$E$ and $F$ measurable compact sets of $\mathbb{R}$ such $\mu(E)=1$ and $\mu(F)=3$. There exist $K$ compact such $\mu(K)=2$. Let $\mu$ be the Lebesgue measure and, $E$ and $F$ compact subsets of $\mathbb{R}$ such that $E\subset F$ and $\mu(E)=1$ and $\mu(F)=3$. Prove there is a compact subset $K$ of $\mathbb{R}$ such that $E \subseteq K  \subseteq F$ and $\mu(K)=2$. First and foremost, as $\mu(E)=1$ and $\mu(F)=3$ are compact subsets of $\mathbb{R}$ there are open subsets of $\mathbb{R}$, $E'$ and $F'$, such as $E \subset E'$  and $F' \subset F$ and $\mu^{\ast}(E'- E) < \epsilon$ and  $\mu^{\ast}(F'- F) < \epsilon$. Where $\mu(E')=\mu(E)=1$ and $\mu(F')=\mu(F)=3$. Obviously, $\mu^{\ast}$ stands for  Lebesgue outer measure. So far, I´ve been run out of ideas and aproaches to solve this. I have googled properties about Lebesgue measure of compact sets and there is nothing about them that I have found helpful, for instance, all compact sets are Lebesgue measurable and $\mu(E)< \infty$ for every compact set in $\mathbb{R}$. But I dont know how I´m supposed to construct this compact set $K$ such as $E \subseteq K  \subseteq F$ and $\mu(K)=2$. Thanks!
 A: Hint:
Consider the sets $[-r,r] \subseteq \mathbb{R}$.
Can you show $r \mapsto \mu(E \cup [-r,r] \cap F)$ is continuous? Can you show for any $r$ the set we're measuring is compact? What does the intermediate value theorem buy us?

I hope this helps ^_^
A: There seems to be a new style of teaching measure theory without mentioning $inner$ measure. Let $m^o$ denote outer measure.
For $S\subseteq \Bbb R,$ let $C(S)$ be the set of compact subsets of S.
The inner measure of $S$ is $m^i(S)=\sup \{m^o(T): T\in C(S)\}=\sup \{m(T): T\in C(S)\}.$
If $m^0(S)<\infty$ then $S$ is measurable iff $m^i(S)=m^o(S)$ iff $m(S)=m^i(S)=m^o(S)$.
If $E,F$ are compact with $E\subset F$ and $m(E)=1$ and $m(F)=3$ then $m(F\setminus  E)=m^i(F\setminus E)=2,$ so take $T\in C( F$ \ $E)$ with $1<m(T)\le 2.$
For $r\in [0,\infty )$ let $f(r)=m(T$ \ $(-r,r)\,).$ Now $f$ is monotonic decreasing with $f(0)=m(T)>2,$ and $f(r)=0$ for some $r>0$ because $F$ \ $E$ is a bounded set.
So let $r_1=\sup \{r>0:f(r)\ge 1\}.$
$(\bullet).\;$ We can now show that $f(r_1)=1.$
So we can let $K=E\cup (T$ \ $(-r_1,r_1)\,).$
Remark:$f$ is Lipschitz-continuous with Lipschitz constant $2$ because if $0\le r<r'$ then $0\le f(r)-f(r')\le m(\,(-r',-r]\cup [r,r')\,)= 2(r'-r).$
