Set $E\subset \mathbb{R}^n$ of positive Lebesgue measure such that the Lebesgue measure of its boundary is zero Let $E\subset \mathbb{R}^n$ have positive Lebesgue measure. What are easily interpretable sufficient conditions on $E$ to guarantee that the difference between the closure $\bar{E}$ and the interior $\operatorname{Int}(E)$ has zero Lebesgue measure? 
In particular, I am interested in the following situation $-$ would the above mentioned condition be satisfied if $E$ is compact and convex?  
 A: 
It is true for compact convex sets. Look here math.stackexchange.com/questions/207609/… -- MotylaNogaTomkaMazura

There are a few  natural conditions that are weaker than convexity condition. I   list them below, and comment on whether they yield the desired conclusion
Star-shaped compact sets
For these, the  boundary may be of positive measure.
Let $C\subset \mathbb{R}$ be a fat Cantor set. Consider $\mathbb{R}$ as  the $x$-axis in $\mathbb{R}^2$. Define a compact set $K\subset \mathbb{R}^2$ as the union of all line segments beginning at $(0,1)$ and ending at a point of $C$. This is a set of compact set of positive measure with empty interior; all these properties are inherited from $C$. It is star-shaped with respect to the point $(0,1)$. 
Quasiconvex compact sets
For these, the  boundary may be of positive measure.
A set $K$ is quasiconvex if there is a constant $c$ such that any two points   $a,b\in K$ can be joined by a curve of length at most $c|a-b|$ lying in $K$. Unfortunately, such a set can still have the boundary of positive measure. Consider the following modification of the Sierpiński carpet: at $k$th step of construction, the square is divided into $n_k\times n_k$ equal squares and the middle one is removed. Here $n_k$ is an odd number $\ge 3$; the original construction has $n_k\equiv 3$. 
The generalized Sierpiński carpet has positive area when $\sum n_k^{-2}<\infty$; specifically, the area is $\prod_k(1-n_k^{-2})$. It has empty interior. The quasiconvexity follows from the fact that when going from $a$ to $b$, having to go around a removed square multiplies the traveled path at most by the factor of $2$. 
Sets with a uniform exterior or interior cone condition
For these, the  boundary has measure zero.
I mean the following condition: there exists a circular cone $C$ (radius $r$, height $h$) such that for every point $p\in\partial K$, there is an isometric copy $C_p$ of $C$ that has vertex at $p$ and such that 


*

*$C_p\cap K=\{p\}$, for the exterior cone condition

*$C_p\cap K^c=\{p\}$, for the interior cone condition


Note that every convex set  satisfies the exterior cone condition: one can use any cone $C$ whatsoever, since there is a whole halfspace that lies outside of $K$. 
Either exterior or interior condition implies that the boundary is porous and therefore has zero measure.  
