Maximal dyadic cubes above a level

Call a dyadic cube bad when $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }f\mathrm{\mid }>\lambda$. Every bad cube satisfies $0\le \lambda \mathrm{\mid }Q\mathrm{\mid }<{\int }_Q\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$, so $\mathrm{\mid }Q\mathrm{\mid }<{\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$; in particular there is a scale above which no bad cube lives.

A bad cube is maximal exactly when none of its strictly larger ancestors is bad: by [F1] any intersecting cube is nested, and any containing cube of coarser generation is the unique ancestor of that generation. Every maximal bad cube therefore has a good parent. A good parent alone need not imply maximality; coarser ancestors must also be excluded. Distinct maximal bad cubes are disjoint, since nesting would otherwise make one a strictly larger bad cube containing the other. The family is countable because it is a subset of the dyadic grid parameterized by $\mathbb{Z}\times {\mathbb{Z}}^n$; no selection is required.

Every bad cube is contained in a maximal bad cube. Let $Q$ be bad of generation $k$, and let $J:=\{j\le k:A_j(Q)\text{ is bad}\}\subseteq \mathbb{Z}$, where $A_j(Q)$ is the unique generation-$j$ ancestor of $Q$ from [F1]. The set $J$ is nonempty because $k\in J$, and it is bounded below: if $j\in J$ then $\mathrm{\mid }{A}_j(Q)\mathrm{\mid }=2^{-jn}<{\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$ by step 1.1, so $2^{-j}<({\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1{)}^{1\mathrm{/}n}$ and $j$ exceeds a fixed bound. A nonempty subset of $\mathbb{Z}$ that is bounded below has a least element $j_0$; the ancestor $A_{j_0}(Q)$ is bad by definition, and every strictly larger ancestor has generation $j<j_0$ and is not bad by minimality. Thus $A_{j_0}(Q)$ is maximal by step 1.2, and contains $Q$.

Let $Q$ be a maximal bad cube and $P$ its parent; by step 1.2 the cube $P$ is good, that is, $\mathrm{\mid }P{\mathrm{\mid }}^{-1}{\int }_P\mathrm{\mid }f\mathrm{\mid }\le \lambda$. Since $Q\subseteq P$ and $\mathrm{\mid }P\mathrm{\mid }=2^n\mathrm{\mid }Q\mathrm{\mid }$ by [F1], ${\int }_Q\mathrm{\mid }f\mathrm{\mid }\le {\int }_P\mathrm{\mid }f\mathrm{\mid }\le \lambda \mathrm{\mid }P\mathrm{\mid }=2^n\lambda \mathrm{\mid }Q\mathrm{\mid }$, so the average of every maximal bad cube is at most $2^n\lambda$. For the sum, the maximal bad cubes are pairwise disjoint by step 1.2, so with disjoint additivity and monotonicity of the integral, ${\sum }_Q\lambda \mathrm{\mid }Q\mathrm{\mid }\le {\sum }_Q{\int }_Q\mathrm{\mid }f\mathrm{\mid }={\int }_{\bigcup Q}\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1,$ because each maximal bad cube has average $>\lambda$; hence ${\sum }_Q\mathrm{\mid }Q\mathrm{\mid }\le {\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$, the sums being understood as suprema of finite partial sums over the countable family.

The union of the maximal bad cubes is $\{M_{d}f>\lambda \}$. If $x$ lies in a maximal bad cube $Q$, then [F1] gives $M_{d}f(x)\ge \mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }f\mathrm{\mid }>\lambda$. Conversely, if $M_{d}f(x)>\lambda$, then by definition of the supremum over a nonempty set of real numbers there is a dyadic cube $Q\ni x$ with $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }f\mathrm{\mid }>\lambda$, i.e. $Q$ is bad; step 2.1 provides a maximal bad cube containing $Q$, hence containing $x$.

Steps 1.2 and 2.1 give the countable pairwise disjoint maximal family with the containment property, step 3.1 identifies its union with $\{M_{d}f>\lambda \}$, and step 2.2 gives both the average bound and the sum bound. This proves the lemma.