Conditional block information rate given the infinite past

For $k\ge 0$, $n\ge 1$ and each word $w\in {\mathcal{A}}^n$, define $q_{k,n}(w)$ on each positive-probability atom of $\sigma (X_{-k},\dots ,X_{-1})$ as the conditional average of $1_{\{X_0^{n-1}=w\}}$ there; on null atoms use the uniform vector on ${\mathcal{A}}^n$. Refining the past partition shows that for fixed $n,w$, $(q_{k,n}(w){)}_{k\ge 0}$ is a martingale bounded in $[0,1]$. For rational $0\le u<v\le 1$, hold one unit after the martingale first falls to at most $u$, sell it when the martingale next reaches at least $v$, and repeat. Each finite-horizon gain is a sum of predictable holdings times martingale increments, so its expectation is zero by the martingale property. Every completed upcrossing earns at least $v-u$, and an unfinished holding loses at most one because the martingale stays in $[0,1]$. Thus the expected completed upcrossing count through any horizon is at most $(v-u{)}^{-1}$. The counts increase with the horizon, so [F4] gives finite expected total count and hence finitely many total upcrossings almost surely for each rational pair. A bounded real sequence that does not converge has distinct liminf and limsup, hence has infinitely many upcrossings of some rational interval strictly between them. A countable intersection over rational pairs, words and lengths therefore gives simultaneous almost-sure limits $q_{\mathrm{\infty },n}(w)$.

For each $n$, the limits are nonnegative and sum to one because there are finitely many words and every $q_{k,n}$ is a probability vector. They are ${\mathcal{F}}^{-}$-measurable. Every finite-past cylinder $B$ belongs to the conditioning sigma-algebra for all sufficiently large $k$; [F5], with majorant one on the probability space, then gives ${\int }_B{q}_{\mathrm{\infty },n}(w)dP=P(B\cap \{X_0^{n-1}=w\})$. For fixed $n,w$, the finite measures $B\mapsto {\int }_B{q}_{\mathrm{\infty },n}(w)dP$ and $B\mapsto P(B\cap \{X_0^{n-1}=w\})$ agree on the algebra of finite-past cylinders. Their agreement class is a monotone class by continuity of finite measures, so [F3] extends the identity to every $B\in {\mathcal{F}}^{-}$. Thus $q_{\mathrm{\infty },n}(w)$ is a conditional probability version of the block given the full past.

Fix a positive-probability atom $B$ of the $k$-past partition. Applying [F2] to $B$ and its successive intersections with the realized symbol events factors $P(X_0^{n-1}=w\mid B)$ into the one-symbol conditional probabilities on the refined atoms $B\cap \{X_0^{i-1}=w_0^{i-1}\}$. By stationarity, the factor for $X_i=w_i$ is $p_{k+i}(w_i)\circ T^i$, because the past $X_{-(k+i)}^{-1}$ at the shifted point $T^{i}x$ is exactly $X_{-k}^{i-1}$ at the original point $x$. Therefore, almost surely on the realized block, $q_{k,n}(X_0^{n-1})={\prod }_{i=0}^{n-1}(p_{k+i}(X_0)\circ T^i)\mathrm{.}$ For fixed $n$, each factor converges almost surely as $k\to \mathrm{\infty }$ to $p_{\mathrm{\infty }}(X_0)\circ T^i$, since the convergence in [F1] holds almost surely and each $T^i$ preserves $P$. The left side converges to $q_{\mathrm{\infty },n}(X_0^{n-1})$ by step 1.1. The realized one-symbol limits are positive almost surely by [F1] and stationarity, so $-{\log }_2{q}_{\mathrm{\infty },n}(X_0^{n-1})={\sum }_{i=0}^{n-1}{g}_{\mathrm{\infty }}\circ T^i\text{almost surely}\mathrm{.}$

Apply [F6] to $g_{\mathrm{\infty }}$. By [F1] its integral is $h$, so the right side of step 2.2 divided by $n$ converges almost surely to $h$. This proves the asserted conditional block-information rate and uses no choice axiom.