A question about measure set 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 $\mu(B)\gt0$.
 A: 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.
A: 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.
