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Let $\lambda( \cdot )$ denote Lebesgue measure on $[0,1]$. Let $(A_n)_{n=1}^\infty$ be a decreasing sequence of Borel subsets of $[0,1]$ such that $\bigcap_{n=1}^\infty A_n = \emptyset$. Given $\epsilon > 0$ does there exist an open set $U \subseteq [0,1]$ such that

(i) $\lambda(A_n \setminus U) = 0$ for some $n$,

(ii) $\lambda(U) \leq \epsilon$, and

(iii) $\lambda(\partial U) = 0$,

where $\partial U$ is the boundary of $U$?

I don't have a good intuition whether this should be true or not. One idea might be to cover some $A_n$ by an open set with measure much smaller than $\epsilon$, then hope that we can expand that open set somewhat to make its boundary null.

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  • $\begingroup$ How did this arise? It looks and quacks a lot like homework. $\endgroup$ – Michael Renardy Mar 31 '15 at 16:48
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    $\begingroup$ If this were true, it would have been an easy way to solve a much harder problem. Ergodic averages of circle rotations converge uniformly for open sets with null boundary. I want to know that every set of small measure is contained in a set of small measure for which ergodic averages converge uniformly. So no, this is not homework. $\endgroup$ – burtonpeterj Mar 31 '15 at 16:55
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No, such a $U$ does not exist in general. You can have (i)+(ii), but (iii) is too much to ask. The problem is that the $A_n$ might all be dense in $[0,1]$ so that $\partial U = \overline{U} \setminus U =[0,1]\setminus U$ has measure $\geq 1-\epsilon$ by (ii).

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  • $\begingroup$ This isn't quite a counterexample, because I only want $\lambda(A_n \setminus U) = 0$, not $A_n \subseteq U$. Perhaps there's some way to tweak it so that it is a counterexample? $\endgroup$ – burtonpeterj Mar 31 '15 at 16:37
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    $\begingroup$ Enumerate the subintervals of $[0,1]$ with rational endpoints as $I_n$. Let $K_n$ for $n\ge 2$ be a sequence of disjoint "fat Cantor sets" (nowhere dense compact sets of positive measure) with $K_n \subset I_n$. Let $A_n = \bigcup_{m \ge n} K_m$. $\endgroup$ – Robert Israel Mar 31 '15 at 16:54
  • $\begingroup$ @burtonpeterj: Yes, I was sloppy there. Nevertheless the argument works, because the $A_n$ could have the property that not only $A_n$ itself is dense but also every $A_n' \subseteq A_n$ with $\lambda(A_n\setminus A_n')=0$ is dense as well (including the set $A_n \cap U$) $\endgroup$ – Johannes Hahn Mar 31 '15 at 18:13

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