False logic from weaking the condition from lemma from Kolmogorov zero-one law

Question

Lemma: Let $B, B_1, B_2, \cdots$ be independent. Then ${B}$ and $\sigma(B_1, B_2, \cdots)$ are independent class, i.e. if $S\in \sigma (B_1, B_2, \cdots)$, then $P(S \cap B)=P(S)P(B)$

What I want to ask is that the following question:

Show that this lemms is false if we require only that $P(B\cap B_n) = P(B)P(B_n)$ for each $n\in \mathbb{N}$, but do not require that the ${B_n}$ be independent of each other.

From A First Look at Rigorous Probability Theory, Jeffrey S. Roseenthal, World Scientific (Lemma 3.5.2, Ex. 3.5.3)

Any hint? Thanks.

Answer 1

I suppose you know the standard construction of three events $B,B_1,B_2$ such that any two of them are independent but the three are not together independent. [See https://math.stackexchange.com/questions/1920473/what-is-the-difference-between-mutually-independent-and-pairwise-independent-eve ]. Take $B_n=\emptyset$ for $n >2$. Then $B$ is not independent of $B_1 \cap B_2$ which belongs to $\sigma (B_1,B_2,\cdots)$ though $B$ is independent of each $B_n$.