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

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 $\mu ([-2,2]^n ) =4^n <\infty.$
