Suppose that $(G_n,\mu_n)$ is a sequence of compact metrizable groups such that each $\mu_n$ is a probability (right-invariant) Haar measure on $G_n$.
Is it true that the product measure $\otimes_n \mu_n$ on $\prod_n G_n$ is also the unique probability (right-invariant) Haar measure on it ?
Edit : we consider compact metrizable groups so the Borel sigma algebra of the product is the product sigma-algebra.
I checked Wikipedia - the entry on product measure didn't say anything about products of more than two spaces. We can go from two to finitely many by induction, but infinite products are different. Is there even a standard notion of product measure for infinite products?
If we're going to have something that works and is compatible with the topology here, it should be done analogously with the product topology - so the basic measurable sets will be products of measurable sets such that all but finitely many terms of the product are the whole space. That can only possibly work out if we arrange things so that, in all but finitely many terms of the product, the whole space has measure $1$ (or, at least, the infinite product converges to something other than $0$ or $\infty$). That won't be a problem in our case - Haar measure on compact groups is normalized to total measure $1$ - but a product with noncompact groups will degenerate into a space where every measurable set has measure zero or $\infty$.
That said, with these rules? Yes. The measure of one of our basic sets is the product measure of something in the first $n$ terms times $1$ for the rest (in which our set covers everything). Translation applies to each component - each of those finitely many nontrivial components has translation-invariant measure, and translating the whole group in the trivial components just results in the whole group back. Verifying that this notion of product measure is a measure on the Borel $\sigma$-algebra of the product space - I haven't gone through it fully, but it should work.