Is the union of an uncountable ascending chain of measurable sets measurable? Suppose that $I$ is an uncountable totally ordered index set and $(X_i)_{i\in I}$ is an ascending chain of measurable sets in some measurable space $(X,\Sigma)$. My question is: Is the union $\bigcup_{i\in I} X_i$ measurable?
One way to solve this would be to show that for every infinite chain $(A_j)_{j\in J}$ of sets ordered by inclusion, its union $A=\bigcup_{j\in J} A_j$ can be expressed as the union of some countable sub-chain. Is this possible perhaps?
For context: I want to make a Zorn argument, where my partially ordered set is the set of $\Sigma$-measurable sets with the inclusion ordering. 
 A: No, the union of a chain of measurable sets doesn't have to be measurable. I will give a counterexample.
I assume that there is a nonmeasurable set, and that all one-element sets are measurable. And of course I assume the axiom of choice, in the form of the well-ordering theorem.
Let $A$ be a nonmeasurable set, and let $\lt$ be a relation that well-orders $A$. For each $a\in A$, let $A_a=\{x\in A:x\lt a\}$.
Case 1. All of the sets $A_a$ $(a\in A)$ are measurable.
If $A$ had a greatest element, call it $a$, then $A=A_a\cup\{a\}$ would be the union of two measurable sets, and so $A$ would be measurable. Since $A$ is not measurable, $A$ has no greatest element, so $A=\bigcup_{a\in A}A_\alpha$ is the union of an ascending chain of measurable sets.
Case 2. Not all of the sets $A_a$ are measurable.
Consider the least element $b\in A$ such that $A_b$ is nonmeasurable, and apply the argument of Case 1 to the nonmeasurable set $B=A_b$.
P.S. Here is a simpler argument using Zorn's lemma instead of the well-ordering theorem. Let $A$ be a nonmeasurable set, and let $P$ be the set of all measurable subsets of $A$, partially ordered by inclusion. If we assume that the union of a chain of measurable sets is always measurable, then Zorn's lemma applies, and $P$ has a maximal element; that is, there is a maximal measurable subset of $A$. But that's absurd, because we can always add one more point.
A: The P.S in bof's answer also answers the following quite interesting question:
Given a chain of sets indexed by a an uncountable family is the intersection equal to the intersection of some countable sub-family of them?
No. Consider the all subsets of $[0,1]$ of Lebesgue measure $1$. If the answer was yes, then by maximality argument we would reach a contradiction ("we can always [subtract] one more point.)
