Uncountable union of positive sets being negative Let $(X,\Sigma,\nu)$ be a signed measure space. A set $S \subseteq X$ is positive/negative  if for all subset $U \subseteq S, U \in \Sigma \rightarrow \nu(U) \geq 0$/$\nu(U) \leq 0$.

Is it true that an uncountable set of positive sets $(S_j)_{j \in M}$, where $|M| &gt; |\mathbb{N}|$, may have a union that is negative?

I know that this is not possible when $M$ is countable, but is it possible to exhibit an example?
In search of closure, here is an answer.  Yes, as the original poster has defined the problem, an uncountable set of positive sets $(S_j)_{j∈M}$ may have a union that is negative.  That is because a positive set $S_j$ is defined by the original poster as one with $\nu(U)\ge 0$ for all $U\subset S_j$, and thus a set $S_j$ with $\nu(U)=0$ for all $U\subset S_j$ qualifies as positive (and also as negative).

As noted a few days ago, an example is where $[0,1]$ is endowed with the negative of Lebesgue measure, and the positive sets are $\{x\}$ for $x\in [0,1]$. Not only is $\nu([0,1])\le 0$, as required by the original poster's definition of negative, but actually $\nu([0,1])=-1$.  Likewise for many subsets $U\subset [0,1]$ we have $\nu(U)&lt;0$, while for others we have $\nu(U)=0$, for example the singletons $\{x\}$ and also, say, $\mathbb{Q}\cap [0,1]$. 

On the other hand, suppose we redefine positive to mean that $\nu(S_j)&gt;0$ and $\nu(U)\ge 0$ for all subsets $U$, and we redefine negative to mean $\nu(S_j)&lt;0$ and $\nu(U)\le 0$ for all subsets $U$.  It's pretty obvious that we can't have an uncountable union of positives that's negative as described, simply because $S_j\subset \bigcup S_k$ and so, if $\bigcup S_k$ is negative as defined, we must have $\nu(S_j)\le 0$.
The answer to the question
'Is it true that an uncountable set of positive sets $(S_j)_{j\in M}$  where $|M|&gt;|N|$, may have a union that is strictly negative?'' 

is  no: Let $X=[-2,-1]\cup [1,2]$. Let $\nu_1$ be a counting measure in $[-2,-1]$ and let $\nu_2$ be - counting measure  in $[1,2]$.  We put $\sum=\{X +Y: X \in [-2,-1]~\&amp;~Y [1,2]\}$. We put $\nu(X +Y)=\nu_1(X)+\nu_2(Y)$. The $\nu$ is signed measure. If $S_j$ is positive set then we have that $S_j \subset [-2,-1]$. Hence $\cup_{j \in M}S_j \subseteq [-2,-1]$. Since  $\cup_{j \in M}S_j$ does not contain any element of $[1,2]$ we  claim that it is not negative.