A chain of measurable sets has union measurable

Question

Consider a measurable space $(\Omega, \mathcal{F})$ and let $I$ be an arbitrary index set.

Is the following true?

If $\left( A_i \right)_{i \in I}$ is a chain in $\mathcal{F}$ – that is, $\forall i \in I$, $A_i \in \mathcal{F}$ and for all $i, j \in I$, we have $A_i \subseteq A_j$ or $A_j \subseteq A_i$ – then $$\displaystyle \bigcup_{i \in I} A_i \in \mathcal{F}.$$

Answer 1

No. Consider Lebesgue measure on the real line. Let $\kappa$ be the minimum cardinality of a non-measurable set, and let $A$ be a non-measurable set of cardinality $\kappa.$ Then $A$ is the union of a chain of sets of cardinality less than $\kappa,$ which are of course measurable sets.

Answer 2

This isn't true in general (assuming the axiom of choice).

Let $\kappa$ be the least cardinal of a non-measurable set of reals; let $f$ map $\kappa$ 1-1 onto such a non-measurable set.

Then $\{f\!"\!\alpha \mid \alpha\lt\kappa\}$ (where $f\!"\!\alpha$ denotes the range of $f$ on $\alpha)$ is a chain of measurable sets whose union is not measurable.

To see that the axiom of choice needs to be used in some way, at least for the case of Lebesgue measure on the reals, note that it's consistent with ZF that every set of reals is Lebesgue measurable.

Answer 3

Within the known axioms of set theory you cannot disprove that $2^{\aleph_0} = \aleph_1.$ Recall that $\aleph_1$ is defined as the cardinality of the set of all countable ordinals and $2^{\aleph_0}$ is the cardinality of $[0,1]$.

Let $A$ be any non-measurable subset of $[0,1]$. Suppose $|[0,1]| = 2^{\aleph_0} = \aleph_1$. Since $A$ must be uncountable and since no cardinality lies between $\aleph_0$ and $\aleph_1$ (that much is provable within conventional set theory), we have $|A|=\aleph_1$, so $$ A = \{ a_i : i \text{ is a countable ordinal} \}, $$ for some indexing $i\mapsto a_i$. Then let $$ A_i = \{ a_j : j \le i\}. $$ Then for each $i$, the set $A_i$ is countable, hence measurable, but $$ \bigcup_i A_i = A $$ is not measurable.

Since you can't disprove that $2^{\aleph_0} = \aleph_1,$ you can't prove that your proposed union must be measurable.