John-Nirenberg exponential inequality

If $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}=0$, then $b$ is constant almost everywhere by [F1], so $\mathrm{\mid }b-b_Q\mathrm{\mid }=0$ almost everywhere for every cube $Q$ and the asserted left-hand side vanishes for every $\lambda >0$; this is the interpretation required by the statement.

Covering by dyadic cubes. Let $m$ be the greatest integer with $2^{-m}\ge \mathrm{\ell }(Q)$ and put $S:=2^{-m}$, so that $\mathrm{\ell }(Q)\le S<2\mathrm{\ell }(Q)$. Every coordinate interval of $Q$ has length at most $S$ and therefore meets at most two of the generation-$m$ dyadic intervals, so $Q$ is contained in the union of the $N\le 2^n$ generation-$m$ dyadic cubes $Q_1,\dots ,Q_N$ that meet it. Each $Q_j$ has side $S$, so $\mathrm{\mid }Q\mathrm{\mid }=\mathrm{\ell }(Q{)}^n\le S^n=\mathrm{\mid }{Q}_j\mathrm{\mid }<2^n\mathrm{\mid }Q\mathrm{\mid }$.

Small levels. If $0<\lambda \le {\mathrm{\Lambda }}_n\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$, then for every $\gamma >0$ one has $\mathrm{\mid }\{x\in Q:\mathrm{\mid }b-b_Q\mathrm{\mid }>\lambda \}\mathrm{\mid }\le \mathrm{\mid }Q\mathrm{\mid }=e^{\gamma {\mathrm{\Lambda }}_n}\mathrm{\mid }Q\mathrm{\mid }{e}^{-\gamma {\mathrm{\Lambda }}_n}\le e^{\gamma {\mathrm{\Lambda }}_n}\mathrm{\mid }Q\mathrm{\mid }{e}^{-\gamma \lambda \mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}}$, because $\lambda \mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}\le {\mathrm{\Lambda }}_n$.

Comparison of the means. Fix $j$ and a point $p\in Q\cap Q_j$; both cubes lie in the cube $R_j$ centred at $p$ of side $4\sqrt{n}S$, because their diameters are $\sqrt{n}\mathrm{\ell }(Q)\le \sqrt{n}S$ and $\sqrt{n}S$ respectively. By [F1], for cubes $E\subseteq R$ one has $\mathrm{\mid }{b}_E-b_R\mathrm{\mid }\le \mathrm{\mid }E{\mathrm{\mid }}^{-1}{\int }_E\mathrm{\mid }b-b_R\mathrm{\mid }\le (\mathrm{\mid }R\mathrm{\mid }\mathrm{/}\mathrm{\mid }E\mathrm{\mid })\mathrm{\mid }R{\mathrm{\mid }}^{-1}{\int }_R\mathrm{\mid }b-b_R\mathrm{\mid }\le (\mathrm{\mid }R\mathrm{\mid }\mathrm{/}\mathrm{\mid }E\mathrm{\mid })\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$. Since $\mathrm{\mid }{R}_j\mathrm{\mid }=(4\sqrt{n}{)}^n\mathrm{\mid }{Q}_j\mathrm{\mid }$ and $\mathrm{\mid }{R}_j\mathrm{\mid }=(4\sqrt{n}S\mathrm{/}\mathrm{\ell }(Q){)}^n\mathrm{\mid }Q\mathrm{\mid }<(8\sqrt{n}{)}^n\mathrm{\mid }Q\mathrm{\mid }$, applying this with $E=Q_j$ and with $E=Q$ gives $\mathrm{\mid }{b}_{Q_j}-b_Q\mathrm{\mid }\le 2(8\sqrt{n}{)}^n\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}=:B_n\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$.

Large levels. Suppose $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}>0$ and $\lambda >{\mathrm{\Lambda }}_n\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$ with ${\mathrm{\Lambda }}_n:=\max \{2B_n,2^{2n+3}\}$, and put $s:=2^{n+1}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}>\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$, ${\lambda }^{\mathrm{\prime }}:=\lambda -B_n\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}>\lambda \mathrm{/}2$, and $k:=\lfloor {\lambda }^{\mathrm{\prime }}\mathrm{/}(2^{n}s)\rfloor$, where $2^{n}s=2^{2n+1}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$. Then $k\ge 1$, $k{2}^{n}s\le {\lambda }^{\mathrm{\prime }}$, and $k>\lambda \mathrm{/}(2^{2n+3}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}})$ because $k>{\lambda }^{\mathrm{\prime }}\mathrm{/}(2^{2n+1}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}})-1>\lambda \mathrm{/}(2^{2n+2}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}})-1\ge \lambda \mathrm{/}(2^{2n+3}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}})$ as $\lambda >2^{2n+3}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$. Apply [F2] on each dyadic cube $Q_j$ at height $s$: the level-$k$ stopping families are pairwise disjoint with ${\sum }_i\mathrm{\mid }{Q}_{j,i}^{(k)}\mathrm{\mid }\le (\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}\mathrm{/}s{)}^k\mathrm{\mid }{Q}_j\mathrm{\mid }=2^{-(n+1)k}\mathrm{\mid }{Q}_j\mathrm{\mid }$, and $\mathrm{\mid }b-b_{Q_j}\mathrm{\mid }\le k{2}^{n}s\le {\lambda }^{\mathrm{\prime }}$ almost everywhere off their union, so step 2.1 gives $\mathrm{\mid }b-b_Q\mathrm{\mid }\le \lambda$ there. Hence $\{x\in Q_j:\mathrm{\mid }b-b_Q\mathrm{\mid }>\lambda \}$ is contained in ${\bigcup }_i{Q}_{j,i}^{(k)}$ up to a Lebesgue-null set and has measure at most $2^{-(n+1)k}\mathrm{\mid }{Q}_j\mathrm{\mid }\le 2^{-(n+1)k}{2}^n\mathrm{\mid }Q\mathrm{\mid }$. Summing over the $N\le 2^n$ cubes of step 1.2 gives $\mathrm{\mid }\{x\in Q:\mathrm{\mid }b-b_Q\mathrm{\mid }>\lambda \}\mathrm{\mid }\le 2^{2n}{2}^{-(n+1)k}\mathrm{\mid }Q\mathrm{\mid }\le 2^{2n}\mathrm{\mid }Q\mathrm{\mid }{e}^{-k\ln 2}\le 2^{2n}\mathrm{\mid }Q\mathrm{\mid }{e}^{-\gamma \lambda \mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}}$ with $\gamma :=\ln 2\mathrm{/}{2}^{2n+3}$.

Assembly. Take $\gamma :=\ln 2\mathrm{/}{2}^{2n+3}$, $B_n:=2(8\sqrt{n}{)}^n$, ${\mathrm{\Lambda }}_n:=\max \{2B_n,2^{2n+3}\}$ and $C_n:=\max \{2^{2n},e^{\gamma {\mathrm{\Lambda }}_n}\}$. Steps 3.1 and 1.3 give $\mathrm{\mid }\{x\in Q:\mathrm{\mid }b(x)-b_Q\mathrm{\mid }>\lambda \}\mathrm{\mid }\le C_n\mathrm{\mid }Q\mathrm{\mid }{e}^{-\gamma \lambda \mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}}$ for every $\lambda >0$ when $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}>0$, covering the large levels and the small levels respectively, and step 1.1 gives the bound when $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}=0$. Both constants depend only on $n$, and the stopping construction inside [F2] is the only place Countable Choice is used.