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.


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 $A=\mathbb R$.

  • $\begingroup$ 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? $\endgroup$ – Alex2021 Feb 7 at 23:39
  • $\begingroup$ 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. $\endgroup$ – Ruy Feb 7 at 23:49
  • $\begingroup$ 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. $\endgroup$ – Alex2021 Feb 7 at 23:52
  • $\begingroup$ Countable sets are always measurable. $\endgroup$ – Ruy Feb 7 at 23:53
  • $\begingroup$ 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. $\endgroup$ – Alex2021 Feb 7 at 23:56

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