Theorem 5.5.11 in Durrett says: "If $p$ is irreducible and has stationary distribution $\pi$, then $\pi(x)=\frac{1}{E_x[T_x]}$." Here $T_x$ is the first positive time the Markov chain visits state $x$, so $E_x[T_x]$ is the expected time to return to state $x$ starting in state $x$. This would imply that the stationary distribution $\pi$ is unique.
However, I noticed the proof of this theorem cites Theorem 5.5.9, which says "If $p$ is irreducible and recurrent (i.e., all states are) then the stationary measure is unique up to constant multiples." So is the recurrence assumption also necessary for Theorem 5.5.11?
I believe I have a counterexample if we do not add that assumption. Equation 5.6.1 says that $$\frac{1}{n}\sum_{m=1}^n p^m(x,y) \to \frac{\rho_{xy}}{E_y[T_y]} \text{ as } n \to \infty$$ where $\rho_{xy}=P_x(T_y<\infty)$. I used this to show that $\pi_x(y):=\frac{\rho_{xy}}{E_y[T_y]}$ defines a stationary distribution for every $x$.* Since $\rho_{xy}$ is not necessarily $1$, even when $p$ is irreducible, I believe this provides a counterexample to Theorem 5.5.11 unless we add the assumption that $p$ is recurrent. Is this correct?
*I have concerns about this step. It seems obviously false for, e.g., simple random walk in on $\mathbb{Z}^2$ where $E_y[T_y]=\infty$ for all $y$, so $\pi_x(y)=0$ for all $x$. My proof is as follows:
\begin{align*} \sum_x \pi_z(x)p(x,y) &= \sum_x \lim_{n\to \infty} \frac{1}{n}\sum_{m=1}^n p^m(z,x) p(x,y)\\ &=\lim_{n \to \infty} \frac{1}{n}\sum_{m=1}^n \sum_x p^m(z,x)p(x,y)\\ &=\lim_{n\to \infty} \frac{1}{n} \sum_{m=1}^n p^{m+1}(z,y)\\ &=\lim_{n \to \infty} \frac{1}{n}\sum_{m=1}^n p^m(z,y)\\ &=\pi_z(y). \end{align*} This proves $\pi_z(\cdot)$ is stationary. To prove it sums to $1$, note for all $n$: $$\sum_x \frac{1}{n} \sum_{m=1}^n p^m(z,x) = \frac{1}{n} \sum_{m=1}^n \sum_x p^m(z,x) = 1$$ so, \begin{align*} \sum_x \pi_z(x)&=\sum_x \lim_{n\to \infty} \frac{1}{n}\sum_{m=1}^n p^m(z,x)\\ &=\lim_{n\to \infty} \sum_x \frac{1}{n}\sum_{m=1}^n p^m(z,x)\\ &=\lim_{n\to \infty} 1\\ &=1. \end{align*}