TITLE: If P(A) = 0 or 1 then A and B are independent
QUESTION [5 upvotes]: The question is: Let A,B and C be events in S then prove if P(A) = 0 A and B are independent.
Here's my working:
For two events to be independent P(A ∩ B) needs to equal P(A)P(B). Since P(A) = 0, P(A ∩ B) = P(Ø) = 0 and P(A)P(B) = 0 therefore P(A ∩ B) = P(A)P(B) = 0 and are independent events.
And for part two it asks to prove A and B are independent if P(A) = 1
My working:
P(A ∩ B) = P(B) since P(A) = 1 and P(A)(B) = P(B) since P(A) = 1 therefore P(A ∩ B) = P(A)P(B).
I'm skeptical about how I dealt with the intersections for example "P(A ∩ B) = P(B) since P(A) = 1" and "Since P(A) = 0, P(A ∩ B) = P(Ø) = 0". Can someone clarify if this is right or wrong and if it's wrong how can I make it correct. 
Thanks

REPLY [8 votes]: There are plenty of events with probability $0$ that are not the empty set. So the argument for the case  $\Pr(A)=0$ has to be modified.  Here is how to do it. Because $A\cap B\subseteq A$, we have $\Pr(A\cap B)\le \Pr(A)$. It follows that $\Pr(A\cap B)=0$.  
For the $\Pr(A)=1$  case, you also need to modify the argument.  Note that $B$ is the disjoint union of $A\cap B$ and $A^c\cap B$. Because $A$ has probability $1$, we have that $A^c$ has probability $0$, and therefore so does $A^c\cap B$.  It follows that $\Pr(B)=\Pr(A)\Pr(B)$.
An alternate approach to the $\Pr(A)=1$ case is to work with $A^c$ and  reuse the result of the probability $0$ case.