The left-hand side of 9.0.1 is by definition

Writing $u(x)$ as an elementary integral in $\lambda$ and then using Fubini, we write for the last display

This proves the lemma.

It clearly suffices to construct finite collections $C(r,k)$ such that

since then the collection $C(r)=\bigcup _{k\in \mathbb{N}}C(r,k)$ has the desired property.

Suppose that $Y\subset B(o,r{2}^k)$ is a collection of points such that for all $y,y^{\prime }\in Y$ with $y\ne y^{\prime }$, we have $\rho (y,y^{\prime })\ge r$. Then the balls $B(y,r/2)$ are pairwise disjoint and contained in $B(o,r{2}^{k+1})$. If $y\in B(o,r)$, then $B(o,r{2}^{k+1})\subset B(y,r{2}^{k+2})$. Thus, by the doubling property 1.0.5,

Thus, we have

We conclude that $|Y|\le 2^{(k+2)a}$. In particular, there exists a set $Y$ of maximal cardinality. Define $C(r,k)$ to be such a set.

If $x\in B(o,r{2}^k)$ and $x\notin C(r,k)$, then there must exist $y\in C(r,k)$ with $\rho (x,y)<r$. Thus $C(r,k)$ has the desired property.

Let the collection $\mathcal{B}$ be given. We first show 2.0.43.

We recursively choose a finite sequence $B_i\in \mathcal{B}$ for $i\ge 0$ as follows. Assume $B_{i^{\prime }}$ is already chosen for $0\le i^{\prime }<i$. If there exists a ball $B_i\in \mathcal{B}$ so that $B_i$ is disjoint from all $B_{i^{\prime }}$ with $0\le i^{\prime }<i$, then choose such a ball $B_i=B(x_i,r_i)$ with maximal $r_i$.

If there is no such ball, stop the selection and set $i^{″}:=i$.

By disjointedness of the chosen balls and since $0\le u$, we have

By 2.0.42, we conclude

Let $x\in \bigcup \mathcal{B}$. Choose a ball $B^{\prime }=B(x^{\prime },r^{\prime })\in \mathcal{B}$ such that $x\in B^{\prime }$. If $B^{\prime }$ is one of the selected balls, then

If $B^{\prime }$ is not one of the selected balls, then as it is not selected at time $i^{″}$, there is a selected ball $B_i$ with $B^{\prime }\cap B_i\ne \mathrm{\varnothing }$. Choose such $B_i$ with minimal index $i$. As $B^{\prime }$ is therefore disjoint from all balls $B_{i^{\prime }}$ with $i^{\prime }<i$ and as it was not selected in place of $B_i$, we have $r_i\ge r^{\prime }$.

Using a point $y$ in the intersection of $B_i$ and $B^{\prime }$, we conclude by the triangle inequality

By the triangle inequality again, we further conclude

It follows that

With 9.0.7 and 9.0.10, we conclude

With the doubling property 1.0.5 applied twice, we conclude

With 9.0.6 and 9.0.12 we conclude 2.0.43.

We turn to the proof of 2.0.44. We first consider the case $p_1=1$ and recall $M_{\mathcal{B}}=M_{\mathcal{B},1}$. We write for the $p_2$-th power of left-hand side of 2.0.44 with Lemma 9.0.1 and a change of variables

Fix $\lambda \ge 0$ and let $x\in X$ satisfy $M_{\mathcal{B}}u(x)\ge 2\lambda$. By definition of $M_{\mathcal{B}}$, there is a ball $B^{\prime }\in \mathcal{B}$ such that $x\in B^{\prime }$ and

Define $u_{\lambda }(y):=0$ if $|u(y)|<\lambda$ and $u_{\lambda }(y):=u(y)$ if $|u(y)|\ge \lambda$. Then with 9.0.15

As $(u-u_{\lambda })(y)\le \lambda$ by definition, we can estimate the last display by

Hence $x$ is contained in $\bigcup ({\mathcal{B}}_{\lambda })$, where ${\mathcal{B}}_{\lambda }$ is the collection of balls $B^{″}$ in $\mathcal{B}$ such that

We have thus seen

Applying 2.0.43 to the collection ${\mathcal{B}}_{\lambda }$ gives

With Lemma 9.0.1,

By definition of $u_{\lambda }$, making a case distinction between $\lambda \ge {\lambda }^{\prime }$ and $\lambda <{\lambda }^{\prime }$, we see that

We obtain with 9.0.14, 9.0.22, and 9.0.23

We split the integral into $\lambda \ge {\lambda }^{\prime }$ and $\lambda <{\lambda }^{\prime }$ and resolve the maximum correspondingly. We have for $\lambda \ge {\lambda }^{\prime }$ with Lemma 9.0.1

We have for $\lambda <{\lambda }^{\prime }$ with Fubini and Lemma 9.0.1

Adding the two estimates 9.0.28 and 9.0.32 gives

With $a\ge 1$ and $p_2>1$, taking the $p_2$-th root, we obtain 2.0.44. We turn to the case of general $1\le p_1<p_2$. We have

Applying the special case of 2.0.44 for $M_{\mathcal{B}}$ gives

This proves 2.0.44 in general.

Now we construct the operator $M$ satisfying 2.0.45 and 2.0.46. For each $k\in \mathbb{Z}$ we choose a countable set $C(2^k)$ as in Lemma 9.0.2. Define

By Lemma 9.0.2, this is a countable collection of balls. We choose an enumeration ${\mathcal{B}}_{\mathrm{\infty }}=\{B_1,\dots \}$ and define

We define

This function is measurable for each measurable $w$, since it is a countable supremum of measurable functions. Estimate 2.0.46 follows immediately from 2.0.44 and the monotone convergence theorem.

It remains to show 2.0.45. Let $B=B(x,r)\subset X$. Let $k$ be the smallest integer such that $2^k\ge r$, in particular we have $2^k<2r$. By definition of $C(2^k)$, there exists $c\in C(2^k)$ with $x\in B(c,2^k)$. By the triangle inequality, we have $B(c,2^k)\subset B(x,4r)$, and hence by the doubling property 1.0.5

It follows that for each $z\in B(x,r)$

This completes the proof. □