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 $x \in \partial E$.