Prove that $λ^∗(A×B)\geq λ^∗(A)λ^∗(B)$ for every pair of sets, $A \subseteq \mathbb{R}^n$ and $B \subseteq \mathbb{R}^m$ Prove that  $ \lambda^*(A\times B)\geq \lambda^*(A) \lambda^*(B)$ for every pair of sets, $A \subseteq\mathbb{R}^n$ and $B \subseteq\mathbb{R}^m$, where $\lambda^*$ denotes the Lebesgue Outer Measure and $\lambda$ the Lebesgue Measure.
Given $\epsilon>0$, there is an open set $G$ such that $A\times B \subseteq G$  and  $ \lambda^*(A\times B) \geq \lambda(G) - \epsilon$.  
Then, naming $G_n$ the first $n$ coordinates of $G$, and $G_m$ the last $m$ coordinates of $G$,  $A \subseteq G_n$ and $B \subseteq{G_m}$,   $G_n$ and $G_m$ are open sets. Thus, $ \lambda(G_n)\geq \lambda^*(A)$ and $ \lambda(G_m) \geq \lambda^*(B)$.
I want to conclude that  $\lambda(G) \geq \lambda (G_n\times G_m)$, but I don't see how. 
Ps: I have already proved: $\lambda(G_n\times G_m) =\lambda (G_n) \lambda (G_m)$.
 A: Proof.  We have $\lambda_{n+m}^*(A\times B)=\inf\{ b_{n+m}(C):A \times B \subseteq C ~ \& ~C \mbox{ is Borel in}~R^{n+m}\}$, where $b_{n+m}$ is a classical Borel measure in $R^{n+m}$(i.e. the restriction of $\lambda_{n+m}$ to the $\sigma$-algebra of Borel subsets of $R^{n+m}$).
Let $C$ be an arbitrary Borel set in $R^{n+m}$ which contains $A \times B$. 
By Fubini Theorem we have
$$
b_{n+m}(C)=\int_{R^n}b_m(C_x)d b_n(x),
$$
where $C_x$ denotes $x$-section of $C$. A function $b_m(C_x):R^n \to R$ is Borel measurable. Hence a set $F:=\{ x : b_m(C_x) \ge \lambda^{*}_m(B)\} $ is Borel  measurable containing the set $A$. 
We have 
$$
b_{n+m}(C)=\int_{R^n}b_m(C_x)d b_n(x) \ge 
\int_{F}b_m(C_x)d b_n(x)\ge
$$
$$
\lambda^{*}_m(B)\times \int_{F}d b_n(x)=
\lambda^{*}_m(B)\times b_n(F)\ge 
\lambda^{*}_n(A) \times  \lambda^{*}_m(B).
$$  
So $C$ was taken arbitrary, we deduce that
$$\lambda^{*}_n(A) \times  \lambda^{*}_m(B) \le \inf\{ b_{n+m}(C):A \times B \subseteq C ~ \& ~C \mbox{ is Borel in}~R^{n+m}\}=\lambda^{*}_{n+m}(A\times B).$$ 
A: Start with the following 
Claim: if $d\geqslant 1$ and $E\subset \Bbb R^d$, then there is a $G_{\delta}$ set $S$ (that is, a countable intersection of open sets), which contains $E$ and and which has the same outer measure as $E$. 
To prove this, note that we can assume the outer measure of $E$ finite (otherwise take $\Bbb R^d$). Then use regularity: for each $n$, there is an open set $O_n$ containing $E$ such that $\lambda^*_d(E)-n^{-1}<\lambda_d(O_n)$.
This allows us to deduce the opposite inequality as what was mentioned in the OP. I'm not sure equality holds. 
A: In general, we can not conclude that $λ_{n+m}(G) \ge \lambda_n(G_n) \times \lambda_m(G_m)$ if 
$\lambda_{n+m}(G)<\lambda_{n+m}^{*}(A \times B)+\epsilon$ and $A \times B \subset G$.
We set $m=n=1$, $\epsilon >0$, $A=(0,1)$, $B=(0,1)$, $G=\{ (x,y): x \in (0,1) \&  0< y < 1+\epsilon \times x \}.$  Then we get $G_1=(0,1)$ and $G_2=(0, 1+\epsilon)$.
We have 
$$\lambda_{2}(G)= 1+\frac{\epsilon}{2} <  1+\epsilon =\lambda_{2}^{*}(A \times B)+\epsilon,$$
but
$$λ_{2}(G)= 1+\frac{\epsilon}{2}  < 1 + \epsilon =\lambda_1(G_1) \times \lambda_1(G_2).$$ 
