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A measurable set $A$ satisfying $T^{-1}A=A$, or equality modulo null sets in the measure-theoretic convention. These sets form a $\sigma$-algebra, and an ergodic measure-preserving transformation gives each invariant set probability zero or one.

Consider $A=\{x:{\omega }_1={\omega }_2\}=[0,1/4)\cup [3/4,1)$. The first digit pair is swapped by the transformation, so the equality of its digits is unchanged: ${\theta }^{-1}(A)=A$. Yet $\mu (A)=1/2$. An ergodic measure-preserving transformation has only invariant events of measure zero or one. ThusIts order-two property also explains why averaging along orbits cannot in general sample the entire interval.

A measure-preserving transformation $T$ on a probability space $(X,\mathcal{A},m)$ is ergodic if every invariant event $A\in \mathcal{A}$, meaningup to an $m$-null set, satisfies $m(A)\in \{0,1\}$. Thus the invariant sigma-algebra is trivial modulo null sets.

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.

On a probability space carrying an infinite IID sequence $(X_n)$ with law $m$, map $\omega$ to $(X_1(\omega ),X_2(\omega ),\dots )$. Its pushforward measure $\mu$ satisfiesfor every cylinder set. Cylinder sets form a pi-system generating $\sigma (\mathcal{C})$, so the sigma-finite uniqueness theorem for measures proves uniqueness. This is the countable product measure $m^{\otimes \mathbb{N}}$.

For a cylinder $A=A_1\times \cdots \times A_N\times \mathbb{R}\times \cdots$,The same generating-class argument extends this equality to every measurable set, so $\theta$ is a measure-preserving transformation.

If $A={\theta }^{-1}A$, then $A={\theta }^{-n}A$ for every $n$, so membership in $A$ is independent of the first $n$ coordinates. Thus $A$ belongs to the tail sigma-algebra. The zero-one law gives $\mu (A)\in \{0,1\}$, proving that the shift is ergodic.

Let $T(x)=x+\alpha (\mathrm{mod}1)$ with irrational $\alpha$. This irrational rotation of the circle preserves Lebesgue measure and is ergodic. Apply the Birkhoff ergodic theorem to the indicator function $1_A$. Its limitis invariant, hence almost everywhere constant by ergodicity. Its integral must equal that of $1_A$, so the constant is $\mu (A)$. Thereforefor Lebesgue measure-almost every $x\in (0,1]$.

Suppose first that the system is ergodic and $K_{F}g=g$. For each real $t$, the level setis invariant modulo a null set, so $\omega (E_t)\in \{0,1\}$. The distribution function of $\mathrm{Re}g$ can therefore jump only once, which makes $\mathrm{Re}g$ constant almost everywhere. The same argument applies to $\mathrm{Im}g$.

Conversely, if $E$ is invariant, then $K_F{1}_E=1_E$. If every invariant $L^2$ function is constant, the indicator function $1_E$ is almost everywhere zero or one, and hence $\omega (E)=0$ or $1$. This proves the invariant-function characterization of ergodicity.