[step:Prove (iii) $\Rightarrow$ (i) by a level-set decomposition followed by the $A_p$ characterisation]
Assume (iii) with constants $C \ge 1$ and $\delta \in (0, 1]$. We show $w \in A_p$ for $p$ sufficiently large.
[claim:Reverse Hölder bound from (iii)]
There exist $\eta > 0$ and $C' > 0$ such that for every cube $Q$,
\begin{align*}
\langle w^{1+\eta} \rangle_Q^{1/(1+\eta)} \le C'\, \langle w \rangle_Q.
\end{align*}
[/claim]
[proof]
Fix a cube $Q$ and write $\bar{w} := \langle w \rangle_Q$. Define the level sets
\begin{align*}
F_t := \{x \in Q : w(x) > t\, \bar{w}\}, \qquad t > 0.
\end{align*}
Each $F_t$ is measurable. Apply (iii) with $E = F_t$ and the same cube $Q$:
\begin{align*}
\frac{w(F_t)}{w(Q)} \le C \left( \frac{|F_t|}{|Q|} \right)^\delta.
\end{align*}
Now we lower-bound $w(F_t)$. By definition of $F_t$,
\begin{align*}
w(F_t) = \int_{F_t} w \, d\mathcal{L}^n \ge t \bar{w} \cdot |F_t|.
\end{align*}
Also $w(Q) = |Q| \bar{w}$. Substituting,
\begin{align*}
\frac{t \bar{w} |F_t|}{|Q| \bar{w}} \le C \left( \frac{|F_t|}{|Q|}\right)^\delta, \qquad t \cdot \frac{|F_t|}{|Q|} \le C \left(\frac{|F_t|}{|Q|}\right)^\delta.
\end{align*}
**Case $\delta < 1$.** The inequality $t \cdot (|F_t|/|Q|) \le C\, (|F_t|/|Q|)^\delta$ rearranges to
\begin{align*}
\frac{|F_t|}{|Q|} \le \left(\frac{C}{t}\right)^{1/(1-\delta)},
\end{align*}
so $|F_t| \le C_1 t^{-1/(1-\delta)}\, |Q|$ for all $t \ge t_0 := C$ (with $t_0 \ge 1$). By the layer-cake formula, for any $\eta > 0$,
\begin{align*}
\int_Q w^{1+\eta} \, d\mathcal{L}^n &= (1+\eta) \int_0^\infty s^\eta \, |\{x \in Q : w(x) > s\}|\, d\mathcal{L}^1(s) \\
&= (1+\eta) \bar{w}^{1+\eta} \int_0^\infty t^\eta\, |F_t|\, d\mathcal{L}^1(t).
\end{align*}
Use $|F_t| \le |Q|$ for $t < t_0$ and $|F_t| \le C_1 |Q|\, t^{-1/(1-\delta)}$ for $t \ge t_0$; splitting at $t_0$,
\begin{align*}
\int_0^\infty t^\eta |F_t|\, d\mathcal{L}^1(t) &\le |Q| \int_0^{t_0} t^\eta\, d\mathcal{L}^1(t) + C_1 |Q| \int_{t_0}^\infty t^\eta\, t^{-1/(1-\delta)}\, d\mathcal{L}^1(t) \\
&= |Q|\, \frac{t_0^{1+\eta}}{1+\eta} + C_1 |Q|\, \frac{t_0^{1+\eta - 1/(1-\delta)}}{1/(1-\delta) - 1 - \eta},
\end{align*}
the second integral converging precisely when $\eta < 1/(1-\delta) - 1 = \delta/(1-\delta)$. Fix any such $\eta > 0$ (this is possible since $\delta > 0$). Both terms are bounded by $C_2(\delta, \eta) |Q|$, so
\begin{align*}
\int_Q w^{1+\eta}\, d\mathcal{L}^n \le (1+\eta) C_2(\delta, \eta)\, \bar{w}^{1+\eta} \, |Q|, \qquad \langle w^{1+\eta} \rangle_Q^{1/(1+\eta)} \le C_2'\, \bar{w} = C_2' \langle w \rangle_Q,
\end{align*}
which is the reverse Hölder bound with $C' := C_2'$ and $\eta = \delta/(2(1-\delta)) > 0$ (say).
**Case $\delta = 1$.** Here the volume bound $|F_t|/|Q| \le C/t$ gives $w$ a quantitative weak-type (1,1) bound on each cube, but the direct exponent computation degenerates as $1/(1-\delta) \to \infty$. We argue instead via $L^1$--BMO duality.
From (iii) with $\delta = 1$ we have $w(E)/w(Q) \le C\, |E|/|Q|$ for every measurable $E \subseteq Q$, which is exactly $w \in A_1$ on the level of cubes: $\langle w \rangle_Q \le C\, \operatorname{ess\,inf}_Q w'$ for the dyadic restriction $w' = w|_Q$. Equivalently, the Hardy--Littlewood maximal function obeys $\mathcal{M}(w \mathbb{1}_Q)(x) \le C\, w(x)$ for a.e.\ $x \in Q$. We then use the [open property of $A_1$](/theorems/???): if $\mathcal{M}w \le C\, w$ uniformly on cubes, then $w^{1+\eta} \in A_1$ for $\eta > 0$ small, in particular $\langle w^{1+\eta}\rangle_Q^{1/(1+\eta)} \le C_3\, \langle w\rangle_Q$.
Since $\delta = 1$ implies the stronger condition (iii) with any $\delta' < 1$ (just weaken the right-hand side from $|E|/|Q|$ to $(|E|/|Q|)^{\delta'}$, which is larger when $|E|/|Q| < 1$), the case $\delta = 1$ in fact reduces to the case $\delta' = 1/2$, say, of the previous paragraph. This gives the reverse Hölder bound with the same exponent $\eta = 1/2$ and a constant $C' = C_2'(1/2, 1/2)$ depending only on $C$ and $n$.
[/proof]
We have just proved the reverse Hölder inequality (claim) from (iii). By the [Reverse Hölder Implies $A_p$ Membership Theorem](/theorems/???) (the converse direction of the $A_\infty$ theory), if $w$ satisfies a reverse Hölder inequality with exponent $1+\eta > 1$, then $w \in A_p$ for $p = p(\eta, C') < \infty$. We verify the hypothesis: the reverse Hölder bound $\langle w^{1+\eta}\rangle_Q^{1/(1+\eta)} \le C' \langle w \rangle_Q$ is in scope, with $\eta > 0$. The conclusion is $w \in A_p \subseteq A_\infty$, establishing (i).
[/step]
