TITLE: Graph of a continuous function has measure zero
QUESTION [7 upvotes]: I need help to solve the following problem:
Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous function. Prove that the graph $G(f)=\{(x,f(x)):x\in\mathbb{R}^n\}$ has measure zero in $\mathbb{R}^{n+1}$.
I suppose that I have to use that f es uniformly continuous, but I don't know what rectangle which sum of volumes is less than $\varepsilon > 0$ should I take.

REPLY [4 votes]: Here's another argument. Assuming the graph is measurable, use Fubini-Tonelli to show that its measure is equal to an iterated integral:
$$ m(G) = \int_{{\mathbb R}^n} \int_{{\mathbb R}} {\bf 1}_{\{f(x)\}}(y) dy dx = \int_{{\mathbb R}^n} 0 dx =0,$$
where the second equality is due to the fact that the Lebesgue measure of the singleton $\{y:y=f(x)\}$ is zero for any $x$. 
Now for the measurability of $G$. It's a closed set.  Why ? Take $(x,y)$ not in $G$. Then $f(x)\ne y$. Therefore by continuity of $f$, there exists a neighborhood $I$ of  $x$ and a neighborhood of  $J$ of $y$ such that for all $x \in I$, $f(x)\not\in J$. That is, $I\times J\subset G^c$.