Relation between two versions of the Second Borel Cantelli lemma

In Durett's book, the second Borel Cantelli lemma is as follows:

Let $$\{F_n\}$$ be a filtration, and $$A_n\in F_n$$ be a sequence of events. Then, $$\{A_n \text{ i.o.}\}=\{\omega:\sum_{n=1}^\infty P(A_n|F_{n-1})=\infty\}$$ a.s..

But the more common version is: Let $$A_n$$ be a sequence of events. If $$\sum_n P(A_n)=\infty$$, and $$A_n$$ are independent, then $$P(A_n \text{ i.o.})=1$$.

How can Durett's version imply the common version? It seems that the independence of $$A_n$$ cannot imply $$A_n$$ is independent with $$F_{n-1}$$.

Assume the $$A_i$$ are independent.

Take $$F_n := \sigma (A_1 , \ldots , A_n)$$. Then $$A_n$$ is independent from $$F_{n-1}$$. Thus $$P(A_n \vert F_{n-1}) = P(A_n )$$.

Now if $$\sum_n P (A_n ) = \infty$$ then

$$P( A_n \text{ i.o.}) = P\left( \sum_n P(A_n \vert F_{n-1}) = \infty\right) = P \left( \sum_n P(A_n) = \infty \right) = P (\Omega)= 1$$

But (for completeness) there is more: The common Borel-Cantelli usually includes also that for arbitrary $$A_n$$, if $$\sum_n P(A_n ) < \infty$$, then $$P(A_n \text{ i.o })=0$$.

By monotone convergence theorem:

$$E[\sum_n P (A_n \vert F_{n-1} )]= \sum_n E[P(A_n \vert F_{n-1})] = \sum_{n} P(A_n) <\infty$$ This implies that $$P(\sum_n P(A_n \vert F_{n-1}) = \infty ) = 0$$, thus $$P(A_n \text{ i-o.}) = 0$$.