Let $X = \mathbb{N}ˆG$, where $G$ is a countable group and, for $K \subset G$ let $X_K = \mathbb{N}^K$. Consider also $\mathcal{B}_K$ the $\sigma$-algebra generated by cylinder sets of the form $[w]= \{x \in X \,:\, x(k) = w(k), \forall k \in K\}$, with $w \in X_K$.
Also, given $K \Subset G$ and $\mu, \nu$ be two probability measures such that $\nu << \mu$. Denote by $f_K$ The Radon-Nikodym derivative of $\mu\vert_K$ with respect to $\nu\vert_K$, where $\mu\vert_K$ and $\nu\vert_K$ denote the restrictions of $\mu$ and $\nu$ to $\mathcal{B}_K$, respectively.
I want to prove that if $E, F \Subset G$ are such that $E \subseteq F$ and $\mu$ and $\nu$ be the ones given. Then $f_{E} = \nu[f_{F} \,\vert\, \mathcal{B}_E]$.
I notice that what I did was wrong. Basically, my idea was to prove that, for every $B \in \mathcal{B}_E$, it holds that \begin{equation*} \int_B f_E \, d\nu = \int_B f_F \, d\nu. \end{equation*}
But for that I used that $\mathcal{B}_E \subseteq \mathcal{B}_F$, if it is the case that $E\subset F$. When I was reviewing, I noticed that what happens is exactly the opposite.
Here is what I did (recalling that I used something that is wrong): from the properties of the Radon-Nikodym derivative, we have that \begin{equation*} \mu(B) = \int_B f_E d\nu, \end{equation*} for all $B \in \mathcal{B}_E$, and that \begin{equation*} \mu(B) = \int_B f_F d\nu, \end{equation*} for all $B \in \mathcal{B}_F$. Thus, for $B \in \mathcal{B}_E \subset \mathcal{B}_F$, \begin{equation*} \int_B f_E d\nu = \mu(B) = \int_B f_E d\nu. \end{equation*}