Let $A_1\supset A_2\supset A_3,\ldots$ and $A_n\rightarrow A$. If $B$ is independent of all $A_n$, then it must be independent of $A$ also, right? I showed this by first noting that $B\cap A_1\supset B\cap A_2,\ldots$, and so continuity from above yields $$\lim\limits_{n\rightarrow\infty} P[B\cap A_n] = P[B\cap A]$$ Using independence, the LHS becomes $$\lim\limits_{n\rightarrow \infty} P[B]P[A_n]=P[B]\lim\limits_{n\rightarrow\infty}P[A_n]=P[B]P[A]$$
The last equality comes from another application of continuity from above. Is this reasoning correct?