# the intersection of a sequence of measurable sets

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?

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This is well known. If the sets are independents on a probability space there is no such subsequence. –  juan Nov 14 '12 at 14:10
I see no reason to close it, since it is not known to non-specialists. More on juan's comment in an answer. –  Gerald Edgar Nov 14 '12 at 15:14
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. –  Emil Jeřábek Nov 14 '12 at 15:29
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! –  Lasse Rempe-Gillen 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 $2^{j-1}$ 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 $2^{-n}$.

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Sorry for vaguely expression, here $\epsilon$ is a positive constant. –  Weeson Dorne Nov 14 '12 at 16:19
Yes, and in this case $\varepsilon=1/2$ ... –  Lasse Rempe-Gillen Nov 14 '12 at 19:17
I get it. Thank you. –  Weeson Dorne Nov 15 '12 at 4:02

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 $\left|\int_\Omega \chi_B div\;\phi\;dx\right| = \left|\int_{\partial B} \phi \cdot n \;d\mathcal{H}^{N-1}\right| \leq \mathcal{H}^{N-1}(\partial B)$.

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.

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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.

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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 $<\varepsilon$.

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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 sub-subsequence, we can assume $\sum_j d(\Omega_{n_j}, \Omega_0) < \epsilon$, and then $$\mu \left(\bigcap_j \Omega_{n_j} \right) \ge \mu(\Omega_0) - \sum_j d(\Omega_0, \Omega_{n_j}) > 0$$