Construct a Cantor set of positive measure in much the same way as you make the `standard' Cantor set but make sure the lengths of the deleted intervals add up to 1/2, say.
Let $U$ be the union of the intervals that are deleted at the even-numbered steps and let $V$ be the union of the intervals deleted at the odd-numbered steps. The Cantor set is the common boundary of $U$ and $V$; their closures are as required.
Yes, we can do this by a small modification of the original argument, see the linked question. I'll describe the whole argument again here, but if you already read those answers, then the short version is that we remove successively less of each $I_n(p)$ as we approach the boundary of that interval. This allows us (for each $x\in I_n(p)$) to replace $I_n(p)$ by its maximal subinterval centered at $x$ while only slightly disturbing the approximate densities.
Construct open sets $G_1\supset G_2 \supset \ldots \supset E$, as follows. Start with any open $G_1\supset E$, and write $G_1 = \bigcup I_1(p)$ as the union of its components. For each $I_1(p)$, we construct an open set $G_2(p)$ with $E\subset G_2(p)\subset I_1(p)$. This $G_2(p)$ will cover only a small portion of $I_1(p)$ (possible since $|E|=0$), and, moreover, this portion will not be unduly concentrated near the boundary of $I_1(p)$ (this is the modification). More specifically, if $I_1(p)=(a,b)$, then we demand that
$$
|(a,a+h)\cap G_2(p)| \le 2^{-1}h \quad\quad\quad (1)
$$
for all $h>0$, and similarly near $b$. We can do this by cutting $I_1(p)$ into pieces $(a+2^{-k}, a+2^{-k+1})$ and treating these separately (if an endpoint is in $E$ here, I very slightly move such a point).
Then we put $G_2=\bigcup G_2(p)=\bigcup I_2(q)$. We repeat this whole procedure, with $2^{-1}$ replaced by $2^{-2}$ in (1), to obtain $G_3$ etc., and we finally set $F=\bigcup (G_{2n-1}\setminus G_{2n})$ (start with $G_1$, remove $G_2$, add $G_3$ etc. etc.).
If $x\in E$, then $x\in I_n(p_n)$ for some (unique) $p_n$ for all $n$. If $n$ is odd and we write $I_n(p_n)=(a,b)$ and $x$ is closer to $a$ than to $b$, then we consider $J=(a,a+2(x-a))$. Then $J\cap F\supset J\setminus G_{n+1}$, so (1) says that $|J\cap F|/|J|$ is almost $1$ if $n$ is large.
Similarly, we find intervals for which this ratio is almost zero from the even $n$'s.
Best Answer
The answer is no. Take countably many disjoint closed balls $B_i$ contained in the square $Q=[0,1]\times [0,1]$ and such that:
(i) Sum of areas of $B_i$ is less than 1
(ii) Sum of perimeters of $B_i$ is finite
(iii) $\bigcup B_i$ is dense in $Q$
Since the series $\sum \chi_{B_i}$ converges in BV norm, the set $E=Q\setminus \bigcup B_i$ has finite perimeter. It also has positive measure and empty interior. Any representative $F$ of the set $E$ also has empty interior and therefore $\partial F$ is not Lebesgue null.
By the way, any Lebesgue measurable set E has a representative F with the property
(*) $0<|F\cap B(x,r)|<|B(x,r)|$ for all $x\in\partial F$ and all $r>0$.
The proof is straightforward: add the points x for which $|E\cap B(x,r)|=|B(x,r)|$ for some r, and throw out all points x such that $|E\cap B(x,r)|=0$ for some $r$. (See Prop. 3.1 in "Minimal surfaces and functions of bounded variation" by E. Giusti.) By virtue of (*) the set $F$ has the smallest (w.r.t inclusion) topological boundary among all representatives of $E$, so if this representative doesn't help you, nothing does.