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Consider a product of two measurable spaces, $(X,\mathcal{A})$ and $(Y,\mathcal{B})$, and a (not necessarily finite) measure, $\varrho$ on the product space $(X \times Y, \mathcal{A} \otimes \mathcal{B})$. I wonder, if there is any general notion for the "marginals" of $\varrho$, so if there is a way to find (or at least prove the existence of) a measure $\mu$ on $(X, \mathcal{A})$ and a measure $\nu$ on $(Y, \mathcal{B})$, for which $\varrho$ equals the product measure $\mu \times \nu$.

I need it the case when $\varrho$ is atom-less and $\sigma$-finite, although I would be interested in the general case as well.

(By the way, I just realised, that if there are such marginals, they can't be unique: actually, if I multiply $\mu$ with a positive constant $c$, and divide $\nu$ with the same $c$ constant, the results will still be marginals of $\varrho$. Although, I guess, they are uniqe, up to this manipulation.)

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Even for "nice" probability spaces, it is not true that a measure on a product space can be factored in the way that you have described. Let $\varrho$ be the probability measure on the measurable space $(\mathbb{R}^{2},\mathcal{B}(\mathbb{R}^{2}))$ (recall that $\mathcal{B}(\mathbb{R}^{n})=\bigotimes_{i=1}^{n}\mathcal{B}(\mathbb{R})$) given by the bivariate normal distribution with zero mean vector and covariance matrix $\Sigma=\begin{bmatrix}1 & 2^{-1/2}\\ 2^{-1/2} & 1\end{bmatrix}$. I.e. $\varrho$ has the density function

$$f(x,y)=\dfrac{\sqrt{2}}{2\pi}\exp\left(-\left[x^{2}+y^{2}-\sqrt{2}xy\right]\right), \ (x,y\in\mathbb{R})$$

Suppose that $\varrho=\mu\otimes\nu$, where $\mu,\nu$ are measures on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. As you noted in your question, $\mu$ and $\nu$ are not unique; we only know that $1=\varrho(\mathbb{R}^{2})=\mu(\mathbb{R})\nu(\mathbb{R})$. But let us insist that both $\mu$ and $\nu$ be probability measures.

I claim that $\mu$ and $\nu$ coincide with marginals of $\varrho$. Indeed, for any $A\in\mathcal{B}(\mathbb{R})$,

$$\mu(A)=\mu(A)\nu(\mathbb{R})=\varrho(A\times\mathbb{R})=\int_{A}\int_{\mathbb{R}}f(x,y)dxdy=\dfrac{1}{\sqrt{2\pi}}\int_{A}e^{-y^{2}/2}dy$$

and analogously for $\nu(A)$. Furthermore, we see that $\mu,\nu$ are standard normal. But the tensor product of standard normals is the bivariate normal distribution with zero mean vector and identity covariance matrix, which is a contradiction.

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You could look at the book of Ambrosio, Fusco, Pallara "Functions of bounded variation and free discontinuity problems" page 56, section 2.5 and in particular Theorem 2.28. If your measures are Radon (and defined in Euclidean spaces), then you can express the measure on the product as an integral with respect to one of the marginals of a Radon measure valued function. This is called measure disintegration and I am not sure whether it works for abstract measure spaces.

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