Proof. Measurability in x follows from joint measurability of K(x,y) \mathbf{1}_{B(x,R_2) \setminus B(x,R_1)}(y) in x and y and part of the proof of Fubini's theorem. Partial continuity in R_1 and R_2 is also a standard lemma in measure theory. It follows for example by splitting the integrand as F_1 - F_{-1} + iF_i - iF_{-i} for positive, disjointly supported functions F_{-} and applying the monotone convergence theorem to each summand. For continuity in \mfa note that |G(x, \mfa, R_1, R_2) - G(x, \mfa', R_1, R_2)| = \left| \int_{R_1 < \rho(x,y) < R_2} K(x,y) f(y) (e(\mfa(y)) - e(\mfa'(y))) \, \mathrm{d}\mu(y) \right| \le \int_{R_1 < \rho(x,y) < R_2} |K(x,y)| |f(y)| |e(\mfa(y) - \mfa'(y) - \mfa(o) + \mfa'(o)) - 1| \, \mathrm{d}\mu(y). By 1-Lipschitz continuity of e^{ix}, the property osccontrol of the metrics d, the kernel upper bound eqkernel-size and |f| \le 1 this is \le \mu(B(x, R_2)) \sup_{R_1 < \rho(x,y) < R_2} \frac{2^{a^3}}{V(x,y)} d_{B(x,\rho(o,x)+R_2)}(\vartheta, \vartheta'). If R_1 < \rho(x,y) < R_2 then there exists n with B(x,R_2) \subset B(x, 2^n \rho(x,y)) and 2^n \le 2 R_2/R_1. Hence, by the doubling property doublingx, V(x,y) = \mu(B(x, \rho(x,y))) \ge 2^{-an} \mu(B(x, R_2)) \ge (2R_2 / R_1)^{-a} \mu(B(x, R_2)). Hence |G(x, \mfa, R_1, R_2) - G(x, \mfa', R_1, R_2)| \le 2^{a^3} \Big(\frac{2R_2}{R_1}\Big)^a d_{B(x, \rho(o,x)+R_2)}(\mfa, \mfa'). Since the topology on \Mf is the one induced by any of the metrics d_B, continuity follows. Finally, for boundedness as a function of x note that we also have by a similar computation using |e(\mfa)|=1 |G(x, \mfa, R_1, R_2)| \le 2^{a^3} \Big(\frac{2R_2}{R_1}\Big)^a.

Proof of Theorem 1.1.1.1. Let Borel sets F, G in X be given. Let a Borel function f: X \to \mathbb{C} with f \le \mathbf{1}_F be given. Let \Mf' \subset \Mf be a countable dense set. By Lemma Lemma 1.3.1 we have \sup_{\mfa\in\Mf} \sup_{0 < R_1 < R_2}\left| \int_{R_1 < \rho(x,y) < R_2} K(x,y) f(y) e(\mfa(y)) \, \mathrm{d}\mu(y) \right| = \sup_{\mfa\in\Mf'} \sup_{0 < R_1 < R_2, R_i \in \mathbb{Q}}\left| \int_{R_1 < \rho(x,y) < R_2} K(x,y) f(y) e(\mfa(y)) \, \mathrm{d}\mu(y) \right|. Consider an enumeration of \Mf' and let \Mf_n be the finite set consisting of the first n+1 functions in the enumeration. Then by the monotone convergence theorem \Big| \int_G Tf \, \mathrm{d}\mu\Big| = \int_G \sup_{\mfa\in\Mf'} \sup_{0 < R_1 < R_2, R_i \in \mathbb{Q}}\left| \int_{R_1 < \rho(x,y) < R_2} K(x,y) f(y) e(\mfa(y)) \, \mathrm{d}\mu(y) \right| \, \mathrm{d}\mu(x) = \lim_{n \to \infty} \int_G \sup_{\mfa\in\Mf_n} \sup_{0 < R_1 < R_2, R_i \in \mathbb{Q}}\left| \int_{R_1 < \rho(x,y) < R_2} K(x,y) f(y) e(\mfa(y)) \, \mathrm{d}\mu(y) \right| \, \mathrm{d}\mu(x). For each n, let Q_n(x) be the measurable function specifying the maximizer in the supremum in \mfa in the previous line. It is possible to construct such a measurable function of x, because the function inside the supremum is measurable in x, being a countable supremum of measurable functions, and because \Mf_n is finite. Then the previous display becomes \le \lim_{n \to \infty} \int_G \sup_{0 < R_1 < R_2} \left| \int_{R_1 < \rho(x,y) < R_2} K(x,y) f(y) e(Q_n(x)(y)) \, \mathrm{d}\mu(y) \right| \, \mathrm{d}\mu(x) = \lim_{n \to \infty} \int_G T_{Q_n} f \, \mathrm{d}\mu. It remains to verify the assumptions of Theorem Theorem 1.1.1.2, which when applied here completes the proof. The assumptions of Theorem Theorem 1.1.1.1 and Theorem Theorem 1.1.1.2 are the same, with the exception of the assumption linnontanbound. The assumption linnontanbound however is weaker than the assumption nontanbound of Theorem Theorem 1.1.1.1. Indeed, setting for fixed x, \mfa the outer radius R_2 in def-tang-unm-op to \min\{R_2', R_Q(\mfa, x')\}, where R_2' is the outer radius in def-lin-star-op, shows that def-lin-star-op is smaller than or equal to def-tang-unm-op. Thus we can apply Theorem Theorem 1.1.1.2, which completes the proof.

Proof of Theorem 1.1.1.2. Let Borel sets F, G in X with finite measure be given. Let a Borel function f: X \to \mathbb{C} with f \le \mathbf{1}_F be given. For each 0 < R_1,R_2,R, we define T_{R_1,R_2,R}f as in TRR. By Lemma Lemma 1.3.1, T_{R_1,R_2,R}f is measurable and bounded, and we clearly have for each x \in X T_Q f(x) = \lim_{n \to \infty} \sup_{2^{-n} < R_1 < R_2 < 2^n} T_{R_1, R_2, 2^n}f(x). For each x and all f, the functions \sup_{2^{-n} < R_1 < R_2 < 2^n} T_{R_1, R_2, 2^n} f(x) are measurable by Lemma Lemma 1.3.1 and form an increasing sequence in n. By the monotone convergence theorem, the claimed estimate linresweak then follows from Lemma Lemma 1.3.2.

Proof. Let F,G,f as in the lemma be given. Let R\in 2^\N be given. By replacing G with G\cap B(o,R) if necessary, a replacement that does not change the conclusion of Lemma Lemma 1.3.2, it suffices to show Rcut under the assumption that G is contained in B(o,R) and thus bounded. We make this assumption. For every x\in G and R_2 < R, the domain of integration in TRR is contained in B(o,2R). By replacing F with F\cap B(o,2R) if necessary, and correspondingly restricting f to B(o, 2R), it suffices to show Rcut under the assumption that F is contained in B(o,2R) and thus bounded. We make this assumption. Using the definition defks of K_s and the partition of unity eq-psisum, we observe that for fixed R_1<R_2 we have K(x,y)\mathbf{1}_{R_1<\rho(x,y)<R_2}=\sum_{s=s_1-2}^{s_2+2} K_s(x,y) \mathbf{1}_{R_1<\rho(x,y)<R_2}, where s_1=\lfloor\log_D 2R_1\rfloor+3 and s_2=\lceil\log_D 4R_2\rceil-2. To obtain this identity, we have used that on the set where R_1<\rho(x,y)<R_2 the kernels K_s vanish unless s is inside the interval of summation. Similarly, we observe \sum_{s=s_1}^{s_2} K_s(x,y) \mathbf{1}_{R_1<\rho(x,y)<R_2} = \sum_{s=s_1}^{s_2} K_s(x,y), because on the support of K_s with s_1\le s\le s_2 we have necessarily R_1<\rho(x,y)<R_2. We thus express TRR as the sum of T_{s_1,s_2}f(x):=\sum_{s_1 \le s\le s_2} \int K_s(x,y) f(y) e(Q(x)(y)) \, \mathrm{d}\mu(y) and \sum_{s=s_1-2,s_1-1, s_2+1, s_2+2} \int_{R_1 < \rho(x,y) < R_2} K_s(x,y) f(y) e(Q(x)(y)) \, \mathrm{d}\mu(y). We apply the triangle inequality and estimate Rcut separately with T_{R_1,R_2,R} replaced by the middle part and by the boundary part. To handle the case of the middle part, we employ Lemma 1.3.3. Here, we utilize the fact that if \frac 1R\le R_1\le R_2\le R, then s_1 and s_2 are in an interval [-S,S] for some sufficiently large S depending on R. To handle the case of the boundary part, we use the triangle inequality and the properties supp-Ks, eq-Ks-size of K_s and |f| \le \mathbf{1}_F to obtain for arbitrary s the inequality \left|\int_{R_1 < \rho(x,y) < R_2} K_s(x,y) f(y) e(Q(x)(y)) \, \mathrm{d}\mu(y)\right| \leq \frac{2^{102 a^3}}{\mu(B(x, D^{s}))} \int_{B(x, D^s)} \mathbf{1}_F(y) \, \mathrm{d}\mu(y) \leq 2^{102 a^3} M\mathbf{1}_F(x), where M\mathbf{1}_F is as defined in Theorem 1.2.6. Now the left-hand side of Rcut, with T_{R_1,R_2,R} replaced by the boundary part, can be estimated using Holder's inequality and Theorem 1.2.6 by 2^{102a^3+2}\int \mathbf{1}_{G}(x) M\mathbf{1}_F(x)\, d\mu(x) \le \frac{2^{102a^3+4a+3}q}{q-1}\mu(G)^{1-\frac{1}{q}} \mu(F)^{\frac{1}{q}}. Summing the contributions from the middle and boundary parts completes the proof.

Proof of Lemma 1.3.3. We reduce Lemma 1.3.3 to Lemma 1.3.4. For each x, let \sigma_1(x) be the minimal element s'\in [-S,S] such that \max_{s'\le s_2\le S} |T_{s',s_2} f(x)| = \max_{-S\le s_1\le s_2\le S} |T_{s_1,s_2} f(x)| =: T_{1,x}. Similarly, let {\sigma}_2(x) be the minimal element s''\in [-S,S] such that |T_{\sigma_2(x), s''} f(x)| = T_{1,x}. With these choices, and noting that with the definition of T_{2, \sigma_1, \sigma_2} from middles1 T_{\sigma_1(x),\sigma_2(x)} f(x)=T_{2,\sigma_1,\sigma_2} f(x), we conclude that the left-hand side of Scut and Sqlin are equal. Therefore, Lemma 1.3.3 follows from Lemma 1.3.4.

Proof of Lemma 1.3.4. Fix \sigma_1, \sigma_2 and \tQ as in the lemma. Applying Theorem 1.2.1 recursively, we obtain a sequence of sets G_n with G_0=G and, for each n\ge 0, G_{n+1} \subset G_n, \mu(G_{n})\le 2^{-n} \mu(G) and \int \mathbf{1}_{G_{n}\setminus G_{n+1}}(x) \left| {T}_{2,\sigma_1,\sigma_2} f(x) \right|\, d\mu(x) \le \frac{2^{442a^3}}{(q-1)^5} \mu(G_n)^{1 - \frac{1}{q}} \mu(F)^{\frac{1}{q}}. Adding the first n of these inequalities, we obtain by bounding a geometric series \int \mathbf{1}_{G\setminus G_{n}}(x) \left| {T}_{2,\sigma_1,\sigma_2}f(x) \right|\, d\mu(x) \le \frac{2^{442a^3+2}}{(q-1)^6} \mu(G)^{1-\frac{1}{q}} \mu(F)^{\frac{1}{q}}. As the integrand is non-negative and non-decreasing in n, we obtain by the monotone convergence theorem \int \mathbf{1}_{G}(x) \left| {T}_{2,\sigma_1,\sigma_2}f(x) \right|\, d\mu(x) \le \frac{2^{442a^3+2}}{(q-1)^6} \mu(G)^{1-\frac{1}{q}} \mu(F)^{\frac{1}{q}}. This completes the proof of Lemma 1.3.4 and thus Theorem 1.1.1.1.