Lebesgue measurable function in $\mathbb{R}^d$ is almost everywhere pointwise limit of step functions (Stein & Shakarchi, Real Analysis)

Question

Please see this SE post for a full quote of Theorem 4.3, which, in summary, states that a measurable function is an almost everywhere pointwise limit of a sequence of step functions. I have a question about this excerpt of the theorem:

To this end, we recall part (iv) of Theorem 3.4, which states that for every $\epsilon$ there exist cubes $Q_1, \ldots, Q_N$ such that $m(E \Delta \bigcup_{j=1}^N Q_j) \leq \epsilon$. 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 rectangles $R_j$ contained in $\tilde R_j$, and slightly smaller in size, we find a collection of disjoint rectangles that satisfy $m(E \Delta \bigcup_{j=1}^M R_j) \leq 2 \epsilon$.

For context, this idea of "extending the sides of these cubes" comes from the proof of Lemma 1.1 in the same textbook:

(Lemma 1.1) If a rectangle is the almost disjoint union of finitely many other rectangles, say $R = \bigcup_{k=1}^N R_k$, then $|R| = \sum_{k=1}^N |R_k|$.

The relevant part of the proof is:

We consider the grid formed by extending indefinitely the sides of all rectangles $R_1, \ldots, R_N$. This construction yields finitely many rectangles $\tilde R_1, \ldots, \tilde R_M$, and a partition $J_1, \ldots, J_N$ of the integers between 1 and $M$, such that the unions $R = \bigcup_{j=1}^M \tilde R_j$ and $R_k = \bigcup_{j \in J_k} \tilde R_j$, for $k = 1,\ldots, N$, are almost disjoint.

My issue is: While I understand this grid extension idea, it seems a bit overkill. In theorem 4.3, all we care about is that we can take a slightly smaller subset of the almost disjoint union of closed cubes so that we have a strictly disjoint union of closed cubes. But there are much simpler ways to achieve this compared to the rectangular grid extension. For example, because we are starting from an almost disjoint union of cubes, then we can just shrink each side of cube $Q_j$ "towards its midpoint," so that the boundary of every cube no longer intersects.

E.g., in 1-D, $[2, 3]$ and $[3, 4]$ are almost disjoint cubes. Suppose I want to decrease the total volume by $\epsilon = 0.4$. If I scale each side length by $1 - \epsilon/2 = 0.8$ "towards its midpoint" (so, now the cubes are $[2.1, 2.9]$ and $[3.1, 3.9]$), then the total volume is $(1 - \epsilon/2) * 2 = 2 - \epsilon$, which is exactly the desired reduction in volume.

This shrinkage extends to $\mathbb{R}^d$ because we know that cube $Q_j$ has sides defined by $[a^{j}_i, b^{j}_i], i = 1,\ldots, d$, with side length $s_j := b^{j}_i - a^{j}_i > 0$, so there always exists some scaling factor $c_j > 0$ so that the resulting closed cubes are well-defined (i.e., non-empty interior and strictly disjoint).

In summary, why bother with the grid extension idea? We kind of get for free an easy way to get a strictly disjoint union of closed cubes with a precise reduction in volume, since the closed cubes are almost disjoint to begin with. Is my thinking correct, or is the grid extension idea necessary to conclude the proof of Theorem 4.3?

Thanks

Answer 1

There is an important difference between rectangles $R_1,...R_n$ in Lemma 1.1 and cubes $Q_1,...Q_n$ in Theorem 4.3. While $R_1,...R_n$ are said to be almost disjoint, $Q_1,...,Q_n$ could overlap each other and their intersection could have measure greater than zero. So there is possibility that for some $\epsilon > 0$ there is no shrinkage of $Q_1,...,Q_n$ to some $S_1,...S_n$ such that $m((\cup{Q_i}) \Delta (\cup{S_i})) < \epsilon$ and $\cap S_i = \varnothing$.

So, in 1-D, you should keep in mind that Theorem 3.4, for example if $E = [1, 6]$ and $\epsilon = 0.1 $, could give you something like $Q_1=[1, 4]$, $Q_2=[3, 6]$ and shrinking those intervals by $\epsilon$ wouldn't result in disjoint intervals.