Support of a probability measure vs support of its margins Let $\mu$ be a Borel probability measure on the product of two metric spaces $(X, d) $ and $(Y, h) $. Assume its marginal probability measures $\mu_X$ on $X $ and  $\mu_Y$ on $Y$ have compact supports. Can we conclude that $\mu$ is compactly supported as well?
 A: Suppose $(x,y)\in X\times Y$ is a member of the support of $\mu.$ Then every open neighborhood of $(x,y)$ has positive $\mu$-measure. For any open neighborhood $U$ of $x$ within $X,$ the set $U\times Y$ is an open neighborhood of $(x,y)$ within $X\times Y.$ Thus $\mu(U\times Y)>0.$ Thus the marginal assigns positive measure to $U.$ So every open neighborhood of $x$ is given positive probability by the marginal. Since $U$ was an arbitrary open neighborhood of $x,$ it follows that $x$ is a member of the support of the marginal.
Thus if $(x,y)$ is a member of the support of $\mu,$ then $x$ is a member of the support of the marginal, and similarly $y$ is a member of the support of the other marginal.
Hence $(x,y)$ is a member of the product of the supports of the marginals.
So the support of $\mu$ is a subset of the product of the supports of the marginals.
The support of any probability distribution is closed. So the support of $\mu$ is a closed subset of the aforementioned product, and the product is compact.
