What is the RN derivative of infinite product measure? 
Suppose $\mu_k$ and $\nu_k$, $k=1,2,...$ are sigma-finite measures on spaces $(S_k,\mathcal F_k)$ such that $\nu_k<<\mu_k$ for each $k$. Let $f_k=\dfrac{d\nu_k}{d\mu_k}$ for each $k$. Then is it true that $\nu:=\prod_{k=1}^\infty \nu_k<<\prod_{k=1}^\infty \mu_k:=\mu$ with $\dfrac{d\nu}{d\mu}(s_1,s_2,...)=\prod_{k=1}^\infty f_k(s_k)$?

The result is true when you have a finite product.
 A: No, not in general.
For a simple counterexample, let all the $\mu_k \sim N(0,1)$ be the standard normal distribution on $\mathbb{R}$, and $\nu_k \sim N(42, 1)$ be a normal distribution with mean 42 and variance 1.  Clearly $\nu_k \ll \mu_k$ for each $k$, since they are both absolutely continuous to Lebesgue measure with strictly positive densities.
Now if $X_i$ are iid $N(0,1)$, the strong law of large numbers says that $\lim_{n \to\infty} \frac{1}{n} \sum_{i=1}^n X_i = 0$ almost surely.  Rephrasing this, it says that if $A \subset \mathbb{R}^\infty$ is the set of sequences $x_i$ satisfying $\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n x_i = 0$, we have $\mu(A) = 1$.  But if $Y_i \sim N(42, 1)$, then $\lim_{n \to\infty} \frac{1}{n} \sum_{i=1}^n Y_i$ equals 42 with probability 1, and thus equals 0 with probability 0; in other words, $\nu(A)=0$.  So we see that $\mu, \nu$ are mutually singular.
More generally, Kakutani's "dichotomy" theorem gives a necessary and sufficient condition for $\nu \ll \mu$: this holds if and only if the infinite product $\prod_{i=1}^\infty \int \sqrt{f_k}\,d\mu_k$ is nonzero.  In this case we will also have $\mu \ll \nu$, and otherwise the measures will be mutually singular.  See for instance Bogachev's Measure Theory, Theorem 10.3.6.  (There is also a Wikipedia article but it has a mistake that I will shortly edit.)
One might naively think that $f(s_1, s_2, \dots) := \prod_{k=1}^\infty f_k(s_k)$ "must" be the desired density.  And so it must, if it is a density at all; but it can happen that $\int f\,d\mu < 1$.  One might naively think that $\int f\,d\mu = 1$ is assured by Fubini's theorem; but Fubini only applies to finite products.  If we let $g_n(s_1, s_2, \dots) = \prod_{k=1}^n f_k(s_k)$, then indeed $\int g_n\,d\mu = 1$ by Fubini and $g_n \to f$ $\mu$-a.e., but as we know, that is not enough to conclude that $\int g_n\,d\mu \to \int f\,d\mu$.  Kakutani's condition provides uniform integrability of the $g_n$ which allows the argument to go through.
