Inclusion for Borelian set with boundary of measure zero Let $\mu$ and $\nu$ be two probability measures on $\mathbb{R}$. We define $\mathcal{B}^0(\mathbb{R}^2)$ to be the set of Borelian sets of $\mathbb{R}^2$ with boundary of $\nu \times \mu$-measure zero. We also define $\mathcal{R}^0(\mathbb{R}^2)$ to be the set of rectangles $A \times B$ of $\mathbb{R}^2$ with boundary of $\nu \times \mu$-measure zero.
Under what condition (possibly minimal) on $\mu$ and $\nu$ is it true that
$$\mathcal{B}^0(\mathbb{R}^2)\subseteq \sigma(\mathcal{R}^0(\mathbb{R}^2))?$$
Here, $\sigma(\cdot)$ denotes the minimal sigma algebra generated by a family of sets
 A: That is always the case. Indeed, $\sigma(\mathcal{R}^0(\mathbb{R}^2))$ is the Borel $\sigma$-algebra of $\mathbb{R}^2$. Note that every open set in $\mathbb{R}^2$ is the countable union of products of open intervals. It, therefore, suffices to show that we can write each such product of intervals as a countable union of elements of $\mathcal{R}^0(\mathbb{R}^2)$.
So let $(a_1,a_2)\times (b_1,b_2)$ be a product of open intervals. For each $n$, let $a_1^n$ and $a_2^n$ be elements of $(a_1, a_1+1/n)$ and $(a_2-1/n, a_2)$, respectively, of $\nu$-measure zero. Since $\nu$ has at most countably many mass points, this can be done. Pick $b_1^n$ and $b_2^n$ similarly. Then for $N$ large enough (so that the intervals are really sub intervals of $(a_1,a_2)$ and $(b_1,b_2)$, respectively), we have
$$(a_1,a_2)\times (b_1,b_2)=\bigcup_{n=N}^\infty [a_1^n,a_2^n]\times[b_1^n,b_2^n].$$
Moreover, the boundary of $[a_1^n,a_2^n]\times[b_1^n,b_2^n]$ is a subset of $\{a_1^n\}\times\mathbb{R}\cup\{a_2^n\}\times\mathbb{R}\cup\mathbb{R}\times\{b_1\}\cup\mathbb{R}\times\{b_2\}$, and each component of this union has $\nu\times\mu$-measure zero by construction.
