Does $P(A \cap B) = 0$ imply that $A \cap B = \emptyset$? Given that $P(A \cap B) = 0$, where  $A$ and $B$ are two events, does this   imply that $A \cap B = \emptyset$ ?
Is it not possible to have a probability zero for an event which is not empty? 
Regards. 
 A: An equivalent simpler way to ask the question is: if $E$ is an event, does $P(E)=0$ mean that $E$ is impossible?
In the discrete case (event $E$ is discrete, finite or countable) the answer is yes.
In the continuous case, the answer is not always. For instance if the distribution is continuous (normal, for example) then the probability of any countable event will be zero. As you may know, for a normal random variable, $P(Z=z)=0$ for any value of $z$ -- even though the event $"Z=z"$ is possible/nonempty.
This also works for probabilities that do not depend on random variables.
A: No, it does not imply it.
 Take $A=\mathbb{Q}$, the rational numbers inside $B=[0,1]$. Then $A\cap B=A$ has probability zero (with respect to the uniform distribution), but obviosuly $A\cap B$ is not empty.
A: No. That depends on the probability space to be honest. For example if your experiment is choosing a point uniformly in the interval $[0,1]$ then the probability of every single point (actually of any countable subset of $[0,1]$) is $0$. 
A: No. For example, if I pick a random number from $[0,2]$, then $P([0,1]\cap [1,2])=0$, however, $[0,1]\cap[1,2]=\{1\}\neq \emptyset$.
A: For continuous case, let $f(x)$ denote the probability density function for random variable $x$ (e.g. height, weight, etc). Then probability of $x$ lying between $a$ and $b$ is defined as
$P(a\le x\le b)=\int_a^b f(x)dx$.
Now what happens if you find $P(x=a)$?
