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.
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.
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