The weighted maximal function of a doubling weight is weak (1,1)

The level set $E_{\lambda }:=\{M^{v}f>\lambda \}$ is open, being the union over $r>0$ of the open sets $\{x:v(B(x,r){)}^{-1}{\int }_{B(x,r)}\mathrm{\mid }f\mathrm{\mid }vd\lambda >\lambda \}$, which are open because the displayed functions are continuous by [F1]. By [F4] its $v$-measure is the supremum of $v(K)$ over compact $K\subseteq E_{\lambda }$.

Let $K\subseteq E_{\lambda }$ be compact. Each $x\in K$ admits $r_x>0$ with ${\int }_{B(x,r_x)}\mathrm{\mid }f\mathrm{\mid }vd\lambda >\lambda v(B(x,r_x))$; finitely many of these open balls cover $K$, and [F3] selects pairwise disjoint balls $B_1,\dots ,B_{\mathrm{\ell }}$ among them with $K\subseteq {\bigcup }_{j}5B_j$ and ${\int }_{B_j}\mathrm{\mid }f\mathrm{\mid }v>\lambda v(B_j)$ for every $j$.

By [F2] and the selection of step 1.2, $v(K)\le {\sum }_{j}v(5B_j)\le c_v^3{\sum }_{j}v(B_j)\le c_v^3{\lambda }^{-1}{\sum }_j{\int }_{B_j}\mathrm{\mid }f\mathrm{\mid }vd\lambda \le c_v^3{\lambda }^{-1}{\int }_{{\mathbb{R}}^n}\mathrm{\mid }f\mathrm{\mid }vd\lambda$, where the last inequality uses the pairwise disjointness of the $B_j$.

Taking the supremum over compact $K\subseteq E_{\lambda }$ in step 2.1 and using the inner regularity of step 1.1 gives $v(\{M^{v}f>\lambda \})\le c_v^3{\lambda }^{-1}\int \mathrm{\mid }f\mathrm{\mid }vd\lambda$, which is the asserted weak $(1,1)$ bound with $C(n,c_v)=c_v^3$.

$M^v$ is sublinear and homogeneous; besides the weak $(1,1)$ bound of step 3.1 with constant $A=c_v^3$, it satisfies the trivial strong $(\mathrm{\infty },\mathrm{\infty })$ bound with constant $B=1$ with respect to the measure $vd\lambda$. Hence [F5] applies and gives, for every $1<q<\mathrm{\infty }$, $\mathrm{\parallel }{M}^{v}f{\mathrm{\parallel }}_{L^q(v)}\le 2(Aq\mathrm{/}(q-1){)}^{1\mathrm{/}q}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^q(v)}$ for all $f\in L^q(v)$, a constant depending only on $n$, $q$ and $c_v$.