The bad part is integrable away from expanded cubes

If $y\in Q$ and $x\mathrm{\notin }{Q}^{\ast }$, then $\mathrm{\parallel }x-c_Q{\mathrm{\parallel }}_{\mathrm{\infty }}>\sqrt{n}\mathrm{\ell }$ and $\mathrm{\parallel }y-c_Q{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\ell }\mathrm{/}2$, so by [F2] $\mathrm{\parallel }x-c_Q{\mathrm{\parallel }}_2\ge \mathrm{\parallel }x-c_Q{\mathrm{\parallel }}_{\mathrm{\infty }}>\sqrt{n}\mathrm{\ell }\ge 2\mathrm{\parallel }y-c_Q{\mathrm{\parallel }}_2$; hence $\mathrm{\mid }x-c_Q\mathrm{\mid }\ge 2\mathrm{\mid }y-c_Q\mathrm{\mid }$, and in particular $x\mathrm{\ne }{c}_Q$ and $x\mathrm{\ne }y$.

For almost every $x\mathrm{\notin }{Q}^{\ast }$ one has, using [F1] and the mean-zero condition, $T{b}_Q(x)=\int k(x-y)b_Q(y)dy={\int }_{{\mathbb{R}}^n}[k(x-y)-k(x-c_Q)]b_Q(y)dy,$ because $b_Q$ vanishes off $Q$, the subtracted term is the constant $k(x-c_Q)$ times $\int b_Q=0$, and $x-c_Q\mathrm{\ne }0$ by step 1.1.

By step 2.1 and nonnegativity, for almost every $x\mathrm{\notin }{Q}^{\ast }$, $\mathrm{\mid }T{b}_Q(x)\mathrm{\mid }\le {\int }_Q\mathrm{\mid }k(x-y)-k(x-c_Q)\mathrm{\mid }\mathrm{\mid }{b}_Q(y)\mathrm{\mid }dy\mathrm{.}$ Integrating this inequality over ${\mathbb{R}}^n\setminus Q^{\ast }$, whose complement has finite measure at every scale and which is $\sigma$-finite, and applying Tonelli's theorem [F3] to the nonnegative product-measurable integrand gives ${\int }_{{\mathbb{R}}^n\setminus Q^{\ast }}\mathrm{\mid }T{b}_Q\mathrm{\mid }\le {\int }_Q\mathrm{\mid }{b}_Q(y)\mathrm{\mid }({\int }_{{\mathbb{R}}^n\setminus Q^{\ast }}\mathrm{\mid }k(x-y)-k(x-c_Q)\mathrm{\mid }dx)dy\mathrm{.}$

For $y=c_Q$ the difference integrand is zero. For every other $y\in Q$ the inner integral is at most $A_2$: by step 1.1 the domain ${\mathbb{R}}^n\setminus Q^{\ast }$ is contained in $\{x:\mathrm{\mid }x-c_Q\mathrm{\mid }\ge 2\mathrm{\mid }y-c_Q\mathrm{\mid }\}$, and the change of variables $u=x-c_Q$, $v=y-c_Q$ turns the integral over that larger set into ${\int }_{\mathrm{\mid }u\mathrm{\mid }\ge 2\mathrm{\mid }v\mathrm{\mid }}\mathrm{\mid }k(u-v)-k(u)\mathrm{\mid }du\le A_2$ by the translation-invariant Hörmander condition of [F1]. Substituting into step 3.1 yields the asserted bound ${\int }_{{\mathbb{R}}^n\setminus Q^{\ast }}\mathrm{\mid }T{b}_Q\mathrm{\mid }\le A_2{\int }_Q\mathrm{\mid }{b}_Q\mathrm{\mid }=A_2\mathrm{\parallel }{b}_Q{\mathrm{\parallel }}_1$.