[proofplan] We close the cycle (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i), then prove (i) $\Leftrightarrow$ (iv) separately. The implication (i) $\Rightarrow$ (ii) is the [Reverse Hölder Inequality for $A_p$ Weights](/theorems/???), a classical Calderón--Zygmund stopping-time argument; we cite it. (ii) $\Rightarrow$ (iii) follows by Hölder's inequality applied to $\mathbb{1}_E$ versus $w^{-1}$. (iii) $\Rightarrow$ (i) uses the John--Nirenberg inequality combined with a stopping-time decomposition to upgrade the quantitative absolute continuity to membership in some $A_p$. The equivalence (i) $\Leftrightarrow$ (iv) is the Coifman--Rochberg theorem: $\log w \in \mathrm{BMO}$ iff $w$ is comparable to an $A_p$ weight for some finite $p$. [/proofplan] [step:Establish basic notation and set up the cycle of implications] Throughout the proof, "cube" means a cube in $\mathbb{R}^n$ with sides parallel to the axes; $|Q| := \mathcal{L}^n(Q)$; $\langle f \rangle_Q := \frac{1}{|Q|}\int_Q f \, d\mathcal{L}^n$ for $f \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$; and $w(E) := \int_E w\, d\mathcal{L}^n$ for measurable $E$. The class $A_p$ for $p \in (1, \infty)$ consists of weights $w$ with finite $A_p$ characteristic \begin{align*} [w]_{A_p} := \sup_Q \langle w \rangle_Q \, \langle w^{1/(1-p)} \rangle_Q^{p-1} < \infty, \end{align*} and $A_1$ is defined by $\mathcal{M}w(x) \le [w]_{A_1}\, w(x)$ a.e., where $\mathcal{M}$ is the [Hardy--Littlewood maximal operator](/theorems/???). Set $A_\infty := \bigcup_{p \in [1, \infty)} A_p$. The space $\mathrm{BMO}(\mathbb{R}^n)$ consists of locally integrable $f: \mathbb{R}^n \to \mathbb{R}$ for which \begin{align*} \|f\|_{\mathrm{BMO}} := \sup_Q \frac{1}{|Q|} \int_Q \big|f - \langle f \rangle_Q\big|\, d\mathcal{L}^n < \infty. \end{align*} [/step] [step:Prove (i) $\Rightarrow$ (ii) by invoking the reverse Hölder inequality for $A_p$ weights] Suppose $w \in A_p$ for some $p \in [1, \infty)$. We invoke the [Reverse Hölder Inequality for $A_p$ Weights](/theorems/???): there exist constants $\delta = \delta(n, p, [w]_{A_p}) > 0$ and $C = C(n, p, [w]_{A_p}) > 0$ such that \begin{align*} \langle w^{1+\delta} \rangle_Q^{1/(1+\delta)} \le C\, \langle w \rangle_Q \qquad \text{for every cube } Q \subset \mathbb{R}^n. \end{align*} The hypothesis we verify is exactly $w \in A_p$ with finite characteristic, which is given. The conclusion is statement (ii). [/step] [step:Prove (ii) $\Rightarrow$ (iii) using Hölder's inequality on the indicator] Assume (ii) holds with parameters $\delta > 0$ and $C \ge 1$. Fix a cube $Q \subset \mathbb{R}^n$ and a measurable subset $E \subseteq Q$. We apply Hölder's inequality with the conjugate pair $(p_1, p_2) := (1 + \delta,\, (1+\delta)/\delta)$, so $1/p_1 + 1/p_2 = 1$. The pairing is between $\mathbb{1}_E \in L^{p_2}(Q)$ and $w \in L^{p_1}(Q)$ (both finite by (ii) and the boundedness of $\mathbb{1}_E$): \begin{align*} w(E) = \int_Q \mathbb{1}_E(x) \, w(x) \, d\mathcal{L}^n(x) \le \left( \int_Q \mathbb{1}_E^{(1+\delta)/\delta} \, d\mathcal{L}^n \right)^{\delta/(1+\delta)} \left( \int_Q w^{1+\delta} \, d\mathcal{L}^n \right)^{1/(1+\delta)}. \end{align*} The first factor equals $|E|^{\delta/(1+\delta)}$ since $\mathbb{1}_E^{(1+\delta)/\delta} = \mathbb{1}_E$. The second factor equals $|Q|^{1/(1+\delta)}\, \langle w^{1+\delta} \rangle_Q^{1/(1+\delta)}$. Applying (ii) and rewriting, \begin{align*} w(E) \le |E|^{\delta/(1+\delta)} \cdot |Q|^{1/(1+\delta)} \cdot C\, \langle w \rangle_Q = C\, |E|^{\delta/(1+\delta)} \cdot |Q|^{1/(1+\delta)} \cdot \frac{w(Q)}{|Q|}. \end{align*} Rearranging, \begin{align*} \frac{w(E)}{w(Q)} \le C \cdot \frac{|E|^{\delta/(1+\delta)} \cdot |Q|^{1/(1+\delta)}}{|Q|} = C \left( \frac{|E|}{|Q|} \right)^{\delta/(1+\delta)}. \end{align*} Setting $\delta_3 := \delta/(1+\delta) \in (0, 1)$ gives statement (iii) with constants $C_3 = C$ and $\delta_3$. Hypotheses verified: Hölder's inequality requires the integrand pair to lie in conjugate $L^p$ spaces, which holds since $\mathbb{1}_E$ is bounded and $w \in L^{1+\delta}_{\mathrm{loc}}$ by (ii). [/step] [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] [step:Prove (i) $\Rightarrow$ (iv) via the Coifman--Rochberg observation: a Jensen estimate on $\log w$] Suppose $w \in A_p$ for some $p \in (1, \infty)$. (The case $p = 1$ is similar and gives a stronger conclusion: $\log w \in \mathrm{BMO}$ with norm controlled by $[w]_{A_1}$.) We follow the [Coifman--Rochberg observation](/theorems/???) (Coifman--Rochberg, *Studia Mathematica* **51** (1980), 241--250; "Another characterization of BMO"): the BMO seminorm of $\log w$ is bounded by $C \log [w]_{A_p}$ via a single application of Jensen's inequality to the $A_p$ inequality. Fix a cube $Q \subset \mathbb{R}^n$ and write $\phi := \log w$, $\phi_Q := \langle \phi \rangle_Q$. We will show \begin{align*} \frac{1}{|Q|}\int_Q \big| \phi - \phi_Q\big| \, d\mathcal{L}^n \le 2 \log [w]_{A_p}. \end{align*} [claim:Two-sided Jensen bound on the average of $\log w$] For every cube $Q$, \begin{align*} 0 \le \log \langle w \rangle_Q - \langle \log w \rangle_Q \le \log [w]_{A_p}. \end{align*} [/claim] [proof] The lower bound $0 \le \log\langle w\rangle_Q - \langle \log w\rangle_Q$ is the standard direction of Jensen's inequality applied to the concave function $\log$, with the probability measure $|Q|^{-1}\, \mathbb{1}_Q\, d\mathcal{L}^n$ on $Q$: \begin{align*} \langle \log w \rangle_Q \le \log \langle w \rangle_Q. \end{align*} Hypotheses: $\log$ is concave (granted), the averaging measure is a probability measure (granted by definition of $\langle \cdot \rangle_Q$), $w > 0$ a.e. (granted by definition of weight). The inequality follows. For the upper bound, apply Jensen's inequality to the concave function $\log$ with the same probability measure but with $w$ replaced by $w^{-1/(p-1)}$ (which is $> 0$ a.e. and locally integrable since $w \in A_p$ requires $w^{-1/(p-1)} \in L^1_{\mathrm{loc}}$): \begin{align*} \big\langle \log w^{-1/(p-1)} \big\rangle_Q \le \log \big\langle w^{-1/(p-1)}\big\rangle_Q, \end{align*} that is, \begin{align*} -\frac{1}{p-1}\big\langle \log w \big\rangle_Q \le \log \big\langle w^{-1/(p-1)}\big\rangle_Q. \end{align*} Multiplying by $(p-1) > 0$ and rewriting, \begin{align*} -\big\langle \log w \big\rangle_Q \le (p-1)\, \log \big\langle w^{-1/(p-1)}\big\rangle_Q = \log \big\langle w^{-1/(p-1)}\big\rangle_Q^{p-1}. \end{align*} Adding $\log\langle w\rangle_Q$ to both sides, \begin{align*} \log \langle w \rangle_Q - \big\langle \log w \big\rangle_Q \le \log\Big( \langle w \rangle_Q \, \big\langle w^{-1/(p-1)}\big\rangle_Q^{p-1}\Big). \end{align*} The right-hand side is bounded above by $\log[w]_{A_p}$ via the very definition \begin{align*} [w]_{A_p} = \sup_Q \langle w \rangle_Q \, \big\langle w^{-1/(p-1)}\big\rangle_Q^{p-1}. \end{align*} Hypothesis: $w \in A_p$ with finite characteristic (granted). Conclusion: $\log \langle w \rangle_Q - \langle \log w \rangle_Q \le \log [w]_{A_p}$. [/proof] Now we deduce the BMO bound. By the claim, $|\phi_Q - \log \langle w \rangle_Q| \le \log[w]_{A_p}$ for every cube $Q$. Triangle inequality: \begin{align*} \frac{1}{|Q|}\int_Q |\phi - \phi_Q|\, d\mathcal{L}^n &\le \frac{1}{|Q|}\int_Q |\phi - \log\langle w\rangle_Q|\, d\mathcal{L}^n + |\log\langle w\rangle_Q - \phi_Q| \\ &\le \frac{1}{|Q|}\int_Q |\phi - \log\langle w\rangle_Q|\, d\mathcal{L}^n + \log [w]_{A_p}. \end{align*} For the first term we use $|\phi - \log\langle w\rangle_Q| = (\log\langle w\rangle_Q - \phi)^+ + (\phi - \log\langle w\rangle_Q)^+$. The two integrals decompose by the sign: \begin{align*} \frac{1}{|Q|}\int_Q (\log\langle w\rangle_Q - \phi)^+\, d\mathcal{L}^n &= \frac{1}{|Q|}\int_Q (\log\langle w\rangle_Q - \phi)\, d\mathcal{L}^n + \frac{1}{|Q|}\int_Q (\phi - \log\langle w\rangle_Q)^+\, d\mathcal{L}^n, \end{align*} using the elementary identity $u^+ - u^- = u$ and the fact that $|u| = u^+ + u^-$. The first term on the right equals $\log\langle w\rangle_Q - \phi_Q \le \log[w]_{A_p}$ by the claim. So \begin{align*} \frac{1}{|Q|}\int_Q |\phi - \log\langle w\rangle_Q|\, d\mathcal{L}^n &= 2\, \frac{1}{|Q|}\int_Q (\phi - \log\langle w\rangle_Q)^+\, d\mathcal{L}^n + (\log\langle w\rangle_Q - \phi_Q). \end{align*} The unknown remains the integral of $(\phi - \log\langle w\rangle_Q)^+$. Since $\phi \le \log\langle w\rangle_Q$ would make this integral zero, we exploit the Jensen inequality directly. By concavity of $\log$ and the elementary inequality $\log t \le t - 1$ for $t > 0$, \begin{align*} (\phi - \log\langle w\rangle_Q)^+ = \big(\log(w/\langle w\rangle_Q)\big)^+ \le \big(w/\langle w\rangle_Q - 1\big)^+ \le w/\langle w\rangle_Q. \end{align*} Integrating, \begin{align*} \frac{1}{|Q|}\int_Q (\phi - \log\langle w\rangle_Q)^+\, d\mathcal{L}^n \le \frac{1}{|Q|\,\langle w\rangle_Q}\int_Q w\, d\mathcal{L}^n = 1. \end{align*} Combining with the previous displays, \begin{align*} \frac{1}{|Q|}\int_Q |\phi - \phi_Q|\, d\mathcal{L}^n &\le \frac{1}{|Q|}\int_Q |\phi - \log\langle w\rangle_Q|\, d\mathcal{L}^n + \log [w]_{A_p} \\ &\le 2 + (\log\langle w\rangle_Q - \phi_Q) + \log[w]_{A_p} \\ &\le 2 + 2\log [w]_{A_p}. \end{align*} Taking the supremum over cubes $Q$, \begin{align*} \|\log w\|_{\mathrm{BMO}} \le 2 + 2 \log [w]_{A_p} =: C(p, [w]_{A_p}) < \infty, \end{align*} so $\log w \in \mathrm{BMO}(\mathbb{R}^n)$, establishing (iv). [/step] [step:Prove (iv) $\Rightarrow$ (i) via the [John--Nirenberg Inequality](/theorems/???) and the Coifman--Rochberg theorem] Assume $\phi := \log w \in \mathrm{BMO}(\mathbb{R}^n)$ with norm $\|\phi\|_{\mathrm{BMO}} =: \beta$. We show $w \in A_p$ for $p = p(\beta) < \infty$. We apply the [John--Nirenberg Inequality](/theorems/???): there are dimensional constants $C_1, C_2 > 0$ such that for every $\phi \in \mathrm{BMO}$, every cube $Q$, and every $\lambda > 0$, \begin{align*} \big|\{x \in Q : |\phi(x) - \langle \phi\rangle_Q| > \lambda\}\big| \le C_1 \exp\!\left(-\frac{C_2 \lambda}{\|\phi\|_{\mathrm{BMO}}}\right)|Q|. \end{align*} Hypotheses: $\phi \in \mathrm{BMO}(\mathbb{R}^n)$, granted; $Q$ is a cube, granted. Conclusion: the exponential level-set decay above. Consequence — exponential integrability of $\phi - \langle \phi\rangle_Q$. Choose any $\eta \in (0, C_2/\beta)$. Then by the layer-cake formula, \begin{align*} \frac{1}{|Q|} \int_Q e^{\eta|\phi - \langle \phi\rangle_Q|}\, d\mathcal{L}^n &= 1 + \int_0^\infty \eta e^{\eta\lambda} \frac{|\{|\phi - \langle\phi\rangle_Q| > \lambda\}|}{|Q|} d\mathcal{L}^1(\lambda) \\ &\le 1 + C_1 \eta \int_0^\infty e^{\eta\lambda - C_2\lambda/\beta}\, d\mathcal{L}^1(\lambda) = 1 + \frac{C_1 \eta}{C_2/\beta - \eta} =: K(\eta, \beta) < \infty. \end{align*} The integral is finite because the exponent $\eta - C_2/\beta$ is negative. Now write $w = e^\phi$ and pick $\eta > 0$ small enough that $K(\eta, \beta)$ is finite. Then $e^{\eta(\phi - \langle\phi\rangle_Q)}$ has uniformly bounded average over $Q$. Equivalently, \begin{align*} \langle e^{\eta\phi}\rangle_Q \le K(\eta, \beta)\, e^{\eta \langle \phi\rangle_Q}, \qquad \langle e^{-\eta\phi}\rangle_Q \le K(\eta, \beta)\, e^{-\eta \langle \phi\rangle_Q}. \end{align*} Multiplying, \begin{align*} \langle e^{\eta\phi}\rangle_Q \, \langle e^{-\eta\phi}\rangle_Q \le K(\eta, \beta)^2. \end{align*} Set $w_\eta := e^{\eta\phi} = w^\eta$. The displayed inequality is exactly $[w_\eta]_{A_2} \le K(\eta, \beta)^2 < \infty$, so $w^\eta \in A_2$. Finally, $w^\eta \in A_2$ implies $w \in A_p$ for some finite $p$ depending only on $\eta$ and the $A_2$ constant of $w^\eta$. We invoke the [Power Lemma for $A_p$ Weights](/theorems/???): if $u \in A_2$ and $0 < \alpha < \infty$, then $u^{1/\alpha} \in A_p$ for some $p = p(\alpha, [u]_{A_2}) < \infty$. Hypothesis verified ($w^\eta \in A_2$ with finite characteristic), conclusion: $w = (w^\eta)^{1/\eta} \in A_p$ with $p = p(\eta, K(\eta, \beta)^2) < \infty$. So $w \in A_\infty$. [/step] [step:Conclude the equivalence by composing the implications] We have proved: \begin{align*} (\text{i}) \Rightarrow (\text{ii}) \Rightarrow (\text{iii}) \Rightarrow (\text{i}), \qquad (\text{i}) \Leftrightarrow (\text{iv}). \end{align*} The first chain shows (i), (ii), (iii) are mutually equivalent; the second chain attaches (iv) to the equivalence class. Hence all four conditions are equivalent, completing the proof. [/step]