If $\Omega$ is a bounded open set in $R^n$, $\Omega\_j\subset\Omega$, and $|\Omega\_j|\geq\epsilon$, which $\epsilon$ is a constant. Can we say there is a subsequence $\Omega\_{j\_k}$ of $\Omega\_j$, such that $|\bigcap\_k\Omega\_{j\_k}|>0$ ? We know this is true when $\Omega\_j$ are all balls. So, considering what assumption， this is true?



the intersection of a sequence of measurable sets 
Here is an explicit example that elaborates on Juan's and Gerald's answers. Let $\Omega=[0,1]$.


Let $\Omega\_j$ be the union of $2^{j-1}$ intervals of length $2^{-j}$, with the first interval starting at $0$, and all intervals being spaced distance $2^{-j}$ apart.


That is,
$\Omega\_1 = (0,1/2)$;
$\Omega\_2 = (0,1/4) \cup (1/2,3/4)$, etc.


Then this sequence is a counterexample to your statement, as the intersection of $n$ of these sets has measure exactly $2^{-n}$.


