Is equivalence of probability measures preserved under infinite products?

Question

For all $n\in \Bbb N$, let $\mu_n$ and $\nu_n$ be equivalent probability measures on a measurable space $(\Omega_n,\mathcal{F}_n)$. Are

$$ \mu:=\bigotimes_{n=1}^\infty\mu_n \quad \text{and} \quad \nu:=\bigotimes_{n=1}^\infty\nu_n$$

also equivalent? I highly suspect that they aren't in general, unless the following example (inspired by this question) contains a mistake. But I'd be interested in simpler examples, if there are any.

---

For all $n\in \Bbb N$, let $(\Omega_n,\mathcal{F}_n)=(\Bbb R,\mathcal{B}(\Bbb R))$ and $\mu_n=\mathcal{N}(0,1)$ and $\nu_n=\mathcal{N}(1,1)$. Let

$$ X\colon \Bbb R^\Bbb N \to \Bbb R^\Bbb N, \quad \omega=(\omega_1,\omega_2,\ldots) \mapsto \omega $$

denote the canonical variable. Its components $X_1,X_2,\ldots$ are then i.i.d. $\mathcal{N}(0,1)$ random variables under $\mu$, and they are i.i.d. $\mathcal{N}(1,1)$ random variables under $\nu$. By the Law of Large Numbers,

$$ \frac{1}{n}\sum_{k=1}^nX_k\xrightarrow[\mu-a.s.]{n\to\infty} 0, \quad \frac{1}{n}\sum_{k=1}^nX_k\xrightarrow[\nu-a.s.]{n\to\infty} 1,$$

so $\mu$ and $\nu$ being equivalent would yield $0=1$.

Answer 1

As indicated in the comments, there is a complete solution to this problem provided that each $\mathcal{F}_n$ is the Borel-$\sigma$-algebra with respect to some topology on $\Omega_n$. In this case, Kakutani's Theorem yields the following:

1.) $\mu$ and $\nu$ are either equivalent or mutually singular.

2.) Set $$ \rho_n:=\int_{\Omega_n} \sqrt{ \frac{\mathrm{d} \mu_n} {\mathrm{d} \nu_n} } \mathrm{d} \nu_n.$$ Then $\mu$ and $\nu$ are equivalent if and only if $$ \prod_{n \in \mathbb{N}} \rho_n >0,$$ which is equivalent to convergence of the series $$ \sum_{n \in \mathbb{N}} \log \rho_n.$$

A proof can be found here.