I am confused on a part of the proof of Theorem 4.3 from Stein and Shakarchi's Real Analysis:
Theorem 4.3 Suppose $f$ is measurable on $\mathbb{R}^d$. Then there exists a sequence of step functions $\{\psi_k\}_{k=1}^{\infty}$ that converges pointwise to $f(x)$ for almost every $x$.
Proof: By the previous theorem, there are simple functions $\{\varphi_k\}$ so that $\lim_{k \to \infty} \varphi_k(x) = f(x)$ for all $x$. To approximate each $\varphi_k$ by a step function, we recall part (iv) of Theorem 3.4, which states that if $E$ is a measurable set of finite measure, then for every $\epsilon > 0$ there exist cubes $Q_1,\ldots,Q_N$ such that $m(E \triangle \bigcup_{j=1}^{N} Q_j) \leq \epsilon.$ By considering the grid formed by extending the sides of these cubes, we see that there exist almost disjoint rectangles $\tilde{R}_1,\ldots,\tilde{R}_M$ such that $\bigcup_{j=1}^{N} Q_j = \bigcup_{j=1}^{M} \tilde{R}_j.$ By taking closed rectangles $R_j$ contained in $\tilde{R}_j$ and slightly smaller in size, we find a collection of disjoint closed rectangles that satisfy $m(E \triangle \bigcup_{j=1}^{M} R_j) \leq 2 \epsilon$. It follows from this observation and the definition of a simple function that for each $k$, there exists a step function $\psi_k$ and a measurable set $F_k$ so that $m(F_k) < 2^{-k}$ and $\varphi_k(x) = \psi_k(x)$ for all $x \notin F_k$.
If we define $F = \bigcap_{\ell = 1}^{\infty} \bigcup_{k > \ell}^{\infty} F_k$, then $m(F) = 0$, because $m(\bigcup_{k > \ell}^{\infty} F_k) \leq \sum_{k \geq \ell} m(F_k) \leq 2^{-\ell}$. For $x \notin F$, there exists $k_0$ so that $x \in \bigcap_{k > k_0} F_k^c$, so for all $k > k_0$ one has $$|f(x) - \psi_k(x)| \leq |f(x) - \varphi_k(x)| + |\varphi_k(x) - \psi_k(x)| = |f(x) - \varphi_k(x)|,$$ and since $\lim_{k \to \infty} \varphi_k(x) = f(x)$ we conclude that $\lim_{k \to \infty} \psi_k(x) = f(x)$ for all $x \notin F$, as desired. $\quad \square$
My confusion: I don't understand the point of constructing a finite union of almost disjoint rectangles from the finite union of cubes. Why is this step necessary? Does it not suffice to just consider the collection of (not necessarily disjoint) cubes?
For completeness, here are the book's definitions of step functions and simple functions:
A step function is a function of the form $$f = \sum_{k = 1}^{N} a_k \chi_{R_k},$$
where $a_1,\ldots,a_N$ are constants (real numbers) and $R_1,\ldots,R_N$ are rectangles (in $\mathbb{R}^d$).
A simple function is a function of the form $$f = \sum_{k=1}^{N} a_k \chi_{E_k}$$ where $a_1,\ldots,a_N$ are constants and $E_1,\ldots,E_N$ are measurable sets with finite measure.
To approximate each $\varphi_k$ with step functions, note that it suffices to approximate $f = \chi_E$, the characteristic function of some set $E$. If the $R_j$ were not disjoint, then $$ f(x) = \sum_{j=1}^M \chi_{R_j}(x) $$ will evaluate to a value more than $1$ for a point in the intersection of two cubes for instance, and this is a poor approximation to $\chi_E$ which takes values only in $\{0,1\}$. The point is that at most one of the $\chi_{R_j}(x)$ in the sum should register the value $1$.