prove that the set $\{A\in \mathcal{B}(\mathbb{R})|A\times \mathbb{R}\in\mathcal{B}(\mathbb{R^{2}})\}$ is a $\sigma$-algebra prove that the set $\mathcal{F}=\{A\in \mathcal{B}(\mathbb{R})|A\times \mathbb{R}\in\mathcal{B}(\mathbb{R^{2}})\}$ is a $\sigma$-algebra
I am not sure about the solution here. I need to show that the whole space is in the set,   complement of any set in $\mathcal{F}$ is also in  $\mathcal{F}$ and finally $\mathcal{F}$ is closed under countable unions
Proof
1) $\mathbb{R^{2}}\in \mathcal{F}$ as $\mathbb{R}\in \mathcal{B}(\mathbb{R})$ because  $\mathcal{B}(\mathbb{R})$ is Borel $\sigma$ algebra and $\mathbb{R}\times \mathbb{R}=\mathbb{R^{2}}$
2) if I pick $A\in \mathcal{B}(\mathbb{R})$ then  $A^{c}\in \mathcal{B}(\mathbb{R})$ by def of $\sigma$ algebra again  and I don't know what is $A^{c}\times \mathbb{R}=(\mathbb{R}\setminus A)\times \mathbb{R}=???$ 
How to show the closure under the countable sums?
 A: The easiest proof would be to note that $\mathcal F$ is just $\mathcal B(\mathbb R)$ itself.
Namely, at each step in the construction of $\mathcal B(\mathbb R)$, every set $A$ that we have found to be Borel so far has the property that $A\times \mathbb R$ is Borel in $\mathbb R^2$, for exactly the same reason that $A$ was Borel in $\mathbb R$ -- because for every open $O\subseteq \mathbb R$, $O\times\mathbb R$ is open in $\mathbb R^2$.
A: You need to show that the set is (i) non-empty, (ii) closed under complementation and (iii) closed under countable intersections.
The solution really boils down to manipulating set products.
(i) $\mathbb{R}^2 \in \mathcal{B}(\mathbb{R^{2}})$, so we have $
\mathbb{R} \in \mathcal{F}$.
(ii) Suppose $A \in \mathcal{F}$, then $A \times \mathbb{R} \in \mathcal{B}(\mathbb{R^{2}})$, and $(A \times \mathbb{R})^c = A^c \times \mathbb{R} \in \mathcal{B}(\mathbb{R^{2}})$, so $A^c \in \mathcal{F}$.
(iii) Suppose $A_n \in \mathcal{F}$, then $A_n \times \mathbb{R} \in \mathcal{B}(\mathbb{R^{2}})$, and so $\cap_n (A_n \times \mathbb{R}) = (\cap_n A_n ) \times \mathbb{R} \in \mathcal{B}(\mathbb{R^{2}})$, so $\cap_n A_n \in \mathcal{F}$.
A: Consider the function $\pi:\mathbb{R}^2\rightarrow \mathbb{R}$ that sends $(x,y)$ to $x$. $\pi$ is a measurable function. Verify that $\{A\in \mathbb{R} | \pi^{-1}[A]\in \mathbb{R}^2\}$ forms a $\sigma$-algebra on $\mathbb{R}$. Finally note that $\mathbb{\pi}^{-1} [A]$ is nothing but $A\times \mathbb{R}$.

The facts that were used in the above solution:
1) $\pi:\mathbb{R}^2\rightarrow \mathbb{R}$ is continuous
2) Let $f:(X,\tau_1)\rightarrow (X,\tau_2)$ be a continuous function between 2 topological spaces. Let $\scr{M}$ be the sigma algebra of $X$ that is generated  by $\tau_1$. Then: $f:(X,\scr{M})\rightarrow$$(Y,\tau_2)$ is a measurable function
3) Let $f:(X,\scr{M})$$\rightarrow (Y,\tau)$ be a measurable function from a measurable space to a topological space, then $\{E\subseteq Y| f^{-1}[E]\in \scr{M}\}$ is a $\sigma$-algebra on $Y$
