Graph of a continuous function has measure zero 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.
 A: It is sufficient to show that the set $G'=\{(x,f(x)) | x \in [0,1)^n \}$ has measure zero. 
Let $\epsilon>0$, then since $f$ is uniformly continuous on the compact set
$[0,1]^n$, there is some $\delta>0$ such that if $\forall x,x' \in [0,1): \|x-x'\|_\infty < \delta$ (note convenient choice of norm) then 
$|f(x)-f(x')| < \epsilon$.
Now choose $n$ such that ${ 1\over n} < \delta$ and, with $k = (k_1,...,k_n)$, let $x_k = {k \over n}$ and
$R_k = \{x_k\} + [0,{1 \over n})^n$,
where each of the $k_i$ range through $0,...,n-1$. Note that
$\sum_k m R_k = m [0,1)^n = 1$.
Note that
for $x \in R_k$, we have $|f(x)-f(x_k)| < \epsilon$, hence
$\{ (x,f(x)) \}_{x \in R_k} \subset R_k \times [f(x_k)-\epsilon, f(x_k) + \epsilon]$
and so
$m \{ (x,f(x)) \}_{x \in R_k} \le m R_k \cdot m [f(x_k)-\epsilon, f(x_k) + \epsilon]  = 2 \epsilon \, m R_k$.
Hence $m G' \le 2 \epsilon \sum_k  \, m R_k = 2 \epsilon$.
Since $\epsilon>0$ was arbitrary, we have $m G' = 0$.
A: 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, $I\times J\subset G^c$. 
A: I thought of a more simpler solution I think. So assume $\epsilon > 0$ to be given. Choose $0<\epsilon_0<\dfrac{\epsilon}{b-a}$. Now, by uniform continuity,
$$
\forall \epsilon_0 > 0,  \exists \delta(\epsilon_0): \forall x,y \in [a,b], |x-y| < \delta(\epsilon_0) \rightarrow |f(x)-f(y)| < \epsilon_0
$$
so for our chosen $\epsilon_0$, we get a corresponding $\delta(\epsilon_0)$ that satisfies the above.
Now, we want to stack countable number of rectangles on our graph so that they cover the graph and have a total volume less than $\epsilon$. To do this, we would need $n=\dfrac{b-a}{\delta(\epsilon_0)}$ number of rectangles of width $\delta(\epsilon_0)$ arranged edge to edge. Now, we choose the height of each rectangle to be $\epsilon_0$.
Then, all these would cover the entire graph and would have a total volume
$$=n \cdot \epsilon_0 \cdot \delta(\epsilon_0)= \dfrac{b-a}{\delta(\epsilon_0)} \cdot \delta(\epsilon_0) \cdot \epsilon_0 = (b-a)\cdot \epsilon_0 < \epsilon$$
Where the last inequality works by our choice of $\epsilon_0$. This satifies the definiton of measure zero. Hence the graph has measure zero.
