Stationary irreducible Markov shift is ergodic

Uniqueness of the invariant probability: [F3] supplies an invariant probability ${\pi }_{\ast }$. Let $\rho$ be any invariant probability and fix an arbitrary $y\in E$. Applying [F11] to each of ${\pi }_{\ast }$ and $\rho$ gives ${\pi }_{\ast }(y)=1\mathrm{/}{\mathbb{E}}_y{T}_y^{+}=\rho (y)$. Since this holds for every $y$, $\rho ={\pi }_{\ast }$ pointwise. Write $\pi$ for this unique invariant probability.

Let ${\mathbb{P}}_{\pi }$ be the canonical path law of the chain with initial law $\pi$; by [F2] the left shift preserves ${\mathbb{P}}_{\pi }$. Fix a measurable $A\subseteq E^{{\mathbb{N}}_0}$ with ${\theta }^{-1}A=A$ and define $h(x):={\mathbb{P}}_x(A)$ for $x\in E$; then $0\le h\le 1$.

The function $h$ is harmonic, $Ph=h$: since ${\theta }^{-1}A=A$, the indicator satisfies ${\mathbf{1}}_A(z)={\mathbf{1}}_A(\theta z)$ for every path $z$, so the bounded product-measurable functional $H:={\mathbf{1}}_A$ obeys $H(X_1,X_2,\dots )={\mathbf{1}}_A(\theta \mathrm{\Phi })={\mathbf{1}}_A(\mathrm{\Phi })$ with $\mathrm{\Phi }=(X_0,X_1,\dots )$; applying [F5] with $n=1$ and taking expectations gives $h(x)={\mathbb{E}}_x[{\mathbf{1}}_A]={\mathbb{E}}_x[h(X_1)]={\sum }_{y\in E}p(x,y)h(y)$ for every $x$.

Since $h$ is bounded and harmonic, [F6] makes $(h(X_n){)}_{n\ge 0}$ a bounded martingale under every ${\mathbb{P}}_x$.

Fix $x,y\in E$ and $n\ge 0$. By [F9] the time $T_y\wedge n$ is a stopping time bounded by the deterministic $n$, so [F7] applied to the bounded martingale of step 4.1 with $\sigma =0$ and $\tau =T_y\wedge n$ gives ${\mathbb{E}}_x[h(X_{T_y\wedge n})]=h(x)$.

Positive recurrence makes every state recurrent, and irreducibility [F10] gives $x\to y$, so [F4] gives ${\mathbb{P}}_x(T_y<\mathrm{\infty })=1$. Hence $T_y<\mathrm{\infty }$ almost surely, $X_{T_y\wedge n}=y$ for all $n\ge T_y$ by [F9], and therefore $h(X_{T_y\wedge n})\to h(y)$ almost surely. The functions are measurable since $h$ is measurable by [F5], and bounded by $1$; applying [F12] under ${\mathbb{P}}_x$ with integrable majorant $1$ gives ${\mathbb{E}}_x[h(X_{T_y\wedge n})]\to h(y)$. Step 5.1 identifies these expectations with the constant $h(x)$. Thus $h(x)=h(y)$, and since $x,y$ were arbitrary, $h\equiv c$ for a single constant $c\in [0,1]$.

Under ${\mathbb{P}}_{\pi }$, [F5] gives ${\mathbb{E}}_{\pi }[{\mathbf{1}}_A\mid {\mathcal{F}}_n]=h(X_n)=c$ almost surely for every $n$. The natural filtration satisfies ${\mathcal{F}}_{\mathrm{\infty }}=\sigma ({\bigcup }_n{\mathcal{F}}_n)=$ the product sigma-algebra on $E^{{\mathbb{N}}_0}$, which contains $A$, so [F8] gives ${\mathbf{1}}_A={\mathbb{E}}_{\pi }[{\mathbf{1}}_A\mid {\mathcal{F}}_{\mathrm{\infty }}]={\lim }_n{\mathbb{E}}_{\pi }[{\mathbf{1}}_A\mid {\mathcal{F}}_n]=c$ ${\mathbb{P}}_{\pi }$-almost surely; as an indicator takes only the values $0,1$ almost surely, $c\in \{0,1\}$ and ${\mathbb{P}}_{\pi }(A)=c$.

Every measurable $A$ with ${\theta }^{-1}A=A$ therefore has ${\mathbb{P}}_{\pi }(A)\in \{0,1\}$, and [F1] says exactly that $\theta$ is ergodic for ${\mathbb{P}}_{\pi }$; $\theta$ preserves ${\mathbb{P}}_{\pi }$ by step 2.1 and $\pi$ is the unique invariant probability by step 1.1, so the corollary holds and no aperiodicity hypothesis was used.

Boundary and axiom cases: if $E$ is a singleton the chain is trivially irreducible and positive recurrent, ${\mathbb{P}}_{\pi }$ is the point mass at the constant path, and every shift-invariant event has measure $0$ or $1$, consistent with steps 7.1–5.1; $A=\mathrm{\varnothing }$ and $A=E^{{\mathbb{N}}_0}$ give $c=0$ and $c=1$; the uniqueness claim and the ergodicity claim are both proved, so the two parts of the statement are not riding on an unproved equivalence; the argument uses the strictly invariant sigma-algebra exactly as in [F1] and never replaces it by the mod-null version; and AC [A1] enters through the chain-law, statewise Kac, and conditional-expectation suppliers [F3]–[F8], [F11], whose statements assume Choice, not through irreducibility itself.