Union of Uncountablely many subsets of a measurable set

Question

I am going through measure theory in self study mode.In lebesgue measure I come across a problem that asking me to Prove or disprove the following:

union of uncountablely many measurable subsets of [a,b] is again a measurable set.

• Now I am guessing that it is not true. I want to give the counterexample in reference to Vitali's construction for Non-measurable sets.
• Vitali's construction is given in

https://en.m.wikipedia.org/wiki/Vitali_set

• Now my argument is that Every Vitali set is constructed by the help of the "axioms of choice", and I saw each element as a subset of the mother closed interval say for example $[0,1]$.

As Vitali's set is uncountable so taking the unions of the singleton element will also give uncountable union which the original Vitali's set ,a subset of the mother closed interval $[0,1]$. As Vitali's set is non-measurable, so I am done.

Please give the desired counterexample or the proof. Thanks in advance and sorry for the long description.

Answer 1

This looks like a correct proof. In general, given any non-measurable subset $S \subseteq [a,b]$, we can write $$ S = \bigcup_{x \in S} \{x\} $$ which is a union of an uncountable number of measurable subsets. You may want to verify that

1. The singletons $\{x\}$ are measurable.
2. There are uncountably many singletons needed to get all of $S$ (or equivalently, $S$ is necessarily a union of uncountably many singletons).

By assumption $S$ was non-measurable and so we have exhibited a counterexample ($S$) to the claim that "every union of every uncountably many measurable subsets of $[a,b]$ is measurable".