Let $X=A^{\mathbb{Z}}$ for finite set $A$ of cardinality $\text{card}(A)=k<\infty$.
Let $\tau$ denote the topology on $X$ with basis given by the set $C$ of cylinders $$ C_n(c_0,c_1,\ldots,c_m):=\{x\in X: x_n=c_0,\ldots,x_{n+m}=c_m\} $$ and $B(\tau)$ the Borel $\sigma$-algebra with the uniform Bernoulli-measure, $$ \mu(C_n(c_0,\ldots,c_m)):=k^{-(m+1)}, $$ on it.
Suppose $Y\subseteq X$ is a $T$-invariant subset, i.e. $T(Y)\subseteq Y$. Consider $S:=T_{|Y}$, the restriction of $T$ on $Y$. Similarly as above, let $\tau'$ be the topology on $Y$ with basis given by the set $C'$ of the cylinders $$ Y\cap C_n(c_0,c_1,\ldots,c_m). $$ Let $B(\tau')$ be the Borel $\sigma$-algebra with uniform Bernoulli-measure $\lambda$ on it.
I have two questions concerning measure theory/ dynamics:
(1) Am I right that $B(\tau')\subseteq B(\tau)$ and $\lambda=\mu_{|B(\tau')}$?
(2) For the measure-theoretical entropy, do we have a statement or property telling that we can "split" it like $$ h_{\mu}(T)=h_{\lambda}(S)+h_{\mu_{|B(\tau')^C}}(T_{|Y^C})? $$
(1) I think this follows by $\tau'\subseteq \tau$, hence $B(\tau')=\sigma(\tau')\subseteq\sigma(\tau)=B(\tau)$. Moreoever, both $(X,\tau)$ and $(Y,\tau')$ are separable, hence $B(\tau)=\sigma(C)$ and $B(\tau')=\sigma(C')$, meaning that the Borel $\sigma$-algebras coincide with the product $\sigma$-algebras (= cylinder $\sigma$-algebras). Moreover, $\mu=\lambda$ on $C'$ and $C'$ is invariant under intersection. By uniqueness theorem, $\mu=\lambda$ on $\sigma(C')$.
(2) The reason why I am asking this is because I have an example where I can argue that $h_{\lambda}(S)=0$ and for further investigation I would like to know if this implies that $h_{\mu}(T)$ can be determined by neglecting "what $S$ and $\lambda$ do". I have no idea...
In case the splitting above is not correct, maybe there is another formula connecting $h_{\mu}(T)$ and $h_{\lambda}(S)$?