This theorem states that, given ANY $f\in L^1(\Bbb R^n)$ and ANY $\alpha>0$, there exists a sequence of (mutual disjoint open with sides parallel to the axis) cubes $\{Q_k\}_{k\ge1}$ such that $$ \alpha<\frac1{|Q_k|}\int_{Q_k}|f|\le2^n\alpha\;\;\;\forall k\ge1 $$ (the thesis of the theorem says other more things but this is the one I'm interested in).
Studying the proof I noticed a point in which we deal with certain cubes $Q$ which have the property of $$ \alpha<\frac1{|Q|}\int_{Q}|f| $$ WITHOUT justifying their existence. So I tried to do it by myself, and observing that if $x$ is a point of $\Bbb R^n$ which belongs to an infinite strictly decreasing family of cubes $\{P_k\}_k$, there is a Theorem by Lebesgue which ensure that $$ \lim_k\frac1{|P_k|}\int_{P_k}|f|=|f(x)| $$ and from this I deduced "maybe an hypothesis of boundedness is missing". In fact if $f$ is bounded, there exists $B>0$ s.t. $|f|\le B$ a.e. thus $$ \alpha<\frac1{|Q|}\int_{Q}|f|\le \frac1{|Q|}\int_{Q}B=B $$ and this is true for every cube $Q$, thus taking $\alpha=2B$ we reach a contradiction.
Where is my mistake?
As I understand it, there is a case in the Calderón-Zygmund lemma in which the set of cubes $Q_k$ is empty because $f(x)\le\alpha$ a.e. The existence of a cube $Q$ meeting the inequality $\alpha<\frac1{|Q|}\int_{Q}|f|$ is presumably obtained assuming a set of measure $>0$ in which $f(x)>\alpha$. Your contradiction arises from not taking into account the possibility that the set of cubes is empty. (I unfortunately know only the version of the lemma from Hörmander v. 1, 4.5.5, which is sort of different from what you've stated.)