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A subset of $\mathbb{R}$ is a Borel set when it belongs to the Borel sigma-algebra, the smallest sigma-algebra containing every open set of $\mathbb{R}$.

Take the unit circle $X=\mathbb{R}/\mathbb{Z}$ with its Borel sigma-algebra and normalized Lebesgue measure, and letThis is the doubling map. Its two inverse branches halve length, so Lebesgue invariance of the doubling map gives $\mu (T^{-1}A)=\mu (A)$. Moreoverfor every point of the circle.

The event $\{X_n=1\}$ is $T^{-n}(0,1/2)$. Since $T^n$ has $2^n$ affine branches of slope magnitude $2^n$, this preimage is, up to endpoints, a disjoint union of $2^n$ intervals of length $2^{-(n+1)}$ and has measure $1/2$.

More generally, for every binary word $(a_0,\dots ,a_{N-1})$, its itinerary cylinderis, up to finitely many endpoints, one monotonicity interval of $T^N$ and has measure $2^{-N}$. Thereforeso the $X_n$ are independent and identically distributed random variables, each with the Bernoulli distribution of parameter $1/2$.

The finite itinerary cylinders have diameters at most $2^{-N}$. Their endpoints form a dense set, so their unions generate the Borel sigma-algebra after completion by null sets. Thusin the standard probability-space convention that σ-algebras are identified modulo null sets.

There is a small literal endpoint defect in the uncompleted formulation of the question: $0$, $1$, and $2/3$ all have the all-zero itinerary, so $\sigma (X_0,X_1,\dots )$ cannot separate these Borel singletons. Consequently the displayed equality is false as an equality of raw Borel σ-algebras with the stated open interval, but it is true after completion and modulo null sets, which is the version used in the ergodicity argument.