Reverse Holder from a distribution estimate for a doubling weight

Since ${\alpha }_k\ge {\alpha }_0$ for every $k\ge 0$ and the top cube $Q_0$ has $\mu$-average ${\alpha }_0$, the cube $Q_0$ is not bad at any level $k$. For every bad subcube $R$ the ancestors of $R$ inside $Q_0$ form a finite chain ending at $Q_0$; hence there is a topmost bad ancestor, and the family of maximal bad subcubes (those with no bad proper ancestor inside $Q_0$) is well defined, pairwise disjoint and at most countable. Let $U_k$ be their union.

Let $R$ be a maximal bad subcube at level $k$ with parent $P$: then $⟨h{⟩}_P^{\mu }\le {\alpha }_k$ by maximality, and $\mu (P)\le C_n\mu (R)$ by [F1] since $P$ is a cube of twice the side length containing $R$; hence ${\int }_{R}hd\mu \le {\int }_{P}hd\mu \le C_n{\alpha }_k\mu (R)$, that is, $⟨h{⟩}_R^{\mu }\le C_n{\alpha }_k$. Moreover $U_{k+1}\subseteq U_k$ by the same parent argument, and $h\le {\alpha }_k$ $\mu$-almost everywhere on $Q_0\setminus U_k$: for a point $x$ outside $U_k$ and outside the $\mu$-null exceptional set of the differentiation lemma, no dyadic subcube $R\ni x$ has $⟨h{⟩}_R^{\mu }>{\alpha }_k$, since such an $R$ would lie in a maximal bad cube containing $x$; the subcubes containing $x$ shrink nicely to $x$, so their $\mu$-averages converge to $h(x)$ by that lemma, and the limit satisfies $h(x)\le {\alpha }_k$.

Decay of the integrals. For a maximal bad subcube $R$ at level $k$, the set $S:=R\cap U_{k+1}$ is measurable and contained in $R$; since $U_{k+1}$ is the disjoint union of its maximal bad subcubes, on each of which the $\mu$-average of $h$ exceeds ${\alpha }_{k+1}$, ${\alpha }_{k+1}\mu (S)\le {\int }_{S}hd\mu \le {\int }_{R}hd\mu \le C_n{\alpha }_k\mu (R)$, so $\mu (S)\le \alpha \mu (R)$ because $C_n{\alpha }_k=\alpha {\alpha }_{k+1}$. The hypothesis applied to $S\subseteq R$ therefore gives ${\int }_{S}hd\mu \le \beta {\int }_{R}hd\mu$; summing over the pairwise disjoint maximal cubes at level $k$ yields ${\int }_{U_{k+1}}hd\mu \le \beta {\int }_{U_k}hd\mu$, hence ${\int }_{U_k}hd\mu \le {\beta }^k{\int }_{Q_0}hd\mu$ by iteration.

Integral bound. If ${\int }_{Q_0}hd\mu =0$, then $h=0$ $\mu$-a.e. on $Q_0$ and the conclusion is immediate. Otherwise ${\alpha }_0>0$ as above. Summing the bounds $\mu (R\cap U_{k+1})\le \alpha \mu (R)$ from step 3.1 shows $\mu (U_{k+1})\le \alpha \mu (U_k)$; hence $\mu ({\bigcap }_k{U}_k)=0$. The sets $Q_0\setminus U_0$ and $U_k\setminus U_{k+1}$ are disjoint measurable pieces covering $Q_0$ up to a $\mu$-null set; by step 2.1, $h\le {\alpha }_0$ on $Q_0\setminus U_0$ and $h\le {\alpha }_{k+1}$ on $U_k\setminus U_{k+1}$, all $\mu$-a.e. Hence, for every $\gamma >0$, ${\int }_{Q_0}{h}^{1+\gamma }d\mu \le {\alpha }_0^{\gamma }{\int }_{Q_0\setminus U_0}hd\mu +{\sum }_{k\ge 0}{\alpha }_{k+1}^{\gamma }{\int }_{U_k}hd\mu \le {\alpha }_0^{\gamma }\left({\int }_{Q_0}hd\mu \right)[1+(C_n{\alpha }^{-1}{)}^{\gamma }{\sum }_{k\ge 0}((C_n{\alpha }^{-1}{)}^{\gamma }\beta {)}^k]$. Choose $\gamma >0$ so small that $r:=(C_n{\alpha }^{-1}{)}^{\gamma }\beta <1$; then the geometric series converges.

Dividing the display of step 4.1 by $\mu (Q_0)$ and using ${\alpha }_0=\mu (Q_0{)}^{-1}{\int }_{Q_0}hd\mu$ gives $(\mu (Q_0{)}^{-1}{\int }_{Q_0}{h}^{1+\gamma }d\mu {)}^{1\mathrm{/}(1+\gamma )}\le c\mu (Q_0{)}^{-1}{\int }_{Q_0}hd\mu$ with $c=(1+(C_n{\alpha }^{-1}{)}^{\gamma }\mathrm{/}(1-r){)}^{1\mathrm{/}(1+\gamma )}<\mathrm{\infty }$; since $Q_0$ was arbitrary, the reverse Hölder inequality holds with $q=1+\gamma >1$ and this $c$, both depending only on $n$, the doubling constant of $v$, $\alpha$ and $\beta$.