I am reading the book Markov Chain and Stochastic Stability by Meyn and Tweedie, and I came across the definition of petite sets that was used later in chapter 15 to derive some ergodicity result of Markov processes (along with the drift condition).
I wonder if the argument applies to the trivial deterministic Markov process (whether it includes the Banach fixed point theorem as a special case). For example, $X_{n+1} = \frac{1}{2}X_n$ with $X_0 \in [-1,1]$. It is clear that this process will converge to the Dirac measure at 0. Does this Markov chain have a petite set?
The definition of petite sets is that: We will call a set $C \in \mathscr{B}(\mathrm{X})$ $\nu_a$-petite if the sampled chain satisfies the bound. $$ K_a(x, B) \geq \nu_a(B), $$ for all $x \in C, B \in \mathscr{B}(\mathrm{X})$, where $\nu_a$ is a non-trivial measure on $\mathscr{B}(\mathrm{X})$.
In the above definition, $K_a(x, B)$ is defined as follows: Let $a=\{a(n)\}$ be a distribution, or probability measure, on $\mathbb{Z}_{+}$, and consider the Markov chain $\boldsymbol{\Phi}_a$ with probability transition kernel $$ K_a(x, A):=\sum_{n=0}^{\infty} P^n(x, A) a(n), \quad x \in \mathrm{X}, A \in \mathscr{B}(\mathrm{X}) $$
I tried to validate the case when C is the unit interval $[-1, 1]$ as a compact set: Take any measurable set B as a subset of $[-1,1]$. However, when I consider $\inf_{x\in[-1,1]} K_a(x, B) $, it seems to be always $0$ except for $B$ such that $ 0 \in B$, because $P(x, B) = I_{\{\frac{x}{2}\in B\}}$? Consequently, the only $\nu_a$ I can think of is $\nu_a(B) =I_{\{0\in B\}}$.
Is there anything wrong with the above reasoning? Does it mean any compact interval containing $0$ is a petite set for this degenerate case?
Right, that $\nu_a$ doesn't work if $C = [-1,1]$. If you take $B = \{0\}$, then, unless $x = 0$, $ K_a(x, B) = \sum_{n=0}^{\infty} a(n) I(2^{-n}x \in B) < 1 = \nu_a(B).$ The left hand side is either $0$ or some $a(k) < 1$.
Then again, if you let $C = \{0\}$, then
$$ \inf_{x \in C} K_a(x, B) = K_a(0, B) = \delta_0(B) $$ which works. This $C$ is petite but it's not always reachable, depending on your starting measure or state space, but that’s another story.