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On a one-sided sequence space $E^{\mathbb{N}}$ with product measure, the Bernoulli shift isIt preserves the product measure. Every invariant event lies in the tail sigma-algebra of the coordinate process, so the Kolmogorov zero-one law makes the shift an ergodic transformation.

The tail sigma-algebra isIts events depend only on arbitrarily late coordinates. The Kolmogorov zero-one law states that for a sequence of independent random variables, every tail event has probability zero or one:Identical probability distributions and integrability are not required for the Kolmogorov zero-one law.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

The Kolmogorov zero-one law says that for independent random variables $(Y_n{)}_{n\ge 0}$, every event in the tail sigma-algebrahas probability zero or one.

In the canonical model of an independent and identically distributed sequence, the sample space is a sequence space $E^{\mathbb{N}}$ with the product law, $Y_n$ is the $n$th coordinate, and the left Bernoulli shift $S$ satisfies $Y_n\circ S=Y_{n+1}$. If $A$ is shift-invariant, then for every $N$,so $A\in \mathcal{T}$. The zero-one law therefore says that the canonical one-sided i.i.d. shift is ergodic.

Yes. Let $A$ be $T$-invariant. Since the itinerary variables generate $\mathcal{A}$ modulo null sets, $A$ can be regarded as an event in the canonical sequence model. For every $N$, invariance and $X_n\circ T=X_{n+1}$ implyThus $A$ belongs to the tail sigma-algebra of the i.i.d. sequence $(X_n)$. By the Kolmogorov zero-one law, $m(A)$ is zero or one. Hence the tent map is an ergodic measure-preserving transformation.

Let $(X_n)$ be independent random variables and letbe their tail sigma-algebra. The Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Indeed, $A$ is independent of $\sigma (X_1,\dots ,X_n)$ for every $n$. The union of these finite-coordinate sigma-algebras generates $\sigma (X_1,X_2,\dots )$, and independence extended from generating pi-systems shows that $A$ is independent of that entire sigma-algebra. Since $A$ itself belongs to it,so $\mathbb{P}(A)\in \{0,1\}$.

A tail event is measurable with respect to a tail sigma-algebra, so changing finitely many coordinates does not affect whether it occurs.

For a sequence of finite real random variables, ${\mathrm{limsup}}_n{n}^{-1}{\sum }_{j=1}^n{X}_j$ is measurable with respect to its tail sigma-algebra: deleting finitely many summands changes each sample mean by a quantity tending to zero. When the random variables are independent random variables, the Kolmogorov zero-one law makes every finite such limit constant almost surely.