4
$\begingroup$

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.

  1. $E$ can be expressed as the finite union of disjoint boxes.
  2. 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.

$\endgroup$
3
  • $\begingroup$ Ok, because all I did was define $X$ and $Y$ to equal to $E$ by definition, I wasn't really proving anything. What I am trying to prove, if I am not mistaken, is the following: If $\bigcup_{i=1}^{k}B_i = E = \bigcup_{j=1}^{k'}B'_j$, then $\sum_i|B_i|=\sum_j|B'_j|$. I know that for each $i$ and $j$, $B_i \bigcap B'_j$ is a box. But I do not know what step to take next. $\endgroup$
    – Jamil_V
    Sep 1 '14 at 1:35
  • $\begingroup$ Hi Jamil, have you figured out this proof? $\endgroup$ Oct 7 '14 at 0:45
  • $\begingroup$ @TylerHilton, I have not figured out the proof. Moreover, since I was also working on other problems, this one sadly got left behind. $\endgroup$
    – Jamil_V
    Oct 7 '14 at 13:33
4
$\begingroup$

$$\sum_{i = 1}^k m(B_i) = \sum_{i = 1}^k \sum_{j = 1}^k' m(B_i\cap B'_j) = \sum_{j = 1}^k' \sum_{i = 1}^k m(B_i\cap B'_j) = \sum_{i = 1}^k' m(B_j)$$

$\endgroup$
2
  • $\begingroup$ This solution is missing the most important part, the exercise is implicitly stating the "that arise from taking Cartesian products of elements from finite collections of disjoint intervals.", i.e., we need to demonstrate that the assertion is true for the partition of the box. Thus, we need to demontstrate that $$\left| \bigcup_{j=1}^{k'} B_i \cap B'_j \right| = \sum_{j=1}^{k'} B_i \cap B'_j $$ $\endgroup$ Oct 2 '19 at 13:01
  • $\begingroup$ It seems to me that we can't use the $m$ notation before showing that the measure is actually independent of the partition. Because before that, the measure is not well-defined. $\endgroup$ May 12 '21 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.