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I am working on something which needs to understand covering lemmas and Calderón–Zygmund decomposition. These type of lemmas are as in the following link

http://en.wikipedia.org/wiki/Calder%C3%B3n%E2%80%93Zygmund_lemma

or in some books. Let me recall their here:

Lemma 1: Let $f: \mathbf{R}^{d} \to \mathbf{C}$ be integrable and $\alpha$ be a positive constant. Then there exist sets $F$ and $Ω$ such that: 1) $\mathbf{R}^d = F \cup \Omega$ with $F\cap \Omega = \varnothing$;

2) $|f(x)| \leq \alpha$ almost everywhere in $F$;

3) $\Omega$ is a union of cubes, $\Omega = \cup_k Q_k$, whose interiors are mutually disjoint, and so that for each $Q_k$, $\alpha < \frac{1}{m(Q_k)} \int_{Q_k} f(x)\, dx \leq 2^d \alpha.$

As I read in its proof, there exists such decompostions, cubes $Q_k$ only in case when the set $\{x\in\mathbb R^d:\;Mf(x)>\alpha\}$ is not an emptyset. If $\{x\in\mathbb R^d:\;Mf(x)>\alpha\}$ is an emptyset, of cause we could not find such $Q_j$. So how should I understand this lemma in this case ?. Does we need to give the addition hypothesis "{x\in\mathbb R^d:\;Mf(x)>\alpha} is not an emptyset"?. In almost books which I know, the athours never disccuss on this case.

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$|f(x)| \leq Mf(x)$ a.e. by the Lebesgue differentiation theorem. So if $\{x \in {\mathbb R}^d: Mf(x) > \alpha\} = \emptyset$, then $|f(x)| \leq Mf(x) \leq \alpha$ for a.e. $x$. So $F$ is just ${\mathbb R}^d$ here.

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    you consider the union of cubes to be a union over no sets, I suppose. Imagine if $0 \leq f(x) \leq {\alpha \over 2}$ for all $x$. You won't get 3) to hold no matter what. So in these situations F is everything2011-12-19