A question about the definition of Markov measure

Question

I have one question about this definition Why is it enough to note equation (3.30) to see that the measure is well defined? is it already additive so you can put the measure that way? if not, what is done to complete this?

Answer 1

There are standard steps for constructing a measure that every ergodic theorist/probabilist has to verify at some point in their life, at least in the setting in which they are most interested.

Take the space $X:=\Sigma^\mathbb{N}$ of sequences with values from a finite alphabet $\Sigma$.

1. Together with the empty set, the cylinder sets in $X$ form a semi-algebra $\mathscr{S}$, hence the finite unions of cylinders form an algebra $\mathscr{A}$.

2. The $\sigma$-algebra $\mathscr{F}$ generated by $\mathscr{A}$ (or by $\mathscr{S}$) is the product $\sigma$-algebra on $X$, which is the same as the Borel $\sigma$-algebra for the product topology.

3. Every countably-additive mapping $\pi:\mathscr{S}\to[0,1]$ with $\pi(X)=1$ extends uniquely to a countably-additive mapping $\pi:\mathscr{A}\to[0,1]$ with $\pi(X)=1$. [General measure theory]

4. Every countably-additive mapping $\pi:\mathscr{A}\to[0,1]$ with $\pi(X)=1$ extends uniquely to a probability measure $\pi:\mathscr{F}\to[0,1]$. [General measure theory; called Caratheodory's extension theorem or Kolmogorov's extension theorem, depending on whom you ask]

5. For this particular space, every additive mapping $\pi:\mathscr{S}\to[0,1]$ is in fact countably-additive. [This follows from the compactness of $X$ and the fact that cylinder sets are both open and closed. Namely, you cannot find infinitely many disjoint cylinder sets whose union is again a cylinder set.]

(Some people prefer to follow slightly different steps, for example via Dynkin systems.)

Back to your question, your equation (3.30) implies that $\mu$ is additive on $\mathscr{S}$ with $\mu(X)=1$.