Some questions regarding a famous example of Sierpinski's. As is well known that, assuming the Continuum Hypothesis, there is a set $\mathcal  S$$\subset$$\mathscr R^2$ such that $\mathcal S$ is countable on every horizontal line and co-countable on every vertical line.  Consider also the relative complement of $\mathcal S$, $\mathcal S^{'}$=$\mathscr R^2$$\setminus$$\mathcal S$.  Since it is known that $\mathcal S$ is nonmeasurable, is $\mathcal S^{'}$ nonmeasurable as well?  Also are the following correct:
|$\mathcal S$|=|$\mathcal S^{'}$|=$\mathfrak c$?
Furthermore, for $\mathcal S^{'}$$\subset$$\mathscr R^2$, is $\mathcal S^{'}$ countable on every vertical line and co-countable on every horizontal line?
Apologies in advance for the triviality (and\or stupidity) of these questions (in case any of these assertions turn out be false).  
Thanks in advance for your tolerance.
 A: Since the measurable sets form a $\sigma$-algebra, they are closed under complements, yes. If $X\subseteq\Bbb R^n$ is non-measurable, then $\Bbb R^n\setminus X$ is also non-measurable.
Moreover, every countable set is in fact measurable. Since we assume the continuum hypothesis, it follows that every non-measurable set has size $\aleph_1=2^{\aleph_0}=\frak c$. In particular $\Bbb R^2\setminus S$ and $S$.
A: The set $S$ could be defined so as to include the diagonal or not. To settle that: Say $\le$ is a well-ordering of $\Bbb R$ such that every real has at most countably many predecessors (and note of course that this has nothing to do with the standard order, for example we haven't specified whether $2<3$ or $3<2$), and let $$S=\{(x,y)\,:\,x<y\}.$$ Instead of saying $S'$ is the complement of $S$, let's say $$S'=\{(x,y)\,:\,x>y\}.$$
So the sets $S$, $S'$ and $D$ form a partition of $\Bbb R^2$, where $D$ is the diagonal $x=y$. It's clear that $S\cup S'$ has cardinality $c$. But there is an obvious bijection between $S$ and $S'$, hence $S$ has cardinality $c$.
And yes to your other questions; $S'$ is the same as $S$ reflected in the diagonal (and my $S'$ is the same as your $S'$ modulo uninteresting details).
