Theorem 6.1.1. of Durrett - Proof that $X_n$ is a Markov chain given a probability measure induced on sequence space.

Question

I have difficulty following the proof of Theorem 6.1.1. from Durrett's Probability Theory and Examples.

Before proceeding to the theorem, below are the definitions used in the text.

Now follows the proof. I don't understand the paragraph starting "To do this,...". How do we set $f$ as an indicator function to replace $p(x_n,B_n)$ when $p$ depends on two variables and the ps in the integral are of the form $p(x_{n-1},dx_n)$? I'm very confused about this part. I would greatly appreciate any help.

Answer 1

How do we set $f$ as an indicator function to replace $p(x_n,B_n)$ when $p$ depends on two variables and the ps in the integral are of the form $p(x_{n-1},dx_n)$?

• We are only replacing $p(x_n, B_{n+1})$ by $f(x_n)$. (There is a small typo.) We are not replacing the $p(x_i,dx_{i+1})$.
• $B_{n+1}$ is fixed, so we are thinking of the function $x_n \mapsto p(x_n,B_{n+1})$ generally as some function $f(x_n)$.

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Demonstration for a characteristic function: essentially uses (6.1.1). \begin{align} &\int_{B_0} \mu(dx) \int_{B_1} p(x_0,dx_1) \cdots \int_{B_n} p(x_{n-1}, dx_n) 1_{(x_n \in C)}\\ &= \int_{B_0} \mu(dx) \int_{B_1} p(x_0,dx_1) \cdots \int_{B_n \cap C} p(x_{n-1}, dx_n)\\ &= P_\mu(X_0\in B_0,\ldots, X_{n-1} \in B_{n-1}, X_n \in B_n \cap C)\\ &= P_\mu(A \cap \{X_n \in C\})\\ &=\int_A 1_{(X_n \in C)} \mathop{dP_\mu} \end{align}