Let $(A_M^{\mathbb{Z}_+}, \Omega, P, \lambda)$ be a Markov Shift where $A$ is a finite alphabet set, $M$ is the admissibility matrix, $P = [P_{i, j}]_{i, j\in A}$ is a stochastic matrix that is indexed by letters from $A$ and determines the transition probability, $\lambda = (\lambda_i)_{i\in A}$ is a row vector that determines the initial distribution, and $\Omega$ is the $\sigma$-algebra generated by cylinder sets. Let $T$ be the shift transformation. For each $B\in\Omega$ and $i, j\in A$, define $B_i = \{\textbf{a}\in A_M^{\mathbb{Z}_+}\,\vert\,a_0 = i\}$, $B_{i, j} = \{\textbf{a}\in A_M^{\mathbb{Z}_+}\,\vert\, a_0 = i, a_1 = j\}$. If I use $\mu$ to denote the probability measure determined by $P$ and $\lambda$, then I am asked to show:
$$ \frac{\mu\big[T(B_{i, j})\big]}{\mu(B_{i, j})} = \frac{\lambda_j}{\lambda_i P_{i, j}} $$
It is clear that if $B$ has the form like $\{\textbf{a}\in A_M^{\mathbb{Z}_+} \,\vert\, a_0 = i_0, \cdots a_n = i_n\}$ (or $B$ is an elementary cylinder set) for $n\in\mathbb{N}$ and $\{i_0, \cdots, i_n\}\subseteq A$, then the equation above is obviously true because, in this case:
$$ \mu(B) = \lambda_{i_0}\prod_{0\leq i < n}P_{i, i+1} $$
Then I fail to prove that this is also true for a general cylinder set, which is always a disjoint union of finitely many elementary cylinder sets. The equation almost "makes sense" to me immediately but unfortunately I cannot prove it. Any hints will be appreciated.
side question: Given $B\in\Omega$, if $\{C_n\}$ is a sequence of cylinder sets such that $\mu(C_n\,\Delta\,B)\rightarrow 0$, is there a necessary condition for $\mu\big[T(B)\,\Delta\,T(C_n)\big]\rightarrow 0$?