Let $(A_i)_{i\in N}$ be a sequence of events which converge to an event $A$. Let $(B_i)_{i\in\mathbb{N}}$ be a sequence of events which converge an event $B$. Additionally assume $\mathbb{P}(A_i) > 0$ for all $i \in \mathbb{N}$ and $\mathbb{P}(B) > 0$.
Prove: $\lim\limits_{i\rightarrow\infty}\mathbb{P}(A|B_i) = \mathbb{P}(A|B)$
Proof: Either $B_i \subseteq B_{i+1}$ or $B_i \supseteq B_{i+1}$ is true.
For $B_i \subseteq B_{i+1}$: \begin{align*} \lim\limits_{i\rightarrow\infty}\mathbb{P}(A|B_i) &= \lim\limits_{i\rightarrow\infty}\frac{\mathbb{P}(A \cap B_i)}{\mathbb{P}(B_i)}&\\ &= \frac{\mathbb{P}\left(A \cap \left(\bigcup_{i\in\mathbb{N}}B_i\right)\right)}{\left(\bigcup_{i\in\mathbb{N}}B_i\right)}&\\ &= \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}&\\ &= \mathbb{P}(A|B)&\\ \end{align*}
For $B_i \supseteq B_{i+1}$: \begin{align*} \lim\limits_{i\rightarrow\infty}\mathbb{P}(A|B_i) &= \lim\limits_{i\rightarrow\infty}\frac{\mathbb{P}(A \cap B_i)}{\mathbb{P}(B_i)}&\\ &= \frac{\mathbb{P}\left(A \cap \left(\bigcap_{i\in\mathbb{N}}B_i\right)\right)}{\left(\bigcap_{i\in\mathbb{N}}B_i\right)}&\\ &= \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}&\\ &= \mathbb{P}(A|B)&\\ \end{align*}
It seems to me, as if I have missed some caveat, but I am not sure. Is it problematic that in some cases of the series $\mathbb{P}(B) = 0$ might be true?
We already proved that $\lim\limits_{i\rightarrow\infty}\mathbb{P}(B_i) = \mathbb{P}\left(\bigcup_{i\in\mathbb{N}}B_i\right)$ for $B_i \subseteq B_{i+1}$ and vice versa.