Let $J\subset\mathbb{R}^n$ be a rectangle (i.e. the cartesian product of $n$ non-empty intervals, open, closed or whatever). Let $A\subset J$ such that if $K$ is a compact subset of $J$ with positive Lebesgue measure, $K\cap A\neq\emptyset$. Prove that $m^*(A)=v(J)$, where $m^*$ denotes the exterior measure and $v$ denotes the volume of the rectangle in the usual sense, as the product of the length of its sides.
Here's my take. Obviously, $m^*(A)\leq v(J)$ by definition of exterior measure, sine $J$ covers $A$. Now, assume $J$ is closed and $m^*(A)<v(J)$. Then, there would exist open rectangles $I_j$ in $J$ so that $A\subset\cup_{j=1}^\infty I_j$ and $\sum_{j=1}^\infty v(I_j)<v(J)$. Then, the set $S=J\setminus\cup_{j=1}^\infty I_j$ is compact as it is bounded and closed and $m^*(S)>v(J)-\sum_{j=1}^\infty v(I_j)>0$ but one easily checks that $A\cap S=\emptyset$, contradicting the hypotheses, therefore concluding that $m^*(A)= v(J)$ if $J$ is closed. My problem is: how can I extend this to the general case, where $J$ is not necessarily closed? Moreover, I'm sure there's an easier path to take, any hints? Thank you in advance.
If $J$ is open,let $\epsilon>0$ and then take a smaller closed rectangle $Q$ such that $Q\subseteq J$ such that $v(J) \leq v(Q)+\epsilon$
Then by hypothesis you have that $A \cap Q$ intersects every compact subset of $Q$ with positive measure. So by the case of closed sets and you have that $$v(J) \geq m^*(A\cap Q)=v(Q) \geq v(J)-\epsilon$$
Send $\epsilon \to 0$ and you have $v(J)=m^*(Q \cap (A)) \leq m^*(A)$
Apply this argument to half-open rectangles too.