I am struggling with what seems like a very simple problem from Terrence Tao's Introduction to Measure Theory book (which is available for free online by the way). What I am trying to prove is the following: Give an alternate proof of Lemma 1.1.2(ii) by showing that any two partitions of $E$ into boxes admit a mutual refinement into boxes that arise from taking Cartesian products of elements from finite collections of disjoint intervals.
The referenced Lemma is provided below:
Lemma 1.1.2 (Measure of an elementary set). Let $E \subset \mathbb{R}^d$ be an elementary set.
- $E$ can be expressed as the finite union of disjoint boxes.
- If $E$ is partitioned as the finite union $B_1 \cup \ldots \cup B_k$ of disjoint boxes, then the quantity $m(E):=|B_1|+ \ldots + |B_k|$ is independent of the partition. In other words, given any other partition $B'_1 \cup \ldots \cup B'_{k'}$ of $E$, one has $|B_1|+ \ldots + |B_k| = |B'_1|+ \ldots + |B'_{k'}|$.
The proof that is provided in the text uses a discretization argument that I do not understand, but the problem at hand is to show the same result holds regardless of the partition used. My approach was to let $X=B_1 \cup \ldots \cup B_k$ and $Y=B'_1 \cup \ldots \cup B'_{k'}$ and then show that $X=Y$. I can rewrite both of these sets as $X=\bigcup\limits _{i=1}^kB_i$ and $Y=\bigcup\limits_{j=1}^{k'}B'_j$, but then I am confused on how to proceed to show their measures are equivalent. The problem states to use Cartesian products, but I notice that my attempt does not seem to use it which is why I am starting to think that I am on the wrong path. Any assistance, suggestions, and/or advice on this would be greatly appreciated. Many thanks in advance.
$$\sum_{i = 1}^k m(B_i) = \sum_{i = 1}^k \sum_{j = 1}^k' m(B_i\cap B'_j) = \sum_{j = 1}^k' \sum_{i = 1}^k m(B_i\cap B'_j) = \sum_{i = 1}^k' m(B_j)$$