Suppose that a sequence of sets $\{A_n:n\in \Bbb N\}$ is increasing, and $A=\bigcup_{n=1}^\infty A_n$. If $A$ is measurable, $\mu(A)\gt 0$ and $\mu$ is an atomless measure, do there exist an $n\in \Bbb N$ and a measurable set $B$ such that $B\subset A_n$ and $\mu(B)\gt0$.
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Welcome to math.SE. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Michael Albanese Jul 13 '15 at 03:57
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what does an increasing sequence of sets mean? Like $A_i \subseteq A_{i+1}$? – user217285 Jul 13 '15 at 04:00
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Yes, like Ai⊆Ai+1. – Xiang Lin Jul 13 '15 at 04:14
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1@XiangLin: I think that the previous answer using Vitali sets would have worked. Even though Vitali sets do not have outer measure zero, I think that it is still true that they do not contain any sets of positive Leb measure. And similarly for any finite union of rational translates of such sets. I could be wrong though... – shalop Jul 13 '15 at 08:04
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However, it is well known that the union of two non-measurable subsets may be measurable. Do you give me a proof that A_n is non-measurable in your answer? – Xiang Lin Jul 13 '15 at 08:47
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If you accept the axiom of choice, then the construction of Vitali gives a counter example.
Namely, consider the unit interval $X:=[0,1)$ with the Lebesgue measure $\mu$. For $a,b\in X$, let us write $a\sim b$ if $a-b$ is rational. Then $\sim$ is an equivalence relation and partitions $X$ into disjoint parts. Choose a set $C\subseteq X$ that contains exactly one element from each part of this partition (using the axiom of choice). Now the cosets $q+C \pmod{1}$ (for $q\in X$ rational) are disjoint and form another partition of $X$. This new partition has a countable number of parts that are translations of one another.
Let $q_1,q_2,\ldots$ be an enumeration of the rationals in $X$ and set $C_i:=q_i+C$. Let $A_n:=\bigcup_{i=1}^n C_i$. So, $\{A_n\}$ is increasing and $A:=\bigcup_{n=1}^\infty A_n=X$, hence $\mu(A)=1>0$. Now, suppose that $B\subseteq A_n$ is a measurable set with $\mu(B)>0$. Let $B_k:=q_k+B \pmod{1}$ be the translations of $B$ by rationals. By translation-invariance of the Lebesgue measure, all the sets $B_k$ must have the same positive measure $c:=\mu(B)>0$. This gives $\sum_{k=1}^\infty \mu(B_k)=\infty$. On the other hand, each element of $X$ appears in at most $n$ elements of the sequence $\{B_k\}$, which implies $\sum_{k=1}^\infty \mu(B_k)\leq n\mu(X)=n<\infty$. We have a contradiction, which means no such set $B$ can exist.
Blackbird
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I think this should work, if the $A_i$ are all measurable. You can find an equivalent collection $A_n'$ so that the $A_n'$ are pairwise disjoint and $\cup A_n' =\cup A_n$. Then $m(\cup A_n')>0= \Sigma m(A_n')>0; m(A_i)\geq 0 \forall i$. Then some $A_n'$ mst have nonzero measure.
Gary.
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Ignorance reigns in MSE. Common, say it with me downvoter: you're ignorant and you're proud. – Gary. Jul 13 '15 at 06:22
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There is one issue with your answer: In the problem statement, the $A_i$ are not assumed to be measurable (only their union is). If the $A_i$ were all assumed to be measurable, then the problem would be easy because we would have that $\mu(A) = \lim_i \mu(A_i)$. – shalop Jul 13 '15 at 07:45
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Thank you Gray. You say is not problem, but it is not my answer in my question. – Xiang Lin Jul 13 '15 at 08:49
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@Shalop: Still, eas or not, a valid answer under the restrictions I assumed. Not a full answer, but not wrong either. – Gary. Jul 13 '15 at 12:29