= Solution
For a probability distribution $m$ on $\mathbb R$, take the canonical sample space $\Omega=\mathbb R^{\mathbb N}$, the product of the Borel $\sigma$-algebras, and the <product measure> $\mathbb P=m^{\otimes\mathbb N}$. The coordinate functions $X_j(\omega)=\omega_j$ have joint cylinder probabilities
$$
\mathbb P(X_1\in A_1,\ldots,X_r\in A_r)=\prod_{j=1}^rm(A_j).
$$
The <Kolmogorov extension theorem> supplies this measure from the consistent finite-dimensional distributions, and uniqueness follows from the generating cylinder sets. Thus the coordinates are <independent and identically distributed>; the same construction works for a given standard Borel state space.
The one-sided <Bernoulli shift> is $T(\omega_1,\omega_2,\ldots)=(\omega_2,\omega_3,\ldots)$. It is measurable and preserves all cylinder probabilities. The uniqueness theorem for probability measures extends this to the product $\sigma$-algebra, proving measure preservation.
To prove <ergodicity>, let $A$ be invariant modulo null sets and approximate it in probability by an event $C$ depending on only the first $r$ coordinates, with $\mathbb P(A\triangle C)<\epsilon$. Such approximations follow from the fact that finite-coordinate cylinder events generate the product $\sigma$-algebra. For $n\geq r$, independence gives $\mathbb P(C\cap T^{-n}C)=\mathbb P(C)^2$. Invariance and measure preservation give
$$
\mathbb P(A)=\mathbb P(A\cap T^{-n}A),\qquad |\mathbb P(A)-\mathbb P(C\cap T^{-n}C)|\leq2\epsilon.
$$
Also $|\mathbb P(A)^2-\mathbb P(C)^2|\leq2\epsilon$. Letting $\epsilon\downarrow0$ proves $\mathbb P(A)=\mathbb P(A)^2$, so \b[$\boxed{\mathbb P(A)\in\{0,1\}}$]. This is exactly ergodicity.
Back to article page
