A graph of any function has content zero?

Question

I know that if $f$ is integrable them his graph have content zero, but if we consider $f : \mathbb{R} \rightarrow \mathbb{R}$ as $f(x) = 1$ if $x \in \mathbb{Q}$ and $f(x) = 0$ if $x \in \mathbb{R} - \mathbb{Q}$ then his graph have content zero and $f$ is not integrable. My question is: exist a function on $\mathbb{R}^{n}$ that his graph did'nt have content zero?

Answer 1

You can define functions that are dense in the plane. Their graphs will not be Lebesgue measurable. In this question, I show one from $(0,1)$ to $(0,1)$. You can stretch both axes with your favorite continuous bijection $(0,1) \leftrightarrow \Bbb R$ to get a graph dense in the plane.

My favorite, though it takes a little cleaning in the corners is this one. Let $x\in (0,1)$ and express it in base $3$. If there are an infinite number of $2$'s in the expansion, set $f(x)=x$ and ignore it. Otherwise, cut off all the leading digits through the last $2$ and read the resulting number in binary to get $f(x)$. Given any $y \in > (0,1)$ expressed in binary and an interval $(a,b)$ we can find $c \in > (a,b)$ expressed in ternary ending in $2$ that we can append $y$ to and stay in $(a,b)$