some notational doubt on probability distribution on metric space. $P: \mathcal{M}_{f i n}(X) \rightarrow \mathcal{M}_{f i n}(X)$ be a Markov operator, $\mathcal{M}_{f i n}(X)$ be the space of all be the set of all Borel finite
measures on $X$. $\{X_n\}_{n\ge 0}$ be a homogeneous Markov chain on $X$, such that $PL(X_n)=L(X_{n+1})$, where $L(X_n)$ denotes the distribution of $X_n$.


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*Could anyone tell me what does the last sentence means mathematically? does it mean if that if $L(X_n)$ is given by some probability measure say $\mu_n$ then $P\mu_n=\mu_{n+1}$?

*Could anyone explain to me the action of $P$ in a bit detail?

*Why $P\mu_n=\mu_{n+1}$? $\mu_n, \mu_{n+1}\in \mathcal M_1(X)$. $X$ is a polish space. 

*On page 3, what is the assumption $H$ means geometrically or analytically in simple words, means what is the picture behind the assumption? Thanks!
based on page $1,2,3$ of 
1Rafał Kapica, Maciej Ślȩczka: Law of large numbers for random iteration; DOI:10.1080/10236198.2017.1332054
 A: 1). Yes. Yes.
2). I'm not sure what you're asking. Maybe this will help. Given the Markov chain $(X_n)_n$ (with initial distribution $\mu_0$ for $X_0$), we can ask what the distribution of $X_1$ is. If $\overline{P}$ is the transition matrix for the Markov chain, i.e. $\mathbb P(X_1=x | X_0 = y) = \mu_0(y)\overline{P}_{xy}$, then the distribution of $X_1$ is given by $\mu_1(x) = \sum_y \mu_0(y)\overline{P}_{xy}$, so the map $P$ is given explicitly as $P(\mu)(x) = \sum_y \mu(y)\overline{P}_{xy}$. The point is that, if we are given the distribution of $X_n$, i.e. the probability distribution of being at certain states of the Markov chain at time $n$, then $P$ finds the probability distribution of being at certain states of the Markov chain at time $n+1$.
3). I have no idea what you're trying to ask here, but hopefully my answer to (2) helps.
4). $\mu_*$ is the stationary distribution of the Markov chain; you should think of it as the "perfectly mixed" distribution (though that is sometimes misleading). For example, if the Markov chain is a random walk on a finite, regular graph, then $\mu_*$ is the uniform distribution. What assumption $H$ is saying is that you have rapid convergence to the stationary distribution: if you know you start off at a given state $x$, then after $n$ iterations, you are very close to being equally likely to be anywhere. The exact form of having a continuous function $\gamma(x)$ is more for technical reasons (see point (ii) of remark 1 for a nice corollary of that technical part of assumption H), but the main part is that $q^n$, rapid convergence.
