Let $(\Omega, \mathcal{F}, P)$ be a probability space. Let $\mathcal{E}_1, \mathcal{E}_2,...$ be a sequence of finite, measurable partitions of $\Omega$ such that each $\mathcal{E}_{n+1}$ is a refinement of $\mathcal{E}_n$, and assume that $\sigma(\cup_n \mathcal{E}_n) = \mathcal{F}$. Let $E_n(\omega)$ be the cell of $\mathcal{E}_n$ that contains $\omega$. For all $A$ and almost all $\omega$, the Levy 0-1 law says that $$P(A \mid E_n(\omega)) \to \mathbf{1}_A(\omega)\tag{1}$$ In contemporary probability textbooks, this result is usually presented as an immediate consequence of the martingale convergence theorem for conditional expectations.
Interestingly, I have recently discovered a much more elementary proof of this result in Halmos's Measure Theory (Theorem 49.B). Besides the basic properties of conditional probability, the only result that Halmos's proof appeals to is the standard fact that measurable sets can be approximated by sets in a generating algebra. In our context, for all $A \in \mathcal{F}$ and all $\epsilon > 0$, there exists $n$ and $E_1,...,E_k \in \mathcal{E}_n$ such that $$P(A \triangle E)< \epsilon, \tag{2}$$ where $E=E_1 \cup ... \cup E_k$.
From (2), Halmos's straightforwardly derives (1), which leads to my question:
Assuming (1), is there a straightforward way to derive (2)?
We might start by noticing that $P(A \mid E_n) \to \mathbf{1}_A$ a.s. iff $P(A^c \mid E_n) \to \mathbf{1}_{A^c}$ a.s. and for all $E \in \sigma(\mathcal{E}_n)$ $$P(A \triangle E) = P(A \cap E^c) + P(A^c \cap E) = \int_{E^c}P(A \mid E_n)dP + \int_E P(A^c \mid E_n)dP.$$ Then I thought about trying something like, let $F_n = \{\omega: P(A^c \mid E_n(\omega)) < 1- \epsilon\} \in \sigma(\mathcal{E}_n)$ so that $F_n^c = \{\omega: P(A \mid E_n(\omega)) \leq \epsilon\}$. Then, $$P(A \triangle F_n) \leq \epsilon P(F_n^c) + (1-\epsilon) P(F_n).$$ But I'm stuck here and not sure this is going to work. Any hints are appreciated.
Yeah, I think you are on the right track by looking at the set where the conditional probability is large. But you can be more crude.
Let $X_n(\omega) = P(A \mid E_n(\omega))$ be the conditional probability. We are assuming $X_n \to 1_A$ a.s., but all we actually need is convergence in probability. Let $G_n = \{X_n \ge 1/2\}$ which is $\mathcal{E}_n$ measurable and thus a finite union of the sets of that partition. Then on $A \triangle G_n$ we have $|X_n - 1_A| \ge 1/2$, so $P(A \triangle G_n) \le P(|X_n - 1_A| \ge 1/2)$ which goes to zero by the convergence in probability.
Thus we can choose $n$ so large that $P(A \triangle G_n) < \epsilon$, and $G_n$ is the desired set.