Let $E \subset \mathbb{R}$ Lebesgue measurable, finite and let $\epsilon>0$. Show that there exists compact $K\subset E$ s.t. $m(E-K)<\epsilon$, and $K^0= \emptyset $.
For K with not necessary entry interior I know it is a classic result, but with entry interior?
Hint: $E=F\cup \bigcup_m (a_n,b_n)$ where $F=E\setminus E^{0}$ and $(a_n,b_n)$ are disjoint open intervals We can find a fat Cantor set $C_n \subset (a_n,b_n)$ such that the measure of $(a_n,b_n)\setminus C_n$ is less than $\frac {\epsilon} {2^{n}}$. Take $K=\bigcup_n C_n$. Then $m(E\setminus K) \leq \epsilon$. Let $K_0$ be a compact subset of $F$ with $m(F\setminus K_o) <\epsilon$. Note that $K_0$ has empty interior. Now show that $K=\bigcup\limits_{n=1}^{N} C_n \cup K_0$ does the job (with $3 \epsilon$ in place of $\epsilon$) for $N$ sufficiently large.
