Krylov–Bogolyubov existence of an invariant probability

Fix one point $x\in K$ and, for $N\ge 1$, define ${\mu }_N(E)=N^{-1}{\sum }_{j=0}^{N-1}{\delta }_{T^{j}x}(E)$ on the Borel sets. By [F2], each summand is a probability; finite sums preserve countable additivity because a finite sum commutes with the increasing partial sums of a nonnegative series. Thus ${\mu }_N$ is a Borel probability. For indicators its integral is exactly $N^{-1}{\sum }_{j=0}^{N-1}{\mathbf{1}}_E(T^{j}x)$; the simple integral gives this for every nonnegative simple function. Applying [F3] and [F4], with finite sums of increasing limits, gives $\int fd{\mu }_N=N^{-1}{\sum }_{j=0}^{N-1}f(T^{j}x)$ for nonnegative Borel $f$. For bounded real $f$, apply this to $f^{+},f^{-}$ and subtract by [F5]. The starting-point choice is a single existential instantiation, not an axiom of choice.

By [F1] there are $N_r\uparrow \mathrm{\infty }$ and a Borel probability $\mu$ such that $\int fd{\mu }_{N_r}\to \int fd\mu$ for all continuous real $f$. Continuity of $T$ ensures that $f\circ T$ is also continuous. Step 1.1 telescopes to $\int (f\circ T-f)d{\mu }_N=(f(T^{N}x)-f(x))\mathrm{/}N$, of absolute value at most $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}N$. Hence $\int f\circ Td\mu =\int fd\mu$ for every such $f$, by [F5] and passage to the two limits.

Define $\nu (E)=\mu (T^{-1}E)$ for Borel $E$. The class of sets whose inverse images are Borel is a sigma-algebra containing the opens, since $T$ is continuous; thus $\nu$ is defined on all Borel sets. Inverse images commute with complements and disjoint countable unions, so $\nu$ is a Borel probability. For indicators, $\int {\mathbf{1}}_{E}d\nu =\int {\mathbf{1}}_E\circ Td\mu$. The finite simple-integral formula, then [F3] and [F4] on both sides, prove $\int fd\nu =\int f\circ Td\mu$ for every nonnegative Borel $f$, including infinite values. Applying it first to $\mathrm{\mid }f\mathrm{\mid }$ verifies integrability of $f\circ T$ whenever $f$ is $\nu$-integrable; positive/negative decomposition and [F5] then prove the real signed identity. Combining it with step 2.1 gives equality of $\mu$ and $\nu$ on all continuous real integrals.

To pass from these test functions to sets without invoking a stronger-choice LCH theorem, let $F$ be a nonempty closed subset of $K$ and put $h_m(z)=\max (0,1-md(z,F))$ for $m\ge 1$. The infimum defining $d(z,F)$ is 1-Lipschitz by the triangle inequality. It is zero on $F$ and positive outside $F$, since the complement of $F$ is open. Thus $h_m$ is continuous and $1-h_m\uparrow {\mathbf{1}}_{K\setminus F}$. By [F4] and the equal continuous integrals, $\mu (K\setminus F)=\nu (K\setminus F)$; total masses one give $\mu (F)=\nu (F)$. Empty $F$ also has equal measure zero. The closed subsets form a nonempty pi-system containing $K$ and generate the Borel sigma-algebra because their complements are exactly the opens. All hypotheses of [F6] hold, so $\mu =\nu$ on the Borel sets. By the definition of $\nu$, this is precisely [F7]. Countable choice enters through [F1]; the orbit, metric test functions and monotone simple approximants require no further selection principle.