If the probability measure $\mu$ is a left-eigenvector to the eigenvalue $1$ of a stochastic matrix $p$, then $\mu p^n=\mu$

Question

Let

• $E$ be an at most countable set and $\mathcal E$ be the discrete topology on $E$
• $p=\left(p(x,y)\right)_{x,y\in E}$ be a stochastic matrix
• $\mu$ be a probability measure on $(E,\mathcal E)$

Let's define $$\mu p^n\left(\left\{x\right\}\right):=\sum_{y\in E}\mu\left(\left\{y\right\}\right)p^n(y,x)\;\;\;\text{for }n\in\mathbb N\text{ and }x\in E$$ and assume, that $\mu p=\mu$. How can we show, that we need to have $$\mu p^n=\mu\;,$$ too?

Answer 1

Hint. As matrix multiplication is associative, we have $$ \mu p^n = (\mu p^{n-1})p $$ Now start with $\mu p = \mu$ and use induction.