Uncountable chain of negligible sets Let $(E,\mathcal{A},m)$ be a (non empty) measure space, with $m$ a complete measure, it's easy to prove that an uncountable union of negligible sets does not need to be negligible or even measurable. But, now, let $(I,<)$ be a totally ordered set ($I\neq \emptyset$ and I not countable) and $(A_i)_{i\in I}$ be an increasing family of negligible sets (for $m$) (ie $\forall i\in I \ \ m(A_i)=0$ and $\forall (i,j)\in I² \ \ i<j \implies A_i \subset A_j$). Is $A=\bigcup_{i\in I} A_i$ measurable and $m(A)=0$?
Thanks.
 A: Let $\omega _1$ be the smallest uncountable ordinal equipped with the measure $\mu $  defined on the $\sigma $-algebra of all subsets by
$\mu (E) = 0$, if $E$ is countable,  and   $\mu (E) = \infty $,  otherwise.
Then the set
$$
  \omega _1 = \bigcup_{x\in \omega _1} [0, x]
  $$
has infinite measure despite $\mu ([0, x])$ being zero for all $x$.

EDIT.
Assuming the continuum hypothesis,  here is an adaptation of the above example for Lebesgue's measure on $\mathbb R$.
Choose any uncountable subset $A\subseteq \mathbb R$, and observe that  the cardinality $|A|$ satisfies
$$
  \aleph_0<|A|\leq 2^{\aleph_0}.
  $$
As we are assuming the continuum hypothesis, we deduce that
$$
  |A|=\aleph_1=|\omega _1|,
  $$
so there exists a bijective mapping
$$
  \varphi :\omega _1\to A.
  $$
With an eye on the terminology introduced in the question we take the totally ordered index set $I$ to be $\omega _1$, and for
every $i$ in $\omega _1$, we put
$$
  A_i = \varphi ([0,i]).
  $$
It is then clear that each $A_i$ is countable and hence  negligible relative to Lebesgue measure.
On the other hand,  we have
$$
  \bigcup_{i\in \omega _1}A_i =
  \bigcup_{i\in \omega _1}\varphi ([0, i]) =
  \varphi \left(\bigcup_{i\in \omega _1}[0, i]\right) =
  \varphi (\omega _1) = A.
  $$
Choosing the apropriate $A$ we may now  answer the last two questions posed by the OP.

Is $A=\bigcup_{i\in I} A_i$ measurable?

No.  Just choose any non-Lebesgue measurable $A$.

Must  $A=\bigcup_{i\in I} A_i$ have Lebesgue measure zero?

No.  Just choose $A=\mathbb R$.
