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.

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    $\begingroup$ How is your $\sigma$-algebra built ? By taking the Borel sets generated by the product topology on $\Omega^{\mathbb N}$ ? $\endgroup$
    – P. Quinton
    Commented May 11, 2023 at 5:31
  • $\begingroup$ 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 $\endgroup$ Commented May 11, 2023 at 5:52

1 Answer 1


This is a long comment, not at all a definite answer.

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).


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