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the question is:

Let A be a given collection of disjoint measurable subsets of R^d , all of which have positive measure. Show that A is countable.

so I was able to prove that each of those subsets has a compact subset that has positive measure, and I want to prove that I can seperate those compacts subsets with open subset (and this will finish the proof, because it is easy to prove that a collection of disjoint open subsets is countable...). help?

*note: the measure I was speaking about is Lebesgue measure

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    $\begingroup$ If $A$ were uncountable, there would be a positive integer $n$ and an infinite subset $B$ of $A$ such that the measure of each element of $B$ exceeds $1/n$. $\endgroup$ – David Mitra Feb 13 '15 at 14:25
  • $\begingroup$ Sorry. But first show there is a disc $D$ of finite measure, such that uncountably many members of $A$ intersect $D$ in a set of positive measure. $\endgroup$ – David Mitra Feb 13 '15 at 14:56
  • $\begingroup$ Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Martin Sleziak Feb 15 '15 at 8:55
  • $\begingroup$ See also: math.stackexchange.com/questions/975768/… and math.stackexchange.com/questions/115026/… $\endgroup$ – Martin Sleziak Feb 15 '15 at 8:58
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Let $A$ be some index set. Suppose for each $\alpha \in A$, we have a measurable subset $E_{\alpha} \subset \mathbb{R}^d$ such that:

1) $|E_\alpha| > 0$

2) If $\alpha \neq \beta$ then $E_{\alpha} \cap E_{\beta} = \emptyset$.

We want to conclude that $A$ is at most countable. Suppose for the sake of contradiction that $A$ is uncountable. Split $\mathbb{R}^d$ up into cubes of side length one and with vertices on the integer lattice $\mathbb{Z}^d$. By the pigeon hole principle, there's a cube $Q$ such that uncountably many $E_{\alpha}$ meet $Q$ in a set of positive measure. Wlog, we'll assume that $Q = [0,1] \times ... \times [0,1]$.

Now, $1 = |Q| \ge \sup_{A'\subset A: A' countable} \sum_{\alpha \in A'} |Q \cap E_{\alpha}|$

where we use monotonicity of the measure, countable additivity, and the assumption that the sets are pairwise disjoint to obtain the last inequality. What can you say about the last term?

Edit:

I'll add some clarity on two points. First, the pigeon hole portion of the argument is kind of the crux of the whole issue. By refining each $E_{\alpha}$, we may assume that, for any cube $Q$ as above, either $E_{\alpha}\cap Q = \emptyset$ or $|E_{\alpha} \cap Q| >0$. Since every $E_{\alpha}$ had positive measure, its intersection with at least one cube has positive measure, hence we have as many sets as we started with.

Let $\mathcal{Q}$ be the collection of all cubes as above. Set $\mathcal{B} = \{ Q\cap E_{\alpha} : Q \in \mathcal{Q}, \alpha \in A, E_{\alpha} \cap Q \neq \emptyset\}$. Then the cardinality of $\mathcal{B}$ is at least as large as that of $A$. Write $\mathcal{B} = \bigcup_{Q \in \mathcal{Q}} \{E_{\alpha} \cap Q: |E_{\alpha} \cap Q| > 0\}$. If each $Q$ only meets countably many $E_{\alpha}$ in sets of positive measure, then $\{E_\alpha \cap Q: |E_{\alpha} \cap Q| > 0\}$ is countable, for every $Q \in \mathcal{Q}$. But $\mathcal{Q}$ is countable, so $\mathcal{B}$ is countable. So $A$ is at most countable, and we're assuming that it's not. So we must be able to find at least one cube $Q$ meeting uncountably many $E_{\alpha}$.

Now, following the above outline, you'll be done if you can show the following easy proposition:

let $I$ be a set and $x_i \ge 0$ be a real number for every $i \in I$.

If $\sup \{ \sum_{i \in I'} x_i: I' \subset I $ is countable $\}$ is finite then at most countably many $x_i \neq 0$.

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Could a solution simply be that if the collection were contained in a cube of finite side length $1$, there is only a finite number of sets in the collection with measure in $(2^{-n-1}, 2^{-n}]$, $n\in\Bbb N$?

Cut a general collection into a refined collection according to a cube grid, this only increases the number of sets in the collection. Countable union of countable sets gives again a countable set.

(Which should also be the same approach as hinted to in the comment by David Mitra)

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  • $\begingroup$ unfortunately, I don't think they meant to the Borel algebra and your solution is wrong for the Lebesgue algebra because there are sets which have positive measure but don't contain any open subset (but I agree that in the Borel algebra you are right) $\endgroup$ – user213158 Feb 14 '15 at 14:59
  • $\begingroup$ Yes, $[0,1]\setminus \Bbb Q$ has measure $1$ and no interior points. I'll let this stand until there is another, correct answer. $\endgroup$ – LutzL Feb 14 '15 at 16:35

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