Proof. We will employ a common inequality involving the $n$ -step transition probabilities: $$P_{ij}^{m+n}\ge P_{ik}^m P_{kj}^n$$ for any $i,j,k\in I$ and non-negative integers $m,n$ .

Suppose first that $d(i)=0$ . Since $i\leftrightarrow j$ , $\displaystyle{ P_{ij}^n>0}$ and $P_{ji}^m>0$ for some $n,m\ge 0$ . This implies that $P_{ii}^{m+n}>0$ , which forces $m+n=0$ or $m=n=0$ , and hence $j=i$ .

Next, assume $d(i)>0$ , this means that $N(i)\ne \varnothing$ . Since $i\leftrightarrow j$ , there are $r,s\ge 0$ such that $P_{ji}^r>0$ and $P_{ij}^s>0$ , and so $P_{jj}^{r+s}>0$ , showing $r+s\in N(j)$ . If we pick any $n\in N$ , we also have $\displaystyle{ P_{jj}^{r+n+s}\ge P_{ji}^r P_{ii}^n P_{ij}^s>0 }$ , or $r+s+n\in N(j)$ . But this means $d(j)$ divides both $r+s$ and $r+s+n$ , and so $d(j)$ divides their difference, which is $n$ . Since $n$ is arbitrarily picked, $d(j)\mid d(i)$ . Similarly, $d(i)\mid d(j)$ . Hence $d(i)=d(j)$ .