Find a subset of A such that its boundary does not have measure zero Question

Find a subset $A$ of $[0,1]$ such that $A=cl(intA)$ and yet $bd(A)$ does not have measure $0$.

I don't know how to construct it.
I think it should be closed set, cannot be empty by taking interior, and have many points like $Q\cap [0,1]$ so that $\partial A$ cannot be finite or countable.
 A: Let $C=[0,1]\setminus\bigcup (a_n,b_n)$ be a fat Cantor set, obtained by removing disjoint intervals $(a_n,b_n)$ from $[0,1]$.  Choose $c_n$ and $d_n$ such that $a_n<c_n<d_n<b_n$, and let $A=[0,1]\setminus \bigcup (c_n,d_n)=C\cup\bigcup((a_n,c_n]\cup[d_n,b_n))$.  Then the interior of $A$ is $\bigcup ((a_n,c_n)\cup(d_n,b_n))$, which is dense in $A$ since $\bigcup \{a_n,b_n\}$ is dense in $C$.  But the boundary of $A$ contains $C$, which has positive measure.
A: Sorry, I'm using the app just now and I may not find it easy to format it properly.
Choose an $a\in(0,1)$.
Let $\Bbb Q_1=\Bbb Q\cap(0,1)$. Let $f:\Bbb Q_1\to\Bbb N$ be bijection. Let $(x_n)$ be a positive sequence that adds up to $1-a$ (obviously the limit is $0$).
Wrap each $y\in\Bbb Q_1$ into an interval of a size $x_{f(y)}$ and clip if it goes over 0 or 1:
$$A = (0,1)\cap\bigcup\limits_{y\in\Bbb Q}(y-{x_{f(y)}\over2},y+{x_{f(y)}\over2})$$
For start, $A$ is open so it's its own interior. As it contains all rational numbers from $(0,1)$, its closure is $[0,1]$.
Let's consider $B=[0,1]\setminus A$. Each element has a rational number arbitrarily close. So each element has an element of $A$ arbitrarily close yet it doesn't belong to $A$, therefore it belongs to the boundary. In fact, $B$ is the whole boundary of $A$ because all elements of $A$ are in $A$'s interior. But what's the measure of $B$?
The measure of $A$ is less than $1-a$ because it's a union of (overlapping) intervals whose sizes add up to $1-a$. The measure of $B$ is then 1 minus the measure of $A$, which is more than the chosen number $a$. So not only is the measure of boundary non-zero, it can be arbitrarily close to 1.
