# Elementary Sets and their Measures

1. Aug 30, 2014

### jamilmalik

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
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
$|B_1|+ \ldots + |B_k| = |B'_1|+ \ldots + |B'_{k'}|$.

3. The attempt at a solution
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.

Last edited: Aug 30, 2014
2. Aug 31, 2014

### verty

Does this help? Consider the set $\{x \cap y : x \in X \land y \in Y\}$. (Sorry, I meant intersection here.) Try to push this all the way to success.

I think I can't give more help, unfortunately.

Last edited: Aug 31, 2014
3. Aug 31, 2014

### Fredrik

Staff Emeritus
Please don't post the same thing in two different forums. If you feel that you have posted something in the wrong forum, use the report button to ask the moderators to move it. I understand that you wanted to add the template as well, but you could at least have made a comment about it in the original thread.

This is what I said in the other thread:

It would be easier to answer if you explained your notation and terminology more, and linked to the book. How do you define "elementary set"? (Is your point 1 the definition?) What is a "box"? What does the notation |B| mean?

Wouldn't these two by definition be equal to E?

4. Aug 31, 2014

### jamilmalik

Thanks for the responses. As noted above, I should add some more information regarding the terminology being used. A box in $\mathbb{R}^d$ is a Cartesian product $B:=I_i \times \ldots \times I_d$ of $d$ intervals $I_1, \ldots , I_d$. The volume $|B|$ of such a box $B$ is defined as $|B|:= |I_1|\times \cdots \times |I_d|.$ An elementary set is any subset of $\mathbb{R}^d$ which is the union of a finite number of boxes.

As for the link, here it is below:
http://terrytao.files.wordpress.com/2011/01/measure-book1.pdf

5. Aug 31, 2014

### jamilmalik

What do you mean by an intersection was intended there?

6. Aug 31, 2014

### jamilmalik

Yes, I think it would be equal to $E$. However I am skeptical if it was really that simple. I don't think I actually proved anything, but rather assigned a variable for each collection of unions.

7. Aug 31, 2014

### Fredrik

Staff Emeritus
Right, if you prove that X=Y, you have proved nothing, since they're both equal to E by assumption.

You need to prove that if $\bigcup_i B_i =E=\bigcup_j B'_j$, then $\sum_i|B_i|=\sum_j|B'_j|$.

So the equality $\bigcup_i B_i =\bigcup_j B'_j$ is your starting point, not a result to be proved. I would try to use the fact that for each i and each j, $B_i\cap B'_j$ is a box.