Signed measure of uncountable set

Question

I have a question and hope some of you can help me :)

Consider a signed measure $\nu$ on $(\Omega, \bf{A})$ and let be $P_i \in \bf{A}$ positive sets, such that $ \forall B \subset P_i: \nu(B) \geq0 $. Then we state that $\bigcup_{i \in I} P_i$ is also a positive set.

If $I$ is countable ( $|I| \leq |\mathbb{N}|$) this seems to be true. To show this I considered an arbitrary $M \in \bigcup_{i \in I} P_i $. I know that there exists $M_i := M \cap P_i \subset P_i$. So we get $\nu(M) = \nu \left( \bigcup_{i \in I}M_i \right) = \sum_{i \in I} \nu(M_i) \geq 0$.

The last $=$ follows due to sigma-additivity of $\nu$ and the $\geq$ follows due to $M_i \subset P_i$ and $P_i$ is positive.

Now I wonder, what happens, if $I$ is uncountable ( $|I| > |\mathbb{N}|$). I think there must be a counterexample, because the sigma-additivity doesn't hold anymore, but I can't find one.

Does anybody have a counterexample for me, or is my assumption wrong, and one can proof, that the statement is still true for overcountable I's.

Thanks a lot!

Answer 1

Let $[0,1]$ be equipped with negative Lebesgue measure.

All singletons are positive sets but the union of all singletons is not.

Answer 2

Consider $(\Bbb R,\text{Lebesgue})$, $I=[0,1]$ and $P_i=\{i\}$ with the measure $\nu(A)=-\int_A e^{-x^2}\sin x\,dx$.