Proof. The left-hand side of eq-layercake is by definition \int_X u(x)^p \, d\mu(x). Writing u(x) as an elementary integral in \lambda and then using Fubini, we write for the last display =\int_X \int _0^{u(x)} p \lambda^{p-1} d\lambda\, d\mu(x) =p\int _0^{\infty} \lambda^{p-1} \mu(\{x: u(x)\ge \lambda\}) d\lambda. This proves the lemma.

Proof. It clearly suffices to construct finite collections C(r,k) such that B(o, r2^k) \subset \bigcup_{c \in C(r,k)} B(c,r), since then the collection C(r) = \bigcup_{k \in \mathbb{N}} C(r,k) has the desired property.

Suppose that Y \subset B(o, r2^k) is a collection of points such that for all y, y' \in Y with y \ne y', we have \rho(y,y') \ge r. Then the balls B(y, r/2) are pairwise disjoint and contained in B(o, r2^{k+1}). If y \in B(o, r), then B(o, r2^{k+1}) \subset B(y, r2^{k+2}). Thus, by the doubling property doublingx, \mu(B(y, \frac{r}{2})) \ge 2^{-(k+2)a} \mu(B(o, r2^{k+1})). Thus, we have \mu(B(o, r2^{k+1})) \ge \sum_{y \in Y} \mu(B(y, \frac{r}{2})) \ge |Y| 2^{-(k+2)a} \mu(B(o, r2^{k+1})). 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, r2^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.

Proof of Theorem 1.2.6. Let the collection \mathcal{B} be given. We first show eq-besico.

We recursively choose a finite sequence B_i\in \mathcal{B} for i\ge 0 as follows. Assume B_{i'} is already chosen for 0\le i'<i. If there exists a ball B_{i}\in \mathcal{B} so that B_{i} is disjoint from all B_{i'} with 0\le i'<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 \sum_{0\le i<i''}\int_{B_i} u(x)\, d\mu(x) \le \int_X u(x)\, d\mu(x). By eq-ball-assumption, we conclude \lambda \sum_{0\le i<i''}\mu(B_i) \le \int_X u(x)\, d\mu(x).

Let x\in \bigcup \mathcal{B}. Choose a ball B'=B(x',r')\in \mathcal{B} such that x\in B'. If B' is one of the selected balls, then x\in \bigcup _{0\le i< i''}B_i\subset \bigcup _{0\le i< i''}B(x_i,3r_i). If B' 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'\cap B_i\neq \emptyset. Choose such B_i with minimal index i. As B' is therefore disjoint from all balls B_{i'} with i'<i and as it was not selected in place of B_i, we have r_i\ge r'.

Using a point y in the intersection of B_i and B', we conclude by the triangle inequality \rho(x_i,x')\le \rho(x_i,y)+\rho(x',y)\le r_i+r'\le 2r_i. By the triangle inequality again, we further conclude \rho(x_i,x)\le \rho(x_i,x')+\rho(x',x)\le 2r_i+r'\le 3r_i. It follows that x\in \bigcup _{0\le i< i''}B(x_i,3r_i). With 3rone and 3rtwo, we conclude \bigcup \mathcal{B}\subset \bigcup _{0\le i< i''}B(x_i,3r_i). With the doubling property doublingx applied twice, we conclude \mu(\bigcup{\mathcal{B}}) \le \sum _{0\le i< i''}\mu (B(x_i,3r_i)) \le 2^{2a}\sum _{0\le i< i''}\mu (B_i). With eqbes1 and eqbes2 we conclude eq-besico.

We turn to the proof of eq-hlm. 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 eq-hlm with Lemma 1.9.1 and a change of variables \|M_{\mathcal{B}}u(x)\|_{p_2}^{p_2} =p_2\int _0^{\infty} \lambda^{p_2-1} \mu(\{x: M_{\mathcal{B}}u(x)\ge \lambda\}) d\lambda =2^{p_2} p_2\int _0^{\infty} \lambda^{p_2-1} \mu(\{x: M_{\mathcal{B}}u(x)\ge 2\lambda\}) d\lambda.

Using Lemma 1.9.1 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'\in \mathcal{B} such that x\in B' and \int_{B'} u(y)\, d\mu(y)\ge 2\lambda \mu({B'}). Define u_\lambda(y):=0 if |u(y)|<\lambda and u_\lambda(y):=u(y) if |u(y)|\ge \lambda. Then with eqbesi10 \int_{B'} u_\lambda (y)\, d\mu(y) =\int_{B'} u (y)\, d\mu(y)- \int_{B'} (u-u_\lambda) (y) d\mu(y) \ge 2\lambda \mu({B'})- \int_{B'} (u-u_\lambda) (y) d\mu(y). As (u-u_\lambda)(y)\le \lambda by definition, we can estimate the last display by \ge 2\lambda \mu({B'})- \int_{B'} \lambda \, d\mu(y) =\lambda \mu({B'}). Hence x is contained in \bigcup(\mathcal{B}_\lambda), where \mathcal{B}_\lambda is the collection of balls B'' in \mathcal{B} such that \int_{B''} u_\lambda (y)\, d\mu(y)\ge \lambda \mu(B''). We have thus seen \{x: M_{\mathcal{B}}u(x)\ge 2\lambda\}\subset \bigcup \mathcal{B}_\lambda. Applying eq-besico to the collection \mathcal{B}_\lambda gives \lambda \mu(\{x: M_{\mathcal{B}}u(x)\ge 2\lambda\})\le 2^{2a} \int u_\lambda (x)\, dx.

With Lemma 1.9.1, \lambda \mu(\{x: M_{\mathcal{B}}u(x)\ge 2\lambda\})\le 2^{2a} \int_0^\infty \mu (\{x: |u_\lambda (x)|\ge \lambda'\})\, d\lambda'. By definition of u_\lambda, making a case distinction between \lambda\ge \lambda' and \lambda <\lambda', we see that \mu (\{x: |u_\lambda (x)|\ge \lambda'\}) \le \mu (\{x: |u (x)|\ge \max(\lambda,\lambda')\}). We obtain with eqbesi11, eqbesi12, and eqbesi13 \|M_{\mathcal{B}}u(x)\|_{p_2}^{p_2} \le 2^{p_2+2a} p_2 \int_0^\infty \lambda^{p_2-2} \int_0^\infty \mu (\{x: |u (x)|\ge \max(\lambda,\lambda')\}) \, d\lambda'd\lambda. We split the integral into \lambda\ge \lambda' and \lambda<\lambda' and resolve the maximum correspondingly.

We have for \lambda\ge \lambda' with Lemma 1.9.1 \int_0^\infty \lambda^{p_2-2} \int_0^\lambda \mu (\{x: |u (x)|\ge \lambda\}) \, d\lambda'd\lambda =\int_0^\infty \lambda^{p_2-1} \mu (\{x: |u (x)|\ge \lambda\}) d\lambda =p_2^{-1} \|u\|_{p_2}^{p_2}.

We have for \lambda< \lambda' with Fubini and Lemma 1.9.1 \int_0^\infty \lambda^{p_2-2} \int_\lambda^\infty \mu (\{x: |u(x)|\ge \lambda'\}) \, d\lambda'd\lambda =\int_0^\infty \int_0^{\lambda'}\lambda^{p_2-2} \mu (\{x: |u (x)|\ge \lambda'\}) d\lambda d\lambda' =(p_2-1)^{-1}\int_0^\infty (\lambda')^{p_2-1} \mu (\{x: |u(x)|\ge \lambda'\}) d\lambda' =(p_2-1)^{-1} p_2^{-1}\|u\|_{p_2}^{p_2}. Adding the two estimates eqbesi14 and eqbesi15 gives \|M_{\mathcal{B}}u(x)\|_{p_2}^{p_2} \le 2^{p_2+2a} (1+(p_2-1)^{-1})\|u\|_{p_2}^{p_2} = 2^{p_2+2a} p_2(p_2-1)^{-1}\|u\|_{p_2}^{p_2}. With a\ge 1 and p_2>1, taking the p_2-th root, we obtain eq-hlm. We turn to the case of general 1\le p_1<p_2. We have M_{\mathcal{B},p_1}u=(M_{\mathcal{B}} (|u|^{p_1}))^{\frac 1{p_1}}. Applying the special case of eq-hlm for M_{\mathcal{B}} gives \|M_{\mathcal{B},p_1}u\|_{p_2}= \|M_{\mathcal{B}} (|u|^{p_1})\|_{p_2/p_1}^{\frac 1{p_1}} \le 2^{2a} (p_2/p_1) (p_2/p_1-1)^{-1} \|(|u|^{p_1})\|_{p_2/p_1}^{\frac 1{p_1}} =2^{2a} p_2(p_2-p_1)^{-1}\|u\|_{p_2}. This proves eq-hlm in general.

Now we construct the operator M satisfying eq-ball-av and eq-hlm-2. For each k \in \mathbb{Z} we choose a countable set C(2^k) as in Lemma 1.9.2. Define \mathcal{B}_\infty = \{B(c, 2^k) \ : \ c \in C(2^k), k \in \mathbb{Z}\}. By Lemma 1.9.2, this is a countable collection of balls. We choose an enumeration \mathcal{B}_\infty = \{B_1, \dotsc\} and define \mathcal{B}_n = \{B_1, \dotsc, B_n\}. We define Mw := 2^{2a}\sup_{n \in \mathbb{N}} M_{\mathcal{B}_n}w. This function is measurable for each measurable w, since it is a countable supremum of measurable functions. Estimate eq-hlm-2 follows immediately from eq-hlm and the monotone convergence theorem.

It remains to show eq-ball-av. 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 doublingx \mu(B(c, 2^k)) \le 2^{2a} \mu(B(x,r)). It follows that for each z \in B(x,r) \frac{1}{\mu(B(x,r))}\int_{B(x,r)} |w(y)| \, \mathrm{d}\mu(y) \le \frac{2^{2a}}{\mu(B(c,2^k))}\int_{B(c,2^k)} |w(y)| \, \mathrm{d}\mu(y) \le Mw(z). This completes the proof.