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 $|M| > |\mathbb{N}|$, 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?
 A: 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 $\nu(U)<0$, 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 $\nu(S_j)>0$ and $\nu(U)\ge 0$ for all subsets $U$, and we redefine negative to mean $\nu(S_j)<0$ 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 $\nu(S_j)\le 0$.
A: Consider the interval $[-1,1]$ together with the signed measure
$$\nu(U) = \begin{cases}
1 - \lambda(U) & 0 \in U\\
-\lambda(U) & \text{otherwise}
\end{cases}$$
where $\lambda$ is the Lebesgue measure on $[-1,1]$. Now for each $x\in [-1,1]$ define $S_x = \{0,x\}$. Clearly for any $x$ this is a measurable set with measure $\nu(S_x)=1$. However, $\bigcup_{x\in\mathbb R}S_x = [-1,1]$, and $\nu([-1,1])=-1$.
Indeed, you can use any collection of Lebesgue null sets whose union is all of $[-1,1]$ for $S_x$, including such that $|S_x|>|\mathbb N|$ (which I suspect you actually wanted).
A: The answer to the question
'Is it true that an uncountable set of positive sets $(S_j)_{j\in M}$  where $|M|>|N|$, 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 $\sum=\{X +Y: X \in [-2,-1]~\&~Y [1,2]\}$. 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.
