Graph of a continuous function has measure zero

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 $m G' = 0$.

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, $I\times J\subset G^c$.