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Show that there is a sequence of sets $A_n\subseteq [0,1]$, with outer measure 1 ($\mu^*(A_n)=1$ for every $n$), so that $A_1\supseteq A_2\supseteq A_3\supseteq...$ and $\bigcap_{n=1}^{\infty}A_n=\varnothing$.

From these conditions, the sets mustn't be measurable, so my direction is to use $\mathbb{R}/\mathbb{Q}$ in some way (like the standart construction of a non-measurable set), but I can't figure out how.

Any ideas?

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The family $\mathcal F$ of nonempty open subsets of $[0,1]$ of measure $< 1$ has cardinality $c$, so it can be well-ordered so that each member of $\mathcal F$ has fewer than $c$ predecessors.

Since the complement in $[0,1]$ of any member of $\mathcal F$ has cardinality $c$, we can choose by transfinite induction, for each $U \in \mathcal F$, a countably infinite set $S(U) =\{s_1(U), s_2(U), \ldots \} \subset [0,1] \backslash U$ such that $S(U)$ is disjoint from $S(V)$ for each predecessor $V$ of $U$.

Let $$ A_n = \bigcup_{U \in \mathcal F} \{s_j(U) \; : \; j \ge n\}$$

It has outer measure $1$ because for each $U \in \mathcal F$ it contains points of $[0,1] \backslash U$. By construction, $A_{n+1} \subset A_n$. Each member of $A_1$ is $s_j(U)$ for some unique $U \in \mathcal F$ and some $j$, and thus is not in $A_n$ for $n > j$. That says $\bigcap_n A_n = \emptyset$.

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  • $\begingroup$ I'm sorry but I think this is to advanced for me. But I'll ask anyway: why is the complement of each member of $F$ has cardinality c? Obviously it can't be $\aleph_0$, but why c? thank you! $\endgroup$ – 35T41 Aug 8 '16 at 21:39
  • $\begingroup$ To see that a closed set $C$ of $[0,1]$ with positive measure has cardinality $c$, note that there must be $0 < a < b < 1$ such that $C \cap [0,a]$ and $C \cap [b,1]$ both have positive measure. Repeating this decomposition infinitely many times, we get a Cantor-type subset of $C$, and this has cardinality $c$ (in fact the decomposition induces a one-to-one map from sequences of $0$'s and $1$'s into $C$). $\endgroup$ – Robert Israel Aug 8 '16 at 22:31
  • $\begingroup$ (+1) Great! I like the arguments by transfinite induction. $\endgroup$ – Sungjin Kim Jan 28 '17 at 9:19

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