I want to show that for every Lebesgue measurable set $ A\in\mathcal{L}(\mathbb{R})$, the set $A\times\mathbb{R}\subset\mathbb{R}^2$ is in $\mathcal{L}(\mathbb{R}^2)$.
I have found a proof, however there are some gaps which I would like to fill.
Proof:
Since $ A\in\mathcal{L}(\mathbb{R})$ we have by the definition of measurability:
$\lambda_1^*(Q)= \lambda_1^*(Q\cap A) + \lambda_1^*(Q\cap A^c), \ \forall Q\subset \mathbb{R}$ where $\lambda_1^*$ is the outer Lebesgue measure in $\mathbb{R}$.
I need to show that
$\lambda_2^*(Q)\ge \lambda_2^*(Q\cap (A\times \mathbb{R}))+\lambda_2^*(Q\cap (A\times \mathbb{R})^c),\ \forall Q\subset\mathbb{R}^2$
The first issue here is, that I somehow must assume that $Q$ can be written as a cartesian product $Q=Q_1\times Q_2$ to be able to compute the measures and reduce it to the one dimensional measure. So I assumed, that it is enough for me to take rectangular $Q$. I read that this is equivalent to the outer measurability definition where we use arbitrary $Q\subset\mathbb{R}^2$.
So let's suppose we can fill this gap (if my assumption is not correct).
We then have
$\lambda_2^*((Q_1\cap A) \times (\mathbb{R}\cap Q_2))+\lambda_2^*(Q\cap (A\times \mathbb{R})^c)=\lambda_2^*((Q_1\cap A) \times (\mathbb{R}\cap Q_2))+\lambda_2^*(((A^c\times\mathbb{R}^c)\cup(A^c\times \mathbb{R})\cup (A\times \mathbb{R}^c))\cap Q)=\lambda_2^*((Q_1\cap A) \times (\mathbb{R}\cap Q_2)) + \lambda_2^*((A^c\cap Q_1\times \mathbb{R}^c\cap Q_2)\cup(A^c\cap Q_1\times \mathbb{R}\times Q_2)\cup(A\cap Q_1\times \mathbb{R}^c\cap Q_2)) $
I then use that the product measure on $\mathbb{R}^2$ can be split into the product of two L-measures on $\mathbb{R}$ and that the unions are disjoint. I obtain:
$=\lambda_1^*(A\cap Q_1)\lambda_1^*(\mathbb{R}^c\cap Q_2)+ \lambda_1^*(A^c\cap Q_1)\lambda_1^*(\mathbb{R}^c\cap Q_2)+\lambda_1^*(A^c\cap Q_1)\lambda_1^*( \mathbb{R}\times Q_2)+\lambda_1^*(A\cap Q_1)\lambda_1^*(\mathbb{R}^c\cap Q_2)$
We the just factor and use the measurability of $A$:
$=\lambda_1^*(A^c\cap Q_1)(\lambda_1^*(\mathbb{R}^c\cap Q_2)+\lambda_1^*(\mathbb{R}\cap Q_2))+\lambda_1^*(A\cap Q_1)(\lambda_1^*(\mathbb{R}^c\cap Q_2)+\lambda_1^*(\mathbb{R}\cap Q_2))=\lambda_1^*(Q_2)(\lambda_1^*(A^c\cap Q_1)+\lambda_1^*(A\cap Q_1))=\lambda_1^*(Q_1)\lambda_1^*(Q_2)=\lambda_2^*(Q)$
which is exactly what we wanted to show. Now the problem is, that I am unsure if I can actually split the measure in a product or not. The other uncertainty, that I used only rectangles is stated above. These issues need to be fixed.
In general, you can't split $Q$. You need to use some approximation by Borel sets using the fact that $\mathscr B(\mathbb R)\otimes \mathscr B(\mathbb R)=\mathscr B(\mathbb R^2)$.
But in any case, I think we can argue as follows:
$(1).\ $ Suppose $f$ is continuous on $\mathbb R.$ Then, the $\sigma$-algebra generated by $\{B\subseteq \mathbb R:f^{-1}(B)\in \mathscr B\}$ contains the open sets, hence all Borel sets. So, whenever $B$ is a Borel set, so is $f^{-1}(B).$
$(2).\ $ The projection $\pi:\mathbb R\times\mathbb R\to \mathbb R:(x,y)\to x$ is continuous.
$(3).\ $ There are Borel sets $F\subseteq A\subseteq U$ such that $\lambda_1(U\setminus F)=0.$
Then, $(1)$ and $(2)$ imply that $F\times \mathbb R$ is a Borel set and $(3)$ implies that $A\times \mathbb R=(A\setminus F)\times \mathbb R\bigcup F\times \mathbb R$ and $\lambda_2\left((A\setminus F)\times \mathbb R\right )=0.$
The result now follows because we have written $A\times \mathbb R$ as a union of a Borel set and a set of $\lambda_2$-measure zero.