Uniqueness of marginal measures Let $X\neq \{0\}$ and $Y\neq \{0\}$ be complete separable metric spaces, $M(A)$ the set of Borel measures with finite total variation in $A$ and $\Pi(\rho_1,\rho_2)$ the set of all Borel measures in $X\times Y$ with marginals $\rho_1$ in $X$ and $\rho_2$ in $Y$.
Given a measure $\eta\in M(X\times Y)$, is it possible in any case to find two different pair of measures $\mu_1,\nu_1\in M(X)$, $\mu_2,\nu_2\in M(Y)$ such that $\eta\in\Pi(\mu_1,\mu_2)$ and $\eta\in\Pi(\nu_1,\nu_2)$?
 A: I guess the question has already been answered  in the comments but it might be  a good idea to summarize it in a formal
answer.
The fact that the OP refers to metric spaces is perhaps irrelavant, so let us simply assume that $(X, \mathcal A)$ and
$(Y, \mathcal B)$ are measurable spaces, that is, $X$ and $Y$ are  sets, while
$\mathcal A$ and $\mathcal B$ are
$\sigma$-algebras of subsets of  $X$ and $Y$, respectively.
One then defines $\mathcal A\times \mathcal B$ to be the smallest $\sigma$-algebra of subsets of $X\times Y$ containing
all sets of the form $A\times B$, where $A\in \mathcal A$, and $B\in \mathcal B$.
Given any measure $\eta $ on $(X\times Y,  \mathcal A\times \mathcal B)$, be it finite or infinite, probability or not, positive or
signed,
the marginal measures $\mu $ and $\nu $ are defined in
terms of $\eta $ by
$$
  \mu (A) = \eta (A\times Y), \text { for all } A\in \mathcal A,
  $$
and
$$
  \nu (B) = \eta (X\times B), \text { for all } B\in \mathcal B.
  $$
It is easy to see that $\mu $ and $\nu $ are measures on $X$ and $Y$, respectively.   They are clearly  unequivocally
determined by $\eta $, in the same way that the derivative of a smooth function $f$  is determined by  $f$.
Thus, the question of whether there is a measure $\eta $ lying simultaneously in $\Pi(\mu_1,\mu_2)$ and in $\Pi(\nu_1,\nu_2)$, for
two different pairs of measures $\mu_1,\nu_1\in M(X)$, $\mu_2,\nu_2\in M(Y)$ therefore has an immediate  negative answer because both
$\mu _1$ and $\nu _1$ must be the marginal of $\eta $ relative to $X$, so necessarily $\mu _1=\nu _1$, and similarly  $\mu _2=\nu _2$.
This is a bit like asking whether there is a smooth function $f$ whose derivative is both equal to $g_1$ and to $g_2$, for
different functions $g_1$ and $g_2$!
An entirely different (and highly relevant) question is whether or not there are two different measures $\eta _1$ and $\eta _2$
on $X\times Y$ whose marginals on $X$ and $Y$ coincide.  In other words,  whether $\Pi(\mu,\nu)$  contains more than
one measure, once we are  given measures  $\mu $ and $\nu $ on $X$ and $Y$, respectively.
Restricting the discussion to probability measures from now on, and
before discussing uniqueness, it is nice to know that, given  $\mu $ and $\nu $, there
always exists at least one measure in $\Pi(\mu,\nu)$, namely the product measure, variously denoted $\mu \times \nu $ or $\mu \otimes \nu $.
This measure is characterized by the fact that
$$
  (\mu \times \nu )(A\times B) = \mu (A)\nu (B),
  $$
for all $A$ in $\mathcal A$, and $B$ in $\mathcal B$.
Incidentally this property, together with the assumption that $\mu (X)=1=\nu (Y)$, immediately implies that the marginal measures
for $\mu \times \nu $ are $\mu $ and $\nu $.
The concept of product measures sits squarely at the center on the notion of independent random variables:
seeing the projections
$$
  x:X\times Y\to X
  $$
and
$$
  y:X\times Y\to Y
  $$
as random variables (this is specially relevant when $X=Y=\mathbb R$), then the probability
$$
  \mathbb P(x\in A\ \wedge\ y\in B)
  $$
is precisealy the measure of $A\times B$.   So this always  coincides with the product of probabilities
$$
  \mathbb P(x\in A)  \  \mathbb P(y\in B)
  $$
a.k.a. $\mu (A)\nu (B)$, iff the random variables are independent, iff the probability measure on $X\times Y$ is the product measure.
The covariance of $x$ and $y$, namely,
$$
  \text{cov}(x,y) = \mathbb E(xy) -\mathbb E(x)\mathbb E(y),
  $$
can be computed by
$$
  \text{cov}(x,y) =   \int_{X\times Y}xy\,d(\mu \times \nu ) - \left(\int_Xx\,d\mu\right)\left(\int_Yy\,d\nu \right),
  $$
and is easily seen to vanish due to the Fubini Theorem which allows for the iterated integration
$$
  \int_{X\times Y}xy\,d(\mu \times \nu ) =   \int_Y\int_X xy\,d\mu \, d\nu .
  $$
In other words, if $x$ and $y$ are independent, then   $\text{cov}(x,y) =  0$.
Back to the uniqueness question, suppose e.g. that $X=Y=[0,1]$, and that $\mu =\nu =\lambda $, where $\lambda $ is  Lebesgue measure.
As already seen, the product measure $\mu \times \nu $ (incidentally the 2-dimensional Lebesgue measure on the square) admits $\mu $
and $\nu $ as marginals.
So what would be another example of a measure in $\Pi(\mu,\nu)$?  Well, here is one:  given any
Borel measurable subset $E\subseteq [0,1]^2$, set
$$
  \eta (E) =  \lambda (\{x\in [0, 1]: (x, x)\in  E\}).
  $$
It is very easy to see that the marginals of $\eta $ are still $\mu $ and $\nu $, but now the measure of a product set $A\times B$ can  no
longer be computed just in terms of $\mu (A)$ and $\nu (B)$.   To see for yourself, try to prove that
$$
  \eta ([0,1/2]\times [1/2, 1]) = 0
  $$
(the top left-hand quarter of the square has only one point in common with the diagonal!), while
$$
  \eta ([0,1/2]\times [0,1/2]) = 1/2.
  $$
Needless to say, $x$ and $y$ are not independent random variables.  They are in fact so dependent to each other that
$x=y$ almost surely, meaning that  the set
$$
  \{(x,y)\in [0,1]\times [0,1]: x=y \}
  $$
has full measure (according to $η$),
