Borel field in statistics theory

Question

I know this might be basic but I have been struggling to show that $$P(A\cap B^c) \le 1-P(A\cap B)$$ The inequality simply throws me off. How do I show that the above statement is true?

Answer 1

We are aware that $A \cap B$ has no intersection with $A \cap B^{c}$ by definition; we know that $A = (A \cap B) \cup (A \cap B^{c})$. By axiom of probability we have $P(A) = P(A \cap B) + P(A \cap B^{c})$; by axiom we have $P(A) \leq 1$. I believe you can complete it from now on.

By the way, the statement has nothing to do with "Borel field". The only requirement is that $A$ is measurable in a given sense of course for it to be meaningful to write $P(A)$.

Answer 2

Hint: add the rightmost term to both sides, then use the disjointness of $A \cap B$ and $A \cap B^C$ to combine the terms on the left.