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Is there a family of measurable sets $\{E_\alpha\}_{\alpha \in A}$ , disjoint, uncountable and each one with positive measure?

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Yes. Take $\mathbb{R}$ with the counting measure. Then the measure of each point (singleton) will be $1$, and they are obviously disjoint and there is $2^{\aleph_0}$ of them. (Does this answer your question or do you want to add some condition?) – Dejan Govc Feb 29 '12 at 22:57
I am thinking about the Lebesgue measure. – checkmath Feb 29 '12 at 23:02
You should mention that in the question! – Dylan Moreland Feb 29 '12 at 23:06
I'm a little late in getting to this question (I've been off the internet the past several days), but for possible future interest in this question I'll point out that given an uncountable family $C$ of Lebesgue measurable subsets of $\mathbb R$, each with positive measure, then there exists an uncountable subcollection $D$ of $C$ such that each pair of sets belonging to $D$ has a positive measure intersection. This follows from results proved in: Joseph E. Gillis, Some combinatorial properties of measurable sets, Quarterly Journal of Mathematics (Oxford) 7 (1936), 191-198. continued – Dave L. Renfro Mar 5 '12 at 17:09
continued On the other hand, the result I gave fails very badly for non-Lebesgue measurable sets. In fact, Lusin and Sierpinski proved in 1917 that there exists continuum many pairwise disjoint subsets of the interval $[0,1]$ such that each of these subsets has outer Lebesgue measure $1.$ See: Nikolai N. Lusin and Waclaw Sierpinski, Sur une décomposition d'un intervalle en une infnité non dénombrable d'ensembles non mesurables [On a decomposition of an interval into a nondenumerably many nonmeasurable sets], Comptes Rendus Académie des Sciences (Paris) 165 (1917), 422-424. – Dave L. Renfro Mar 5 '12 at 17:13
up vote 14 down vote accepted

It depends on your measure space. For instance, if you take as your measure space $\mathbb{R}$ with counting measure, then the family of singletons works.

However, for a $\sigma$-finite measure space (in particular, for Lebesgue measure), the answer is no. Suppose your measure space is $(X,\mu)$ and you can write $X = \bigcup_{n = 1}^\infty X_n$ where $\mu(X_n) < \infty$. Let $\mathcal{E}$ be any family of pairwise disjoint sets, all of which have positive measure. Let $$\mathcal{E}_{n,k} = \{ E \in \mathcal{E} : \mu(E \cap X_n) \ge 1/k\}.$$ Then every $E \in \mathcal{E}$ is in at least one $\mathcal{E}_{n,k}$, so $\bigcup_{n,k=1}^\infty \mathcal{E}_{n,k} = \mathcal{E}$. Now suppose $E_1, \dots, E_m$ are distinct sets from $\mathcal{E}_{n,k}$. Since these sets are disjoint, we have $$\frac{m}{k} \le \sum_{i=1}^m \mu(E_i \cap X_n) \le \mu(X_n)$$ so $m \le k \mu(X_n)$. Thus the cardinality of $\mathcal{E}_{n,k}$ is at most $k \mu(X_n)$; in particular it is finite. So we have shown that $\mathcal{E}$ is a countable union of finite sets, hence countable.

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This is impossible for Lebesgue measure. For simplicity suppose we're on the interval $[0,1]$. One can find countably many of the sets whose union has measure 1, in which case the complement has measure zero.

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