I am assuming that the product measure is the one induced by the product outer measure. (This avoids any issue with ambiguity of definition.)
Any set of the form $B \times A$, with $B$ Borel is measurable ($A$ is arbitrary).
Take the sets $S_n = \{(1,1)\} \cup\left( \cup_{k=0}^{n-1} [\frac{k}{n},\frac{k+1}{n})^2 \right)$. Clearly each $S_n$ is measurable, and hence $D = \cap_{n \ge 0} S_n $ is measurable.
We must have $(m \times \mu) D = \infty$.
To see this, suppose
$D \subset \bigcup_{n \ge 0} B_n \times A_n$, where $B_n$ is Borel, and $A_n$ is arbitrary. Let $D_n = D \cap (B_n \times A_n)$, and let $\pi_x((x,y)) = x$ and similarly $\pi_y((x,y)) = y$. Since $[0,1] \subset \cup_n \pi_x D_n$, we must have $m^* (\pi_x D_{n'} ) >0$ for some $n'$, where $m^*$ is the Lebesgue outer measure (using $m^*$ avoids having to worry about the measurability of $\pi_x D_{n'}$).
Hence $m B_{n'} \ge m^* (\pi_x D_{n'} ) > 0$. In addition, $\pi_x D_{n'}$ must be uncountable (otherwise $m^* (\pi_x D_{n'} ) $ would be zero). Furthermore, $\pi_y D_{n'} = \pi_x D_{n'}$, $\pi_y D_{n'} \subset A_{n'}$, hence $A_{n'}$ is also uncountable. Hence $m(B_{n'}) \mu(A_{n'}) = \infty$, and so $\sum_{n \ge 0} m(B_{n}) \mu(A_{n}) = \infty$ for any cover of $D$ by measurable rectangles. Hence $(m \times \mu) D = (m \times \mu)^* D =\infty$.
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.
Best Answer
$L_1\otimes L_1$ is generated by $\mathcal{C}=\{A\times B:A,B\in L_1\}$. Since $\mathcal{C}\subset L_2$, $L_1\otimes L_1\subseteq L_2$.
For a set $N\in L_1$ s.t. $N\ne \emptyset$ and $\lambda_1(N)=0$, the set $E\times N\in L_2$ ($\because E\times N\subset [0,1]\times N$ and $\lambda_2([0,1]\times N )=0$) but not in $L_1\otimes L_1$ ($\because$ for any $L_1\otimes L_1$-measurable set $A$, the sections $A^y=\{x\in \mathbb{R}:(x,y)\in A\}$ belong to $L_1$).