A clarification regarding the Borel $\sigma$-algebra on $\mathbb{R}^{2}$ Let $B_{2}$ be the Borel-$\sigma$ algebra on $\mathbb{R}^{2}$, that is, the smallest $\sigma-algebra$ that contains all open subsets of $\mathbb{R}^{2}$. Let $B_{1}$ be the usual Borel-$\sigma$ algebra on $\mathbb{R}$. 
(1) Is $B_{2}$ equal to $B_{1}\times B_{1}$? Prove or Disprove.
(2) What about the completion of $B_{2}$ and $B_{1}\times B_{1}$? Are they same? Prove or Disprove.
I have been trying this for quiet sometime, but have been unable to do it. I have read this product measure from the book "Real analysis for Graduate Students-R. Bass" and this is a problem from the same book.
What I know is given two measure space $(X,A,\mu)$ and $(Y,B,\nu)$, the product sigma algebra denoted by $A\times B$ on the space $X\times Y$, is defined to the smallest sigma algebra containing the measurable rectangles, where the measurable rectangles are subsets of $X\times Y$ of the form $a\times b$(where $a\in A, b\in B)$.
Now I know that the problem with this product measure is that even if $\mu$ and $\nu$ are complete, the product measure may not be complete. 
The book has given a very common example taking $\mathbb{R}^{2}$ and the lebesgue measure $m$ on $\mathbb{R}$. So $m\times m$ is the product measure and he shows that this measure is not complete by displaying a null set that is not in the product sigma algebra. And he remarks that the two dimensional lebesgue measure can easily be constructed by taking the completion of this product measure $m\times m$.
Also I know that Lebesgue sigma algebra in $\mathbb{R}$ is basically the completion of the Borel $\sigma$-algebra.
For the first part I have no idea. But for the second part I think that the completion of $m\times m$ is same as $B_{1}\times B_{1}$ and $B_{2}$. But then I have to prove it which I have no idea about.
I hope to get a elaborate explanation so that I can understand this concept clearly. Thanks in advance.
 A: Comment about your notation: Usually  $\times$ means the Cartesian product. Your third paragraph makes clear what you are asking is: 
Does $\mathcal{B}(\mathbb{R}^2) = \mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R}) = \sigma(\mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R}))$ ?
The second equality is just a statement about what the notation $\otimes$ means.
Answer: Yes. An outline of the proof is:


*

*For $D \subset \mathbb{R}$ open, $D \times \mathbb{R}$ is in $\mathcal{B}(\mathbb{R}^2)$

*The collection of sets $\{D \subset \mathcal{B}(\mathbb{R}) \,\, | \,\, D \times \mathbb{R} \in \mathcal{B}(\mathbb{R}^2)\}$ is a $\sigma$-field, and since it contains the open sets, $D \times \mathbb{R} \in \mathcal{B}(\mathbb{R}^2)$ for any Borel set $D$.

*Write for any rectangle of Borel sets $D_1 \times D_2 = (D_1 \times \mathbb{R}) \cap (\mathbb{R} \times D_2)$. Therefore $\mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R}) \subset \mathcal{B}(\mathbb{R}^2)$, which implies $\mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R})  \subset \mathcal{B}(\mathbb{R}^2)$

*Use the fact that $\mathbb{R}^2$ has a countable base of the form $\{U_1 \times \mathbb{R} \,\,,\,\, \mathbb{R} \times U_2\,\,\, | \,\,\, U_i \text{  open  } \subset \mathbb{R}\quad i = 1,2\}$ to show the open sets of $\mathbb{R}^2$ are in $\mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R})$ --- giving $\mathcal{B}(\mathbb{R}^2) \subset \mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R})$


For your second question, see this answer.
