Is in every set of lebesgue measure $>0$ on $\mathbb R$ a "almost compact" set included

Question

Let $\lambda$ be the Lebesgue measure on $\mathbb R$.

Let $A \subset \mathbb R$ be a measurable set such that $\lambda(A)>0$.

Question:

Can we always find two measurable sets $K_1, K \subset \mathbb R$ such that

• $K_1 \subset A$
• $K \subset \mathbb R$
• $K_1 \subset K$
• $K$ compact
• $\lambda(K_1) = \lambda(K)$
• $\lambda(K_1)>0$

?

Motivation why $K_1$ and $K$:

$K_1$ does not have to be compact... Imagine $A=[0,1] \cap \mathbb Q^c$, then there is no compact set included, but $K_1=A$ and $K=[0,1]$ would satisfy the question...

Answer 1

Yes. You can argue it using the fact Lebesgue measure on $\mathbb{R}$ is inner regular, hence a measurable set $A$ can be approximated by compact subset from inside.