Prove if $P(A | B^c) = P(A | B)$ then the events A and B are independent. So I've started by saying that since $P(A | B^c) = P(A | B)$ we know that $\frac{P(A \cap B^c)}{P(B^c)} = \frac{P(A \cap B)}{P(B)}$. However I'm not sure where to go from there. Any help would be great!

The defining relation is: $$P(A\cap B)=P(A|B)P(B)$$ 
You could also write: $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$ but that asks for an apart treatment of the special case $P(B)=0$. 
Let's say that $P(A\mid B)=c=P(A\mid B^c)$. 
Then:
$$P(A)=P(A\mid B)P(B)+P(A\mid B^c)P(B^c)=cP(B)+cP(B^c)=c=P(A\mid B)$$
So: $$P(A\cap B)=P(A|B)P(B)=P(A)P(B)$$
