Prove that a set of norm p is of measure zero The question is to show that the set $A = $ {$x \in R^n |   (||x||_p)^p \le 1$} is a Jordan set for $0 \le p \le 1$.
($||x||_p$ is the $p$ -norm).
I know i need to prove that the boundary of $A$ is of measure $0$.
And also the boundary is the set $B = $ {$x \in R^n |   (||x||_p)^p = 1$}.
I started by declaring the function $f(x) = (||x||_p)^p -1$.
So $B=$ {$x \in R^n || f(x)=0$}.
My direction was to prove that the graph of the function is of measure zero
 A: The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $\mu (A) =k>0,$ then since $A\subset [-2,2]^n $ and $$\bigcup_{v\in [0,1] } (vA)\subset [-2,2]^n $$ and $(vA)\cap (uA) =\emptyset $ for $u\neq v$ and $\mu (tA) =t^n\mu (A),$ then we get a contradiction with the fact that $\mu ([-2,2]^n ) =4^n <\infty.$
A: To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:k\in \mathbb N.$ Let $f$ be your function. And let $\epsilon>0.$ 
Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, t\in Q_i\Rightarrow |f(s)-f(t)|<\epsilon.$ Then $G(f)\subseteq \bigcup Q_i\times [m_i,M_i]$ where $m_i=\min_{Q_i} f(x)$ and $M_i=\max_{Q_i} f(x).$ 
Now, the measure of the graph of $f$ on $Q$ satisfies $\mu(G_f)\le \epsilon\cdot \text{vol}\ Q =k^n\epsilon.$ Thus, $\mu(G_f)=0$ on $Q$. To conclude, note that $\mathbb R^n$ is a countable union of such cubes $Q$.
