Let $M$ and $S$ be compact smooth manifolds. Let $\mu$ be a Borel probability measure on $M\times S,$ such that $$\mathrm{supp}(\nu) = M\times S. $$
Suppose that $\nu$ can be disintegrated in the following way $$ \mu(\mathrm{d}x\times \mathrm{d }s) = \nu_x(\mathrm{d}s) \rho(\mathrm{d}x);$$ i.e. there exist a family of Borel probability measures $\{\nu_x\}_{x\in M}$ on $S$ and a Borel probability measure $\rho$ on $M$ such that $$\mu(A\times B) = \int_{A} \nu_x(B) \mu(\mathrm{d}x),\ \forall A\times B \in \mathcal B(M\times S), $$ where $\mathcal B(M\times S) $ is a Borel $\sigma$-algebra on $M\times S.$
My question: Let $W\in\mathcal B(M\times S)$ be a set such that $$\mu(W) =1,$$ consider the sets $W_s = \{x\in M; (x,s)\in W\}.$ Is it possible to garantee that there exists at least one $s\in S,$ such that $$\rho(E_s) = 1?$$
If we drop the condition that $$\mathrm{supp}(\nu) = M\times S,$$ then we can easily find a counter-example. In fact, $M = S= \mathbb S^1,$ with $$\mu = \delta_{x}(\mathrm{d}y) \mathrm{Leb}(\mathrm{d} x), $$ and $W = \{(x,x),x\in\mathbb {S}^1\}$ is such a counter-example. But with the assumption $\mathrm{supp}(\nu) = M\times S,$ the result seems to hold.
Can anyone help me?