Compact support of measure.

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Let $\mu$ and $\nu$ two probabilites on $\mathbb{R}^{d}$

Let $\Pi(\mu,\nu)$ is the subset of the probability measure $\pi$ such as $$ \pi (A\times Y) =\mu(A) \text{ and } \pi(X\times B)=\nu(B) $$

Now suppose that the support of $\mu$ and $\nu$ is compact. I would like to deduce that it exists a compact $K$ such that $$ \forall \gamma \in \Pi(\mu,\nu), \text{supp}(\gamma) \subseteq K $$

I think $K$ could be $ \text{supp}(\mu) \times \text{supp}(\nu)$. Do you think it is true ? How to prove it ?

Thanks for your help and regards.

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Well, I think I found a proof using this definition of the support

The support of the measure $μ$ is defined as the set of all points $x$ in $X$ for which every open neighbourhood $N$ of $x$ has positive measure.

Let $x \in \text{supp}(\gamma)$ which we can write $(x_{1},x_{2})$ now we would like to show that $x_{1} \in \text{supp}(\mu)$ and $x_{2} \in \text{supp}(\nu)$ .

Let $B_{i}$ a open set containing $x_{i}$. As $B_{1} \times B_{2}$ is a open set containing $x$ we can deduce $\gamma(B_{1} \times B_{2}) >0$

As $B_{1} \times B_{2} \subseteq R^{d} \times B_{2}$ we have $\nu(B_{2})>0$ and for the same reason $\mu(B_{1})>0$.

We proved that $\text{supp}(\gamma) \subseteq \text{supp}(\mu) \times \text{supp}(\nu)$

Do you agree ?

I used very often the notion of support of a meausre during my project (for the university) but I didn't find any book containing general result. So if you have any general comment about my question or the support of a measure, I would feel glad to hear it.