Not that I know of necessarily , other than calculating it . To see how this can be the case , consider the complement of $E_i$ . It has measure 0 .Now the union of the complements (or the complement of $\cap_{i\in \mathcal{I}} E_i$) can be a very general set considering we have uncountable $i$'s . In particular , for $[0,1]$ with the usual topology , it can be an any subset . To see this , let $S$ be a subset of $[0,1]$ , such that the complement of $S$ , $C_S$ is uncountable , and let $b :[0,1] \rightarrow C_S $ be a bijection . Define $E_i$ , $i \in [0,1] $ , to be $[0,1] - \{ b(i) \} $ . Now $\cap_{i\in [0,1] } E_i$ is $S$ . Notice that if $\Omega$ is countable then you can not obtain non-measurable sets by intersecting $E_i$ . The general condition for this to happen is that the union of an uncountable ($R$ , the cardinality of the continuum ) number of the subsets with measure $0$ has measure $0$ .If this doesn't happen , there will be $E_i$'s such that the intersection is a non-measurable set . In general , for a triplet $(\Omega, \mathcal{F},\mathbb{P})$ , if there exists an uncountable subset $J \subseteq F $ , such that the elements of $J$ are pairwise disjoint , $\cup J= \Omega $ and the probability space $(J,M(J) ,\mathbb{P})$ (where $M(J)$ is the set of measurable subsets of the $powerset$ of $J$ )is isomorphic to
$([0,1],M([0,1]) , \mathbb{P} _{ [ 0,1 ] } )$ , then there exist $E_i$ for $(\Omega, \mathcal{F},\mathbb{P})$ such that the union is non-measurable . In general , the condition for non-masurable unions of such sets not to exist for $(\Omega, \mathcal{F},\mathbb{P})$ is that $P(E_s)$ = 1 , where $E_s$ is the set of elementary events with non-zero probability .