Proving that $m^*(A\times B)=m^*(A)m^*(B)$ for lebesgue measure

Question

So we already know that $m^*(A \times B)\leq m^*(A)m^*(B)$ for an outer measure now I am trying to prove that it is an equality in Lebesgue measure. I know that is true if we consider one of the sets to be open, we need to use the Fubini theorem in the proof, now it is for an arbitrary set. I tried considering an open set $U$ that covers one of them and use the result I already know but I didn't get far. Any help is appreciated, Thanks.

Answer 1

You don't even need Fubini's theorem. Note that if either is infinite or $0$, then it's trivial, so assume that $m^*(A)$ and $m^*(B)$ are positive and finite. Now, fix $\epsilon>0$. By the definition of outer measure, there exists a countable covering of $A$ by open boxes $(A_j)$ so that $$\sum_j\text{vol} (A_j)<m^*(A)+\frac{\epsilon}{2m^*(B)}.$$ Similarly, we get a countable covering of $B$ by open boxes $(B_k)$ so that $$\sum_k\text{vol} (B_k)<m^*(B)+\frac{\epsilon}{2\sum\limits_j \text{vol} (A_j)}.$$ Now, observe that $$m^*(A)m^*(B)\geq m^*(B)\sum\limits_j\text{vol}(A_j)-\epsilon/2\geq\sum\limits_k\text{vol}(B_k)\sum\limits_{j} \text{vol}(A_j)-\epsilon.$$ Further, $(A_j\times B_k)$ is a countable collection of open boxes covering $A\times B$, with $$\sum\limits_{j,k} \text{vol}(A_j\times B_k)=\sum\limits_j \text{vol}(A_j)\sum\limits_k \text{vol}(B_k),$$ and so by countable subadditivity, $$m^*(A\times B)\leq m^*(A)m^*(B)+\epsilon,$$ where $\epsilon>0$ was arbitrary.