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.)



Measurability of sets of pairs
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.


