Alright, another, hopefully quick, probability question.

Let ($\displaystyle A_n$) and ($\displaystyle B_n$) be in A with $\displaystyle A_n$-->A and $\displaystyle B_n$-->B, with P(B)>0 and P($\displaystyle B_n$)>0. Show that:

a) $\displaystyle lim_{n->\infty}P(A_n|B)$=P(A|B)

b) $\displaystyle lim_{n->\infty}P(A|B_n)$=P(A|B)

c) $\displaystyle lim_{n->\infty}P(A_n|B_n)$=P(A|B)

So, I know that:

$\displaystyle P(A_n|B)$=$\displaystyle \frac{P(B|A_n)P(A_n)}{P(B)}$=$\displaystyle \frac{P(B \cap A_n)}{P(B)}$

$\displaystyle P(A|B_n)$=$\displaystyle \frac{P(B_n|A)P(A)}{P(B_n)}$=$\displaystyle \frac{P(B_n \cap A)}{P(B_n)}$

$\displaystyle P(A_n|B_n)$=$\displaystyle \frac{P(B_n|A_n)P(A_n)}{P(B_n)}$=$\displaystyle \frac{P(B_n \cap A_n)}{P(B_n)}$

$\displaystyle P(A|B)$=$\displaystyle \frac{P(B|A)P(A)}{P(B)}$=$\displaystyle \frac{P(B \cap A)}{P(B)}$

My problem really is how to show the limit of $\displaystyle P(B|A_n)$, $\displaystyle P(B_n|A)$, and $\displaystyle P(B_n|A_n)$ is$\displaystyle P(B|A)$. They seem obvious, but I can't figure out how to show it.