 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 probabilitiesThe 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 $T({\omega }_1,{\omega }_2,\dots )=({\omega }_2,{\omega }_3,\dots )$. 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 $\mathbb{P}(A△C)<ϵ$. Such approximations follow from the fact that finite-coordinate cylinder events generate the product $\sigma$-algebra. For $n\ge r$, independence gives $\mathbb{P}(C\cap T^{-n}C)=\mathbb{P}(C{)}^2$. Invariance and measure preservation giveAlso $|\mathbb{P}(A{)}^2-\mathbb{P}(C{)}^2|\le 2ϵ$. Letting $ϵ\downarrow 0$ proves $\mathbb{P}(A)=\mathbb{P}(A{)}^2$, so $\boxed{\mathbb{P}(A)\in \{0,1\}}$. This is exactly ergodicity.