Let $(\Omega,\,P(\cdot))$ denote a probability space and suppose that %B% is an event in this space where $P(B)>0$. Prove $P(A|B)\ge0$ for all events $A\subseteq\Omega$.
So far I've done:
Using Bayes Theorem, $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
The question tells us that $P(B) > 0$, so we just have to prove that $P(A \cap B)\ge0$.
I think intuitively it makes sense that $A \cap B$ is non-negative since $A\subseteq\Omega$ and $B\subseteq\Omega$, but I'm not sure how to formally prove it.
Note that a fraction of real numbers $\dfrac{a}{b}$ with $b\neq 0$ is negative if and only if either $a$ or $b$ is negative but not if both have the same sign.
You have already used Bayes theorem to see that $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$. Since $P(B) > 0$ the only way to obtain a negative fraction would be if $P(A\cap B)<0$. But $P$ is a probability measure so it always takes values in $[0,1]$. Hence the whole fraction is positive and so $P(A|B)$.