Constructing a dense subset $ C \subseteq [0,1] \times [0,1] $ with a special property. Construct a subset $ C \subseteq [0,1] \times [0,1] $, dense in $[0,1] \times [0,1] $ that has the property of every horizontal or vertical line has at most one point of $C$.
I think if I guarantee that the set I'm constructing has a point in every quarter of the square, then in every 16-portion of the squarte, then in every 4^n portion of the square it'll work. But my problem is writing this.
Ant the second part of this problem is proving the following: 
Let $C$ be the created subset,
$ \int_{0}^{1} ( \int_{0}^1 \chi (C) (x,y) dx ) dy =  \int_{0}^{1} ( \int_{0}^{1} \chi (C) (x,y) dy ) dx = 0$, but $\chi(C)$ is not integrable.
Can I use Fubini theorem here? The set is 2-measurable, so the first equality should hold, right? But why is it equal to 0?
Thanks!
 A: Start with a countable dense subset $S$ of the plane. Consider the family $\mathcal L$ of all straight lines determined by two points in $S.$ Inasmuch as there are a continuum of possible rotation angles, while the family $\mathcal L$ is countable, we can rotate the plane in such a way that none of the lines in $\mathcal L$ is horizontal or vertical. In other words, we now have a countable dense subset of the plane with no two points lying on the same horizontal or vertical line.
A: Your approach to constructing $C$ will work, but it’s easier to write up a more general construction.
Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be an enumeration of all sets of the form $(p,q)\times(r,s)$ such that $p,q,r,s\in\Bbb Q\cap[0,1]$; every non-empty open subset of $[0,1]\times[0,1]$ contains some $B_n$, so any set that has non-empty intersection with each $B_n$ is dense in $[0,1]\times[0,1]$. Now just construct $C=\{c_n:n\in\Bbb N\}$ recursively. If $n\in\Bbb N$, and you’ve chosen $c_k=\langle x_k,y_k\rangle$ for $k<n$, let $c_n$ be any point of $B_n$ that does not lie on any of the lines $x=x_k$ or $y=y_k$ for $k<n$; clearly this is always possible.
For the second part, note that for any fixed $y\in[0,1]$ there the function $\chi(C)(x,y)$ either is the constant function $0$ or is the characteristic function of a singleton, and in either case
$$\int_0^1\chi(C)(x,y)dx=0\;.$$
The same sort of thing happens in the second iterated integral.
