# Weak convergence on product space

Let $$\Omega$$ be a metric space and consider $$\Omega^\mathbb{N}$$. Let $$\mu_n,\mu$$ be Borel measures on $$\Omega^\mathbb{N}$$ such that $$\mu_n(C) \to \mu(C)$$ for all cylinder sets. Then is it true that $$\mu_n \to \mu$$ weakly?

If $$\Omega$$ is a discrete set, e.g., $$\Omega=\{0,1\}$$, then it's clear (by Portemanteau) since any open set $$C$$ in $$\Omega^\mathbb{N}$$ is the countable union of cylinder sets. However, if $$\Omega$$ was more general (maybe compact?), I'm not sure if this were true, since measures usually deal with countable unions, while a topology can have uncountable unions.

EDIT. So I think this may be true as long as $$\Omega$$ is second-countable so that it has a countable base, and thus $$\Omega^\mathbb{N}$$ is also second countable (link) with a countable base of cylinder sets generated by the countable base of $$\Omega$$. But I'm a bit fuzzy on the details. Any help would be appreciated.

EDIT 2. I've found a nice reference that confirms my previous edit. See P. Billingsley, Convergence of Probability Measures, Theorem 2.2-2.4.

• How is your $\sigma$-algebra built ? By taking the Borel sets generated by the product topology on $\Omega^{\mathbb N}$ ? Commented May 11, 2023 at 5:31
• I think it should be equivalent, i.e., the product of the Borel algebra of each $\Omega$ is equal to the Borel algebra of the product topology Commented May 11, 2023 at 5:52

Let $$\Sigma_0$$ be the (not $$\sigma$$) algebra generated by the measurable cylinder sets. As $$\Sigma_0$$ is made by a finite number of operation such as intersection, union and complementation of measurable cylinder sets, and because of your assumption that $$\mu_n(A)\to\mu(A)$$ on measurable cylinder sets, we get that $$\mu_n\to\mu$$ on $$\Sigma_0$$. Let $$\Sigma$$ be the $$\sigma$$-algebra generated by $$\Sigma_0$$ is the product Borel $$\sigma$$-algebra. By the Carathéodory's extension theorem, there is a unique extension of $$\lim \mu_n$$ on $$\Sigma$$ and it must therefore be $$\mu$$. Therefore we only need to show that $$\lim\,\mu_n$$ is indeed a measure (and then it would equal $$\mu$$).
The only hard part in showing that $$\lim_n\, \mu_n$$ is a measure is showing $$\sigma$$-additivity, and in general I don't think this is doable because we need to exchange an infinite summation and a limit, but if $$\mu$$ is a finite measure than we are good. It might be that $$\sigma$$-finiteness also works and I am sure there are other conditions that would do (probably $$\Omega$$ being compact).