TITLE: Absolute c0ntinuity of infinite product measure
QUESTION [5 upvotes]: Suppose $\mu, \nu$ are Borel probability measures on $\mathbb{R}$ and that $\mu^\infty$ and $\nu^\infty$ are the infinite products, in the sense of the Kolmogorov extension theorem, of $\mu$ and $\nu$ with themselves respectively. Does $\mu \ll \nu$ imply $\mu^\infty \ll \nu^\infty$? If not, what additional assumptions would make this true?

REPLY [8 votes]: So long as $\mu$ and $\nu$ are distinct, the infinite product measures will be mutually singular.  Indeed, choose an event $B$ with $\mu(B)\not=\nu(B)$. Then
$$
G:=\{x=(x_1,x_2,\ldots)\in {\Bbb R}^\infty: \lim_n n^{-1}\sum_{k=1}^n 1_B(x_k)=\mu(B)\}
$$
satisfies $\mu^\infty(G)=1$ but $\nu^\infty(G)=0$ by the strong law of large numbers.
A definitive answer to this type of question, when the factor measures $\mu_1,\mu_2,\ldots$ and $\nu_1,\nu_2,\ldots$ are allowed to depend on $n$,and satisfy
$\mu_n\sim\nu_n$ for each $n$ (that is, are mutually absolutely continuous) was found by S. Kakutani in the 1940s. See "On Equivalence of Infinite Product Measures" Annals of Mathematics, Vol. 49 (1948) 214-224.