If T is a Markov map with associated partition $\{A_i\}$ and $A_{i_0i_1i_2...i_{n-1}}$ an n-cylinder, i.e. $A_{i_0i_1i_2...i_{n-1}}:=A_{i_0}\cap T^{-1}A_{i_1}\cap T^{-2}A_{i_2}\cap ...\cap T^{-(n-1)}A_{i_{n-1}}.$
It should be true that $$\mu(A_{i_0i_1i_2...i_{n-1}})=\mu(A_{i_0})P_{i_0i_1}P_{i_1i_2}...P_{i_{n-2}i_{n-1}}.$$ Where $P_{ij}$ is the transition matrix associated to T. i.e. $$P_{ij}=\frac{\mu(T^{-1}A_j\cap A_i)}{\mu(A_i)}$$
This would be a necessary part of showing metric conjugacy between the Markov map and the associated subshift of finite type with associated so-called "Markov measure" defined exactly as above.
Is there a simple induction proof of this? Do we need that $\mu$ is $T$-invariant? The base case (and the 1-cylinder case) is clear, and I have tried to use some total probability type arguments, but something is eluding me and I have a sneaking suspicion that I'm missing something silly.