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