First assume that $A$ is compact; then $f$ is uniformly continuous on $A$.
Hence fix an $\epsilon$ and pick a $\epsilon_1$ to be decided later so that for some $\epsilon_2$, we have that any $|x-y| < \delta$ implies that $|f(x)-f(y)|< \epsilon$.
Now, note that the measure of the graph of $f$, denoted by $|\Gamma(f)|$, has bound $|\Gamma(f)| \leq 2 \epsilon |B(0, \delta)| N(\delta)$ Where $N(\delta)$ denotes the number of balls with radius $\delta$ it takes to cover $A$ and $|B(0,\delta)|$ is the measure of the ball of radius $\delta$ in n dimensions.
Recall that $|B(0,\delta)| \leq C \delta^n$. Also, if $A$ has side lengths $l_i$ in dimension $i$, then $N(\delta) \leq C \prod_{i=1}^n \frac{l_i}{\delta}$ (I threw in the constant because I may have been a little sloppy with that bound) Thus $|\Gamma(f)| \leq 2 K \epsilon$ for some constant $K$. But $\epsilon$ was arbitrary, hence the result.
For general $A$, since $|\Gamma(f)| = 0$ on every compact $A_n$, $|\Gamma(f)| = 0$.