Measurability of sets of pairs

Question

Suppose $(X,\mathcal{B},\mu)$ is a measure space, and let $B\subseteq X^2$ be an arbitrary set.

1) Is there a nice characterization of the circumstances under which there is a $\sigma$-algebra $\mathcal{C}\supseteq\mathcal{B}$ and a measure $\nu$ on $\mathcal{C}$ extending $\mu$ such that $B$ is measurable with respect to $\mathcal{C}\times\mathcal{C}$?

2) Is it possible for there to be distinct extensions $\mathcal{C}_1,\nu_1$ and $\mathcal{C}_2,\nu_2$ such that $B$ is measurable with respect to $\mathcal{C}_1\times\mathcal{C}_1$ and $\mathcal{C}_2\times\mathcal{C}_2$, but $\nu_1^2(B)\neq \nu_2^2(B)$?

3) If the answer to 2 is yes, is there a nice characterization of the circumstances under which extensions of $\mathcal{B}$ assign a unique value to the measure of $B$?

(If it helps, $\mathcal{B}$ can be assumed to have or not have nice properties like separability or being a regular Borel measure.)

Answer 1

Without regularity at least, such pathologies can happen. In fact, taking products is not necessary to get them. It can happen that with $A\subset X$ we have $\nu_1(A)\neq \nu_2(A)$. Taking the product of $A$ with $X$ yields counterexamples like the one sought in (2).

First, it is easy to do this in a trivial way by making $X$ a space with two or more points and $\mathcal{B}=\{\emptyset,X\}$. Taking $\mathcal{C}=\{\emptyset,A,A^c,X\}$ where $A,A^c\neq\emptyset$ we can assign $\nu_i(A)$ to be anything less than $\mu(X)$.

This can also be done in nontrivial cases. Let $X=[0,1]$, $\mu$ be Lebesgue measure, and $\mathcal{B}$ the set of Lebesgue measureable sets. Let $A\subset[0,1]$ be a set of Lebesgue inner measure $0$ and outer measure $1$. Let $\mathcal{C}$ be the sigma algebra generated by $\mathcal{B}$ and $A$.

Sets in $C$ are of the form $(Y\cap A)\sqcup (Z\cap A^c)$ where $Y,Z\in\mathcal{B}$. Suppose we can express a set in two different ways in this form: $(Y\cap A)\sqcup (Z\cap A^c) = (Y'\cap A)\sqcup (Z'\cap A^c)$. Then the symmetric difference of $Y$ and $Y'$ is contained in $A^c$, which has Lebesgue inner measure zero. But this symmetric difference is also Lebesgue measurable, so it has Lebesgue measure zero. Thus $\mu(Y) = \mu(Y')$. Similarly $\mu(Z) = \mu(Z')$. Therefore if we define $\nu\left((Y\cap A)\sqcup (Z\cap A^c)\right) = \lambda \mu(Y)+(1-\lambda)\mu(Z)$ for any $\lambda\in[0,1]$ we get a well-defined measure on $\mathcal{C}$ which extends $\mu$ and satisifes $\nu(A) = \lambda$.

This leads me to believe you're not going to get any nice condition on the measure $\mu$ to ensure uniqueness, unless the measures you consider "nice" are very different from Lebesgue measure or you believe in a set-theoretic universe where sets like $A$ don't exist.