# 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 $$\forall (i,j)\in I² \ \ i). Is $$A=\bigcup_{i\in I} A_i$$ measurable and $$m(A)=0$$? Thanks.

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 $$A=\mathbb R$$.

• Well done, thanks a lot. I was so focused on Lebesgue measure that I did not think at other types of examples. Just an extra question: what about my question for Lebesgue measures? – Alex2021 Feb 7 at 23:39
• I guess Lebesgue measure also presents this phenomenon. Assuming the continuum hypothesis, we may identify $\mathbb R$ with $\omega_1$, so $\mathbb R$ itself may be writen as an increasing union of countable sets. – Ruy Feb 7 at 23:49
• But in this case, I'm not sure we can keep measurability: a well-order on $\mathbb{R}$ is not really compatible with Lebesgue measure. – Alex2021 Feb 7 at 23:52
• Countable sets are always measurable. – Ruy Feb 7 at 23:53
• Yes of course, but in this case, I don't see how to use your example because the measure you introduced is very different from Lebesgue measure. – Alex2021 Feb 7 at 23:56