Htykuut

Let $E subset mathbb{R}^n$ be a convex set. I need to prove there exists an open set $A subset mathbb{R}^n$ equivalent to $E$, that means such $mathcal{L}^{n} (A triangle E)= 0$, where $E triangle A = (E setminus A ) cup ( A setminus E) $.

My idea: I know that for all $epsilon >0$ there exist an open set $A supset E$ such that $mathcal{L}^{n} (A setminus E) < epsilon$, but it doesn't help because, in this way, the measure of the symmetric difference $A triangle E$ is less than $epsilon$ but it could be grater than $0$.

My second idea: If I set $A := Int(E)$ then $A$ is open and it holds $mathcal{L}^{n} (A triangle E)= mathcal{L}^{n} (E setminus A) leq mathcal{L}^{n} (partial E)$ (in the case $E$ is closed we have the equality). But in general we do NOT have $mathcal{L}^{n}( partial E) = 0$.

In both attempts I haven't used the fact that $E$ is convext, but I don't know how to use it. Does this fact imply that $mathcal{L}^{n}( partial E) = 0$? If it is true, do you know a simple proof of it?

Thank you

Claim: if $E$ is convex, then $mathcal{L}^n(partial E) = 0$.

Proof: Suppose otherwise. Then there is $x in partial E$ such that $limsup_{r downarrow 0} frac{|partial E cap B_r(x)|}{|B_r(x)|} = 1$. That is, $partial E$ contains most points of small balls around $x$. But by convexity, this means $E$ contains a small ball around $x$. This contradicts $x in partial E$.