Set where the measure theoretic interior and the interior aren't the same

Question

$\DeclareMathOperator{\varint}{int}$I'm studying real analysis from my professor's notes and I'm having some trouble thinking of sets where the measure theoretic interior and the interior aren't the same.

The measure theoretic interior is defined as such: $$\varint_*(E)= \{ x \in \Bbb R^N \mid \Delta_x(E) =1 \} $$ where $$\Delta_x(E) = \lim_{r \to 0+}\frac{m^*(E\cap B(x,r))}{m(B(x,r))}$$ is the density of the set $E$ in the point $x$.

If we consider the interior of a set with the topological definition it seems to me that that $\varint(E)\subset \varint_*(E)$, and so the difference must lie in the boundary of the set, where in some points of the boundary the set must have density equal to $1$ for the two to differ.

I tried thinking of such a set but I ended up nowhere. So I ask you, how would such a set look like?

Answer 1

Consider the standard topology on the real line, and take $A := (-1,0) \cup (0,1)$. Since $A$ is open we have $A = \text{int}(A)$. But $$ Δ_0(A) = \lim_{r \to 0+} \frac{m^*(B(0,r) \cap I)}{m(B(0,r)} = \lim_{r \to 0+} \frac{m(B(0,r) \setminus \{0\})}{m(B(0,r)} = 1. $$ So we have $$ (-1,0) \cup (0,1) = \text{int}(A) \subset \text{int}_*(A) = (-1, 1) $$ as you wished.

It's a quite lame example, since $1 \notin A$. To fix this pause and think of a similar example, but in $ℝ^2$. If you tried, let's check if we came up with the same thing (feel free to post your example in the comments):

Take standard topology on $ℝ^2$. Let $C$ be this guy

Formally $$C := (0,1)^2 \cup (-1,0) \times (0,1) \cup \{(0,0)\}.$$

Now we have

$$ C \setminus \{(0,0)\}= \text{int}(C) \subset \text{int}_*(C) = (-1,1) \times (0,1). $$