I want to show that for any open $A\subset R^{n}$, with $\mu (A)>0$ and any $a \in R$ with $0<a<\mu(A)$, exists a $K \subset A$ compact and $\mu(K)=a$. Where $\mu$ is the Lebesgue measure.
I know that K is compact iff it is bounded and closed, but I have no idea how to construct such an K. Any hints are much appreciated.
Each open set is the countable union of open balls and therefore also the countable union of closed balls, which are compact. Take a sequence of such compact balls that covers $A$. Finitely many of them will have a union of measure larger than $a$. Reducing the radius of these compact balls by the right amount and then taking the union gives you the desired set. To get the radius right, it helps to take the balls using the maximum distance, $d(x,y)=\max\{|x_i-y_i|:i=1,\ldots n\}$, because the volume of such balls is easy to calculate.