Uncountable unions of nested sets in a sigma field I know that in a $\sigma$-algebra, uncountable unions may not exist. However suppose I have a directed system $\{A_i,i\in I\}$ where for each $i\in I$, $A_i\in\mathcal A$ (the $\sigma$-algebra) with the property that for any $i,j\in I$, we have $A_i\subseteq A_j$ or $A_j\subseteq A_i$. Then I believe $\cup_{i\in I}A_i\in\mathcal A$ but I am having difficulty in writing a proof. Notice here that $I$ may be uncountable.
 A: Here is a very general counterexample. Suppose $\mathcal A$ is any proper subcollection of the power set $\mathcal P(E)$ and suppose that all finite subsets of $E$ belong to $\mathcal A.$ Choose an element $X\in\mathcal P(E)\setminus\mathcal A$ of minimum cardinality. By well-ordering the infinite set $X$ we can write it as the union of a chain of subsets of smaller cardinality. Thus $X$ is the union of a chain of elements of $\mathcal A,$ but $X\notin\mathcal A.$
A: I think I have an idea counterexample (not complete). Define $x\sim y :\Leftrightarrow x-y\in\mathbb{Q}$, look at $A:=\mathbb{R}/\sim$ and let $X$ be a representative set of $A$ in the Interval $[0,1]$. Note we can choose $X$ in such a way that $|X\cap[0,1-\frac{1}{n}]|\leq n$.$^1$ For $x\in\mathbb{R}$ let $[x]$ denote the class of $x$ in $A$, and for $a\in A$ let $\langle a\rangle$ be the representative of $a$ in $X$. Set $S_x = \{\langle[x]\rangle\}$ for $x\in[0,1]$. Set $$A_x=\left(\bigcup_{0\leq y\leq x} S_y \right)\cap [0,x]$$
for $x\in [0,1]$. Now the $A_x$ are all finite for $0\leq x < 1$ and therefore Borel-measurable, we have $A_x \subseteq A_y$ for $x\leq y$ and
$$X\setminus S_1 \subseteq \bigcup_{x\in[0,1)} A_x \subseteq X$$
(I'm not sure if $\langle [0]\rangle=\langle[1]\rangle$ is contained in the union), so the uncountable union of the $A_x$ is not Borel-measurable.
Since the Borel-measure is a $\sigma$-algebra over the real numbers, this would answer your question negatively.
$^1$ I assume this is possible using the axiom of choice, but I'm not one hundred percent sure. 
EDIT: Oh no, $^1$ can't be. It would imply that the middle term of $$X\setminus S_1 \subseteq \bigcup_{n\in\mathbb{N}} \left(X\cap[0,1-\tfrac{1}{n}]\right) \subseteq X$$ is countable, which can't be. But it is interesting that given a representive set $X$ there is always some $n\in\mathbb{N}$ such that $A_\frac{1}{n}$ is finite but $A_\frac{1}{n+1}$ is uncountable.
