5

How can I construct a set $E$ such that $E$ is dense in $[0,1]\times [0,1]$, and the intersection of $E$ and any line parallel to the axes has at most one point?

I am finding this set in order to construct a counterexample to show that the Tonelli's Theorem is not true for Riemann integral because we can define a function $$f(x,y)=\begin{cases} 1, &(x,y)\in E,\\ 0, &(x,y)\notin E, \end{cases}$$ then $f\notin R([0,1]\times [0,1])$, but we have the following: $$ \int_0^1dy\int_0^1f(x,y) dx=0=\int_0^1dx\int_0^1f(x,y)dy.$$

Apparently, I also want the measure of $E$ to be $0$.

Bach
  • 5,730
  • 2
  • 20
  • 41
  • @DavidMitra Just make sure they are not on the same vertical or horizontal line. (There is always space for us doing so since each little square is open in $\mathbb R^2$.) – Bach Apr 24 '19 at 08:20
  • 1
    Here's the comment I deleted again: Try this: Subdivide $I\times I$ into four squares. In the interior of each square, place a point. Now partition each of the first four squares into four squares. Place appropriate points in these. Continue... – David Mitra Apr 24 '19 at 08:25
  • For fun: find a dense set $E$ such that every vertical or horizontal line that meets $I\times I$ meets $E$ in exactly one point. – David Mitra Apr 24 '19 at 08:28
  • @DavidMitra So basically, find a bijection $[0,1]\to[0,1]$ with dense graph. You can find one by taking the construction of bof (see the answer below), and fill in the uncountably many gaps in a bijective manner. Also see here. – M. Winter Apr 24 '19 at 08:47

1 Answers1

5

Let $D$ be any countable dense subset of $\mathbb R\times\mathbb R$. The set of all lines intersecting $D$ in at least two points is countable, i.e., there is one line for each pair of points. Rotate $D$ so that none of those lines is horizontal or vertical.

bof
  • 78,265