In other occasions, people have asked here how to prove that the graph of a continuous function defined on a box has measure zero. The arguments given where normally:
1) Use the fact that the function must be uniformly continuous to cover the graph with appropriate balls;
2) "Apply Fubini's Theorem".
My issue is with number 2). I know it might be very silly, but how exactly do you apply Fubini's Theorem to prove this result? Could someone provide full details, please?
Thank you.
Let $f : \Bbb{R}^n \to \Bbb{R}$ and define
$$ G := \{(x,f(x)) \mid x \in \Bbb{R}^n\}, $$
the graph of $f$. Observe that $G$ is closed, hence measurable.
By the Fubini-Tonelli theorem, applied to the characteristic function/indicator function $\chi_G$ of $G$, we see
$$ \lambda_{n+1}(G) = \int_{\Bbb{R}^{n+1}} \chi_G (x,y) d(x,y) = \int_{\Bbb{R}^n} \int_\Bbb{R} \chi_G (x,y) \, dy \, dx = \int_{\Bbb{R}^n} 0 \, dx = 0, $$
where the step before the last used that $\chi_G (x,y) \neq 0$ implies $(x,y) \in G$ and hence $y = f(x)$, so that $\chi_G (x,y) = 0$ for all $y \neq f(x)$ and hence in particular for almost all $y \in \Bbb{R}$, so that $\int_\Bbb{R} \chi_G (x,y) dy = 0$.
EDIT: Note that we did use the continuity of $f$ only to conclude that $G$ is indeed measurable. This is the case as soon as $f$ is (Borel/Lebesgue) measurable.
