Kolmogorov zero-one law

For $m\in \mathbb{N}$, let ${\mathcal{H}}_m:=\sigma (X_0,\dots ,X_m)$ and put $\mathcal{A}:={\bigcup }_{m\in \mathbb{N}}{\mathcal{H}}_m$. Since the family $({\mathcal{H}}_m)$ is increasing, $\mathcal{A}$ is an algebra of events. Define $\mathcal{D}:=\{B\in \sigma (\mathcal{A}):\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\}\mathrm{.}$ Using the continuity statements from [L3], the class $\mathcal{D}$ is closed under increasing unions and decreasing intersections, so it is a monotone class.

For each $m$, [L1] gives ${\mathcal{H}}_m\subseteq \mathcal{D}$. Hence the algebra $\mathcal{A}={\bigcup }_m{\mathcal{H}}_m$ is contained in $\mathcal{D}$.

Because $\mathcal{D}$ is a monotone class containing the algebra $\mathcal{A}$, [L2] yields $\sigma (\mathcal{A})\subseteq \mathcal{D}$. But $\sigma (\mathcal{A})=\sigma (X_n:n\ge 0)$, and [L4] puts every tail event in this full-coordinate sigma-algebra. Therefore $A\in \mathcal{D}$, so $\mathbb{P}(A)=\mathbb{P}(A\cap A)=\mathbb{P}(A{)}^2\mathrm{.}$

Step 2.1 and [L3] show that the probability $p=\mathbb{P}(A)$ satisfies $0\le p\le 1$ and $p=p^2$, so $p\in \{0,1\}$.