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?

2$\begingroup$ This is well known. If the sets are independents on a probability space there is no such subsequence. $\endgroup$ – juan Nov 14 '12 at 14:10

1$\begingroup$ I see no reason to close it, since it is not known to nonspecialists. More on juan's comment in an answer. $\endgroup$ – Gerald Edgar Nov 14 '12 at 15:14

1$\begingroup$ As far as I understand it, the question is not whether the property holds for every such sequence $\Omega_j$ (which is false), but under what additional conditions it holds. $\endgroup$ – Emil Jeřábek Nov 14 '12 at 15:29

$\begingroup$ Since the answer is negative even when the sets $\Omega_j$ are open, it is hard to imagine an assumption where the statement holds and which does not imply that each set contains a ball of some fixed diameter. However, perhaps I am wrong! $\endgroup$ – Lasse RempeGillen Nov 14 '12 at 19:21
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 $2^{j1}$ 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 $2^{n}$.

$\begingroup$ Sorry for vaguely expression, here $\epsilon$ is a positive constant. $\endgroup$ – Weeson Dorne Nov 14 '12 at 16:19

$\begingroup$ Yes, and in this case $\varepsilon=1/2$ ... $\endgroup$ – Lasse RempeGillen Nov 14 '12 at 19:17
NO Based on juan's comment.
Suppose $\Omega$ has measure $1$ and the $\Omega_j$ are independent in the sense of probability with measure $\epsilon$. Then the intersection of any $n$ of the sets has measure $\epsilon^n$, and a countable intersection of the sets has measure zero.
A concrete example: let $\Omega$ be the unit interval $(0,1)$ in $\mathbb R$, let $\epsilon = 1/10$. For each $i$, let $\Omega_i$ be all numbers in $(0,1)$ such that the $i$th decimal digit is $7$. This is a finite union of intervals, leave out the endpoints if you want your sets to be open.
As Emil Jeřábek pointed the question must be under what conditions exists the subsequence. We may always normalize and consider $\Omega=1$.
The answer in general is NO. For a more general condition (more general than independence) it can be proved that for any pair of constants $0<\alpha, \beta <1$ with $\beta<\alpha^2$ there is an integer $N=N(\alpha,\beta)$ such that for any sequence $(\Omega_j)_{j=1}^n$ of measurable sets on a probability measure space such that $\Omega_j\ge\alpha$ for all $j$ and $\Omega_j\cap\Omega_k\le\beta$ for all $j\ne k$, we have $n\le N$.
To prove it consider the inequality $\Vert \sum\chi_{A_j}\Vert_1\le \Vert \sum\chi_{A_j}\Vert_2$.
The only positive conditions that cames to mind is. If $\sum(1\Omega_j)<\infty$, then the subsequence exists. Take an N such that $\sum_{j>N}(1\Omega_j)<\varepsilon$. It is easy to show that $\Omega\smallsetminus\bigcap_{j>N}\Omega_j$ has measure $<\varepsilon$.
To generalize somewhat the case of balls:
Let $(\Omega, \Sigma, \mu)$ be a measure space. Let $\cal M \subset \Sigma$ be a family of sets with the property that the symmetric difference $A \Delta B$ has positive $\mu$measure whenever $A$ and $B$ are distinct members of $\cal M$. Then $\cal M$ is a metric space with the metric $d(A,B) = \mu(A \Delta B)$. Suppose further that $\cal M$ is compact. Then any sequence $\Omega_j$ in $\cal M$ with $\mu(\Omega_j) \ge \epsilon$ has a subsequence whose intersection has positive $\mu$measure. For we can take a subsequence $\Omega_{n_j}$ that converges to $\Omega_0 \in \cal M$. Since $\mu(\Omega_0) \ge \mu(\Omega_{n_j})  d(\Omega_{n_j}, \Omega_0)$ we get $\mu(\Omega_0) \ge \epsilon$. By taking a subsubsequence, we can assume $\sum_j d(\Omega_{n_j}, \Omega_0) < \epsilon$, and then $$ \mu \left(\bigcap_j \Omega_{n_j} \right) \ge \mu(\Omega_0)  \sum_j d(\Omega_0, \Omega_{n_j}) > 0$$
Perhaps if you consider the case where $\Omega_j$ are sets of finite perimeter you can recover the statement. A set $\Omega_j$ is of finite perimeter in $\Omega \subset \mathbb{R}^N$ if $\chi_{\Omega_j} \in L^1(\Omega)$ and $Per(\Omega_j,\Omega):=\sup_\phi \left\vert\int_\Omega \chi_{\Omega_j} div\; \phi\;dx \right\vert <\infty$ where the supremum is taken over $\phi \in C_c^1(\Omega;\mathbb{R}^N)$ such that $\phi_\infty \leq 1$.
In particular, balls are sets of finite perimeter, since we may integrate by parts to show that $\left\int_\Omega \chi_B div\;\phi\;dx\right = \left\int_{\partial B} \phi \cdot n \;d\mathcal{H}^{N1}\right \leq \mathcal{H}^{N1}(\partial B)$.
Then a general compactness result for sets of finite perimeter/$BV$ functions implies that if we have a sequence $\Omega_j$ with $\epsilon \leq \Omega_j \leq C$ and $Per(\Omega_j,\Omega) \leq C$ we may find a subsequence whose characteristic functions converge strongly in $L^1(\Omega)$, and using this subsequence it should be possible to choose another subsequence with the desired property (eventually this subsequence will converge to some limit $A$ with $A\geq \epsilon$ from the strong convergence in $L^1(\Omega)$, and therefore choosing $\Omega_{j_k}$ which are near the limit we can find a countable family with positive measure of the intersection.