Outer Measure of Cartesian Product with Interval (Apologises in advance if this has already been asked, but I looked around and couldn’t find anything that answered my question).
Let $\lambda_m^*$ denote the Lebesgue outer measure on $\mathbb{R}^m$, and $[a,b]$ be an interval of $\mathbb{R}$. If $A$ is a (not necessarily Lebesgue measurable) subset of $\mathbb{R}^n$, is it possible to say that:
$\lambda_{n+1}^*(A \times [a,b]) = \lambda_n^*(A) (b - a)$?
It is pretty straight forward to see that the left hand side is less than or equal to the right hand side (that’s true for arbitrary Cartesian products), and that equality holds if $A$ is Lebesgue measurable. But what about the general case?
I’m not sure what the best way to find either a proof or a counterexample is, so some help would be much appreciated.
 A: For convenience, set $n=1$ and let $\epsilon>0$. There is an open set $A\times [a,b]\subseteq V'\in \mathbb R^2$ such that $\lambda_{2}^*(V')\le \lambda_{2}^*(A \times [a,b])+\epsilon$. Now $V'=\bigcup_{i}U_i\times (\alpha_i,\beta_i)$ where $U_i\subset \tau_{\mathbb R}$. This is just because $V'$ is a union of basis elements in the product topology. Actually, the $U_i$ are just open intervals, but if $n>1,$ they will be open cubes (or disks). But $V:=\bigcup_{i}U_i\times [a,b]\subseteq V'$, and $A\times [a,b]\subseteq V$ so $\lambda_{2}^*(V)\le \lambda_{2}^*(A \times [a,b])+\epsilon.$
Now, by definition of the product outer measure $\lambda_{2}^*(V)=\lambda_{2}^*(\bigcup_{i}U_i\times [a,b])=\lambda_{1}^*(\bigcup_{i}U_i)\cdot \lambda_{1}^*([a,b])$ so $\lambda_{1}^*(A)\cdot \lambda_{1}^*([a,b])\le \lambda_{1}^*(\bigcup_{i}U_i)\cdot \lambda_{1}^*([a,b])\le \lambda_{2}^*(A \times [a,b])+\epsilon.$
This shows that $\lambda_{1}^*(A)\cdot \lambda_{1}^*([a,b])\le \lambda_{2}^*(A \times [a,b])$ and since you have proven the other inequality, the result follows.
