# do the product measures separate “points”?

Suppose $$X,Y$$ to be compact spaces and $$f\in C(X\times Y)$$ different from zero. Is there two probability measures $$\mu_f\in C(X)_+^*$$, $$\nu_f\in C(Y)_+^*$$ such that $$\mu_f\otimes\nu_f(|f|^2)>0$$. In other words, do the product measure separate the elements of $$C(X\times Y)$$? If the answer is yes (as I suppose), I need a reference where such a proof is exhibited. If the answer is no, I need a counterexample.

• Not sure if I am understanding your question correctly, but assuming that $X$ and $Y$ are Hausdorff, how about taking $(x_0, y_0) \in X \times Y$ such that $f(x_0, y_0) \neq 0$ and then setting $\mu_f = \delta_{x_0}$ and $\nu_f = \delta_{y_0}$? – Sangchul Lee Oct 7 '18 at 11:20
• for "points" I mean the (continuous) functions in $C(X\times Y)$ – francesco fidaleo Oct 7 '18 at 11:24
• What I mean is that, for $\mu_f$ and $\nu_f$ chosen as above, we should have $(\mu_f \otimes \nu_f)(|f|^2) = |f(x_0, y_0)|^2 > 0$. I am not sure if this answers your question. And of course, if $X$ or $Y$ is no longer Hausdorff, I have no good idea... – Sangchul Lee Oct 7 '18 at 11:26
• Perfect! This seems the answer. Indeed, I was searching for the noncommutative version of this problem: Let $A$ and $B$ be two unital $C^*$-algebras and $x\in A\otimes B$ (the algebraic tensor product), different from $0$. Is there $\omega\in A^*_+$ and $\varphi\in B^*_+$ such that $\omega\times \varphi(x^*x)>0$. You are suggesting to look at pure states. – francesco fidaleo Oct 7 '18 at 11:33