))=\lambda (B(0,1))((r+\frac{1}{m}{)}^n-(r-\frac{1}{m}{)}^n)\to 0,$ so $\lambda (\mathrm{\partial }B(x,r))=0$.

By [L2] and [L3], [L2, L3, algebra] $\lambda (B(x_k,r_k))=\lambda (B(0,1))r_k^n\to \lambda (B(0,1))r^n=\lambda (B(x,r))\mathrm{.}$ The limit denominator is positive by [L2].

Steps 1.1, 1.2, and 1.3 let us apply [L4] to [step 1.1, step 1.2, step 1.3, L4] $f{\chi }_k$, dominated by $g$, and obtain $\int f(y){\chi }_k(y)d\lambda (y)\to \int f(y)\chi (y)d\lambda (y)\mathrm{.}$ In other words, ${\int }_{B(x_k,r_k)}fd\lambda \to {\int }_{B(x,r)}fd\lambda \mathrm{.}$

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

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

Let $x\in \{Mf>t\}$. By [L1], there is $r>0$ with $A_r\mathrm{\mid }f\mathrm{\mid }(x)>t$. Apply [L1, L2, given, choose] to the locally integrable function $\mathrm{\mid }f\mathrm{\mid }$: continuity of $y\mapsto A_r\mathrm{\mid }f\mathrm{\mid }(y)$ at $x$ gives $\delta >0$ such that $\mathrm{\parallel }y-x{\mathrm{\parallel }}_2<\delta ⟹A_r\mathrm{\mid }f\mathrm{\mid }(y)>t\mathrm{.}$ Hence $B(x,\delta )\subseteq \{Mf>t\}$.

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

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

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

Let $B(y,r)$ be any ball containing $x$. If $z\in B(y,r)$, then [L1, algebra] $\mathrm{\parallel }z-x{\mathrm{\parallel }}_2\le \mathrm{\parallel }z-y{\mathrm{\parallel }}_2+\mathrm{\parallel }y-x{\mathrm{\parallel }}_2<r+r=2r,$ so $B(y,r)\subseteq B(x,2r)$. Hence ${\int }_{B(y,r)}\mathrm{\mid }f\mathrm{\mid }\le {\int }_{B(x,2r)}\mathrm{\mid }f\mathrm{\mid }d\lambda \mathrm{.}$

Step 1.2 and [L2] give [step 1.2, L1, L2, algebra] $\frac{1}{\lambda (B(y,r))}{\int }_{B(y,r)}\mathrm{\mid }f\mathrm{\mid }\le \frac{\lambda (B(x,2r))}{\lambda (B(y,r))}\cdot \frac{1}{\lambda (B(x,2r))}{\int }_{B(x,2r)}\mathrm{\mid }f\mathrm{\mid }=2^n{A}_{2r}\mathrm{\mid }f\mathrm{\mid }(x)\le 2^{n}Mf(x)\mathrm{.}$ Taking the supremum over all balls $B(y,r)$ containing $x$ yields $M^{\ast }f(x)\le 2^{n}Mf(x)$.

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

Put $E_t=\{x:Mf(x)>t\}$. For a fixed $x\in E_t$, [L1] supplies a radius $r>0$ with ${\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }>t\lambda (B(x,r))$. Continuity [L2] with this fixed radius makes the inequality persist in a neighbourhood of $x$, so $E_t$ is open. Let $K\subseteq E_t$ be compact. If $K=\mathrm{\varnothing }$, the required bound for $\lambda (K)$ is immediate. Otherwise take the family of ALL balls $B(x,r)$ with $x\in K$, $r>0$ and the strict witness inequality. This family covers $K$, so it has a finite subcover $B(x_1,r_1),\dots ,B(x_m,r_m)$. No simultaneous choice of a radius for every point of $K$ is made.

Apply [L4] to that finite subcover. There are pairwise disjoint balls [step 1.1, L4, L5, algebra] $B_{i_1},\dots ,B_{i_{\mathrm{\ell }}}$ among $B(x_1,r_1),\dots ,B(x_m,r_m)$ such that $K\subseteq {\bigcup }_{j=1}^{m}B(x_j,r_j)\subseteq {\bigcup }_{k=1}^{\mathrm{\ell }}5B_{i_k}$ and therefore $\lambda (K)\le 5^n{\sum }_{k=1}^{\mathrm{\ell }}\lambda (B_{i_k})\mathrm{.}$ Each chosen ball still satisfies the witness inequality from step 1.1, so $t{\sum }_{k=1}^{\mathrm{\ell }}\lambda (B_{i_k})<{\sum }_{k=1}^{\mathrm{\ell }}{\int }_{B_{i_k}}\mathrm{\mid }f\mathrm{\mid }d\lambda \le {\int }_{{\mathbb{R}}^n}\mathrm{\mid }f\mathrm{\mid }d\lambda =\mathrm{\parallel }f{\mathrm{\parallel }}_1,$ because the chosen balls are pairwise disjoint. Hence $\lambda (K)\le \frac{5^n}{t}\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{.}$

By [L3], $\lambda (E_t)$ is the supremum of $\lambda (K)$ over compact $K\subseteq E_t$. Step 2.1 gives the required upper bound for nonempty $K$, and step 1.1 handles the empty set. Hence $\lambda (E_t)\le (5^n\mathrm{/}t)\mathrm{\parallel }f{\mathrm{\parallel }}_1$. The stated countable choice covers [L3] and the measure clause of [L4]; the finite-subcover argument introduces no stronger selection assumption.

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

Fix $x\in {\mathbb{R}}^n$ and $r>0$. By [L2], [L1, L2, given, algebra] ${\int }_{B(x,r)}\mathrm{\mid }f(y)\mathrm{\mid }d\lambda (y)\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\lambda (B(x,r))\mathrm{.}$ Dividing by $\lambda (B(x,r))$ gives $A_r\mathrm{\mid }f\mathrm{\mid }(x)\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$

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

Fix $t>0$ and $\eta >0$, and put $C:=B+\eta >0$. Since the strong [L1, given, construct, algebra] $(\mathrm{\infty },\mathrm{\infty })$ bound with constant $B$ also holds with the larger constant $C$, split $f=f_t^{>}+f_t^{\le },f_t^{>}:=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }>t\mathrm{/}(2C)\}},f_t^{\le }:=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le t\mathrm{/}(2C)\}}\mathrm{.}$ Sublinearity gives $\mathrm{\mid }Tf\mathrm{\mid }\le \mathrm{\mid }T{f}_t^{>}\mathrm{\mid }+\mathrm{\mid }T{f}_t^{\le }\mathrm{\mid }\mathrm{.}$ Since $\mathrm{\parallel }{f}_t^{\le }{\mathrm{\parallel }}_{\mathrm{\infty }}\le t\mathrm{/}(2C)$, the strong $(\mathrm{\infty },\mathrm{\infty })$ bound with constant $C$ yields $\mathrm{\mid }T{f}_t^{\le }\mathrm{\mid }\le C\mathrm{\parallel }{f}_t^{\le }{\mathrm{\parallel }}_{\mathrm{\infty }}\le t\mathrm{/}2.$ Therefore $\{\mathrm{\mid }Tf\mathrm{\mid }>t\}\subseteq \{\mathrm{\mid }T{f}_t^{>}\mathrm{\mid }>t\mathrm{/}2\}\mathrm{.}$

On $\{\mathrm{\mid }f\mathrm{\mid }>t\mathrm{/}(2C)\}$ one has $\mathrm{\mid }f\mathrm{\mid }\le (2C\mathrm{/}t{)}^{p-1}\mathrm{\mid }f{\mathrm{\mid }}^p$, so $f_t^{>}\in L^1$ because $f\in L^p$. Apply the weak $(1,1)$ bound: [L1, step 1.1, algebra] $\mu (\{\mathrm{\mid }Tf\mathrm{\mid }>t\})\le \mu (\{\mathrm{\mid }T{f}_t^{>}\mathrm{\mid }>t\mathrm{/}2\})\le \frac{2A}{t}\mathrm{\parallel }{f}_t^{>}{\mathrm{\parallel }}_1=\frac{2A}{t}{\int }_{\{\mathrm{\mid }f\mathrm{\mid }>t\mathrm{/}(2C)\}}\mathrm{\mid }f\mathrm{\mid }d\mu \mathrm{.}$

Using [L3] with $g=Tf$ and then step 2.1, [L2, L3, step 2.1, algebra] $\mathrm{\parallel }Tf{\mathrm{\parallel }}_p^p=p{\int }_0^{\mathrm{\infty }}{t}^{p-1}\mu (\{\mathrm{\mid }Tf\mathrm{\mid }>t\})dt\le 2Ap{\int }_0^{\mathrm{\infty }}{t}^{p-2}\left({\int }_{\{\mathrm{\mid }f\mathrm{\mid }>t\mathrm{/}(2C)\}}\mathrm{\mid }f\mathrm{\mid }d\mu \right)dt\mathrm{.}$

Define the positive measure $\lambda (E)={\int }_E\mathrm{\mid }f\mathrm{\mid }d\mu$ by [L4]. Apply [L3] with exponent $p-1>0$, function $\mathrm{\mid }f\mathrm{\mid }$, and measure $\lambda$; then substitute $t=2Cu$ in the one-dimensional nonnegative integral from step 3.1. This gives ${\int }_0^{\mathrm{\infty }}{t}^{p-2}\lambda (\{\mathrm{\mid }f\mathrm{\mid }>t\mathrm{/}(2C)\})dt=\frac{(2C{)}^{p-1}}{p-1}{\int }_X\mathrm{\mid }f{\mathrm{\mid }}^{p-1}d\lambda \mathrm{.}$ By [L4], the last integral is ${\int }_X\mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu$. Hence $\mathrm{\parallel }Tf{\mathrm{\parallel }}_p^p\le \frac{2^{p}Ap}{p-1}{C}^{p-1}{\int }_X\mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu \mathrm{.}$ This uses no sigma-finiteness assumption on $\mu$.

Taking $p$th roots in step 4.1 gives [step 4.1, algebra] $\mathrm{\parallel }Tf{\mathrm{\parallel }}_p\le 2{\left(\frac{Ap}{p-1}\right)}^{1\mathrm{/}p}{C}^{1-1\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{.}$ Because $\eta >0$ was arbitrary, letting $\eta \downarrow 0$ yields $\mathrm{\parallel }Tf{\mathrm{\parallel }}_p\le 2{\left(\frac{Ap}{p-1}\right)}^{1\mathrm{/}p}{B}^{1-1\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{.}$ Thus $T$ is of strong type $(p,p)$ for every $1<p<\mathrm{\infty }$.

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 $\mathrm{\parallel }Mf{\mathrm{\parallel }}_p\le C_{n,p}\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{.}$

Step 1.1 proves the centered estimate. Then [L4] gives [step 1.1, L4, algebra] $\mathrm{\parallel }{M}^{\ast }f{\mathrm{\parallel }}_p\le 2^n\mathrm{\parallel }Mf{\mathrm{\parallel }}_p\le 2^n{C}_{n,p}\mathrm{\parallel }f{\mathrm{\parallel }}_p,$ so the uncentered estimate follows as well.