Disjoint system of sets with positive measure is countable 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
 A: 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 $x_i \neq 0$.  
A: I realise the question is old and already has some answers, but I believe my answer will benefit others who see this question:
Proof outline: 1. First show that it is sufficient to consider unit cubes. 2. Then show that in any unit cube, there cannot be uncountably many disjoint sets of positive measure. 
Since 1. has been discussed in detail, I'll clarify a nice way of doing 2.

Claim: if $\mathcal F$ is a family of disjoint subsets of $[0,1]^d$ with positive measure, then $|\mathcal F|$ is at most countable.

Proof: For each $n$, let $\mathcal F_n = \{E\in \mathcal F: m(E) > \frac{1}{n}\}$. Then clearly $\bigcup_{n}\mathcal F_n \subset \mathcal F$. On the other hand, given any $E \in \mathcal F$, the Archimedean property ensures that we can find an $n$ such that $E \in \mathcal F_n$. This establishes that
$$\bigcup_n \mathcal F_n = \mathcal F.$$
But for each $n$, what is the cardinality of $\mathcal F_n$? Since $m([0,1]^d) = 1$, finite additivity of the Lebesgue measure ensures that $|\mathcal F_n| < n$. Thus $\mathcal F$ is a countable union of finite sets, so $\mathcal F$ is at most countable.
A: 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)
