Kolmogorov zero-one law from martingale convergence

Fix $A\in \mathcal{T}$ and let ${\mathcal{H}}_n$ be generated by the first $n$ random elements. F2, F3 give $\mathbb{E}[1_A\mid {\mathcal{H}}_n]=\mathbb{P}(A)\text{a.s. for every }n\mathrm{.}$

The increasing union of the ${\mathcal{H}}_n$ generates the $\sigma$-algebra of the entire sequence. By F1, $A$ belongs to that $\sigma$-algebra. F4 therefore says the constants in step 1.1 converge almost surely to $\mathbb{E}[1_A\mid \sigma ({\bigcup }_n{\mathcal{H}}_n)]=1_A,$ where the last identity is F3's known-variable clause. Hence $1_A=\mathbb{P}(A)$ almost surely.

An indicator that is almost surely the constant $p=\mathbb{P}(A)$ can take only $0$ or $1$ on a nonnull set, so $p\in \{0,1\}$. This proof does not consume the already-published zero-one theorem that it recovers. AC has only the dependence in F5.