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 $(m \times \mu) D = (m \times \mu)^* D =\infty$.
The article to which you linked actually says that cylindrical $\sigma$-algebra is contained in Borel $\sigma$-algebra. The answer by Michael Greinecker shows the reverse inclusion may fail. Concerning
why would one introduce this product of sigma algebras, instead of the more natural one from topology?
I'd say that from the measure theory point of view the cylindrical $\sigma$ algebra is more natural. They both begin with the same basis of sets, finitely restricted rectangles. From here we either
- take the smallest $\sigma$-algebra containing the rectangles, or
- take the smallest topology containing the rectangles, and then take the smallest $\sigma$-algebra containing that topology.
Clearly, the first approach (which gives cylindrical $\sigma$-algebra) is more direct. The second approach (which gives the Boreal $\sigma$-algebra) involves taking uncountable unions of basis sets in the process of generating topology. Uncountable unions do not play well with measures.
In "small" spaces like $\mathbb R^n$, where the $\sigma$-algebras are the same, we usually think in Borel terms. This is mostly because for analysis it is important to have close connection to the topology of Euclidean space. In huge uncountable products, there is not much that topology can do for us.
Best Answer
First question (title): Sure. It is not hard to show that the sets of the form $B \cup S$, where $B$ is Borel and $S$ is a subset of the Cantor-set constitute a $\sigma$-algebra. There are $2^{\mathfrak{c}}$ subsets of the Cantor set but only $\mathfrak{c}$ Borel sets, hence this $\sigma$-algebra lies strictly between the Borel sets and the Lebesgue-measurable sets.
It's very rare that there is no $\sigma$-algebra strictly between two $\sigma$-algebras, it essentially means that there are atoms (sets that cannot be written non-trivially as union of proper subsets.
I'm not sure I understand the last question. You can take for instance Lebesgue measure plus a Dirac measure if you want something strictly different (i.e. not related via Radon-Nikodym).