$ \lambda_{n+m}( A \times B ) = \lambda_n(A) \cdot \lambda_m(B) $ Actually I have to show two things:
$(a)$  Show that: $\mathbb{B}(\mathbb{R^n}) \bigotimes\mathbb{B}(\mathbb{R^m}) =\mathbb{B}(\mathbb{R^{n+m}}) $  $\mathbb{B}$ means Borel $\sigma$-algebra.
$(b)$ Show that  $ \lambda_{n+m}( A \times B ) = \lambda_n(A) \cdot \lambda_m(B) $ for $A \in \mathbb{B}(\mathbb{R^n}) $ and $ B \in \mathbb{B}(\mathbb{R^m}) $.
So $(a)$ was no big deal. But I'm stucked in $ (b) $. Maybe we could use $(a)$ somehow? 
Thank you for your help.
 A: Denote $\pi$ the set of all bounded rectangles in ${\bf{R}}^{n}$. Now fix an $I\in\pi$, and consider 
\begin{align*}
\mathcal{D}_{I}=\{B\in\mathcal{B}({\bf{R}}^{m}): |I\times B|=|I|\times|B|\}.
\end{align*}
Then $\pi\subseteq\mathcal{D}_{I}$. Note that the set $\pi$ is a $\pi$-system. Now we are to show that $\mathcal{D}_{I}$ is a $\lambda$-system. Clearly ${\bf{R}}^{m}\in\mathcal{D}_{I}$, and whenever $(B_{n})$ is a disjoint sequence of $\mathcal{D}_{I}$, then the union of $(B_{n})$ is still in $\mathcal{D}_{I}$. 
Now, given $B\in\mathcal{D}_{I}$, assume first that $B$ is contained in a bounded rectangle $Q$, then 
\begin{align*}
\left(I\times(Q-B)\right)\cup\left(I\times B\right)=I\times Q,
\end{align*} 
and we have 
\begin{align*}
|I\times(Q-B)|&=|I\times Q|-|I\times B|\\
&=|I|\times|Q|-|I|\times|B|\\
&=|I|\times(|Q|-|B|)\\
&=|I|\times|Q-B|.
\end{align*}
Now express ${\bf{R}}^{m}$ as the union of non-overlapping rectangles $Q_{n}$, then 
\begin{align*}
|I\times B^{c}|&=\left|I\times\left(\bigcup_{n}(Q_{n}-B)\right)\right|\\
&=\left|\bigcup_{n}(I\times(Q_{n}-B))\right|\\
&=\sum_{n}|I\times(Q_{n}-B)|\\
&=\sum_{n}|I|\times|Q_{n}-B|\\
&=|I|\times\left|\bigcup_{n}(Q_{n}-B)\right|\\
&=|I|\times|B^{c}|,
\end{align*}
this shows that $B^{c}\in\mathcal{D}_{I}$, so $\mathcal{D}_{I}$ is a $\lambda$-system. By $\pi$-$\lambda$ theorem, we have $\mathcal{B}({\bf{R}}^{n})\subseteq\mathcal{D}_{I}$.
Now let $A\in\mathcal{B}({\bf{R}}^{n})$ and consider
\begin{align*}
\mathcal{D}_{A}=\{B\in\mathcal{B}({\bf{R}}^{m}): |A\times B|=|A|\times|B|\}.
\end{align*}
Now $\pi\subseteq\mathcal{D}_{A}$. With the same reasoning one can show $\mathcal{D}_{A}$ is a $\lambda$-system, and use the theorem again we get $\mathcal{B}({\bf{R}}^{n})\subseteq\mathcal{D}_{A}$.
