In proving the existence of a countable collection of disjoint dyadic cubes which we denote $B_{j}$ with $f \geq 0, f \in L^{1}(\mathbb{R}^{n}), \lambda > 0$ and we define $E_{t}f:= \sum_{B \in \{\text{dyadic cubes}\}}(\frac{1}{\mu(B)}\int_{B}f)\chi_{B}$ where $\chi$ denotes the indicator function. satisfying:
- $ f \leq \lambda a.e.$ on $(\cup B_{j})^{c}$
- $\mu(\cup B_{j}) \leq \frac{1}{\lambda}||f||_{L^{1}}$ 3.$\lambda < \frac{1}{\mu(B_{j}} \int_{b_{j}}f \leq 2^{n} \lambda. \forall j$
One defines the set $\Omega_{t}:= \{x \in \mathbb{R}^{n}: E_{t}f(x) > \lambda, E_{j}f(x) \leq \lambda, \forall j < t\}$ i.e. the larget dyadic cube whose average is geater than $ \lambda$.
Question: The $\Omega_{t}$ is a disjoint union of elements of $\{\text{dyadic cubes}\}$. What does this mean? I'm only familiar with disjoint union in the sense given in the following link: https://en.wikipedia.org/wiki/Disjoint_union
There is something wrong in your definition of $E_t f$. It does not depend on $t$... I am copying below a proof of Calderon-Zygmund theorem.
Theorem Calderon-Zygmund Let $f\in L^{1}\left(\mathbb{R}^{N}\right) $ be a nonnegative function, and let $\lambda>0$. Then there exists a countable family $\left\{ Q_{n}\right\} $ of open mutually disjoint cubes such that \begin{equation} f\left( x\right) \leq \lambda \quad \text{ for }\mathcal{L}^{N}\text{ a.e. }% x\in\mathbb{R}^{N}\setminus\bigcup_{n=1}^{\infty}\overline{Q_{n}}, \end{equation} and for every $n\in\mathbb{N}$, \begin{equation} \lambda<\frac{1}{\left\vert Q_{n}\right\vert }\int_{Q_{n}}f\left( x\right) \,dx\leq2^{N}\lambda\text{.} \end{equation} Proof: Choose $L>0$ so large enough $$ \int_{\mathbb{R}^{N}}f\left( x\right) \,dx\leq \lambda L^{N}\text{.}% $$ Decompose $\mathbb{R}^{N}$ into a rectangular grid such that each cube $Q$ of the partition has side length $L$. Since $f\ge 0$ we have \begin{equation} \frac{1}{\left| Q\right| }\int_{Q}f\left( x\right) \,dx=\frac{1}{L^N }\int_{Q}f\left( x\right) \,dx\le \frac{1}{L^N } \int_{\mathbb{R}^{N}}f\left( x\right) \,dx \leq \lambda . \end{equation} Fix one such cube $Q$ and subdivide it into $2^{N}$ congruent subcubes. Let $Q^{\prime}$ be one of these subcubes. If $$ \frac{1}{\left| Q^{\prime}\right| }\int_{Q^{\prime}}f\left( x\right) \,dx>\lambda, $$ and in view of the fact that $$ \frac{1}{\left| Q^{\prime}\right| }\int_{Q^{\prime}}f\left( x\right) \,dx\leq\frac{2^{N}}{\left| Q\right| }\int_{Q}f\left( x\right) \,dx\leq2^{N}\lambda, $$ then $Q^{\prime}$ will be selected as one of the good cubes $Q_{n}$. On the other hand, if $$ \frac{1}{\left| Q^{\prime}\right| }\int_{Q^{\prime}}f\left( x\right) \,dx\leq \lambda $$ (note that since $\int_{\mathbb{R}^{N}}f\left( x\right) \,dx\leq \lambda L^{N}$ there is at least one), then we subdivide $Q^{\prime}$ into $2^{N}$ congruent subcubes and we repeat the process.
In this way we construct a family of cubes $\left\{ Q_{n}\right\} $ for which \begin{equation} \lambda<\frac{1}{\left\vert Q_{n}\right\vert }\int_{Q_{n}}f\left( x\right) \,dx\leq2^{N}\lambda\text{.} \end{equation} is satisfied, and it remains to prove $f\left( x\right) \leq \lambda$ for $\mathcal{L}^{N}$ a.e. $ x\in\mathbb{R}^{N}\setminus\bigcup_{n=1}^{\infty}\overline{Q_{n}}$. It can be seen from the construction that the cubes that were not selected to belong to the family $\left\{ Q_{n}\right\} $ form a fine covering $\mathcal{F}$ of $\mathbb{R}^{N}\setminus\bigcup_{n=1}^{\infty}\overline {Q_{n}}$. Therefore if $x\in\mathbb{R}^{N}\setminus\bigcup_{n=1}^{\infty }\overline{Q_{n}}$ is a Lebesgue point for $f$, then by Lebesgue differentiation theorem we have $$ f\left( x\right) =\limsup_{\operatorname*{diam}F\rightarrow0\text{,}% \,F\in\mathcal{F}\text{, }x\in F}\frac{1}{\left| F\right| }\int_{F}f\left( y\right) \,dy\leq \lambda $$ and the proof is completed.