Is it that $P(A\mid\emptyset) = 1?$ I think I get that if $P(B) = 0$, we need not have $P(A\mid B) = 1$ from 1 2 3 4 5 6 7 8
But what if $B = \emptyset$?
'$\omega \in \emptyset \to \omega \in A$' is a vacuously true statement right?
It seems then that $P(A\mid\emptyset) = 1$
 A: The definition of the conditional expectation (applied to the special case of the conditional probability) says: $Y=P(A\mid \mathscr B)$ is a $\mathscr B$ measurable random variable defined with probability $1$ such that
$$\int_B\chi_A(\omega)\ dP(\omega)=\int_B Y(\omega) \ dP(\omega)$$
for all $B\in\mathscr B\subset \mathscr A$ with respect to a probability space $(\Omega,\mathscr A, P).$ So, for all $\omega\in\Omega$ (except a set of measure $0$)
$$Y(\omega)=P(A\mid \mathscr B)(\omega).$$

In our case $\mathscr B=\{\Omega,\emptyset\} $ and we are looking for a $\{\Omega,\emptyset\}$ measurable function $Y$ for which 
$$P(A)=\int_{\Omega}\chi_A(\omega) dP(\omega)=\int_{\Omega}Y(\omega)\ dP(\omega).$$
$Y(\omega)$ cannot change as $\omega$ is changing since it has to be $\{\Omega,\emptyset\}$ measurable. So,
$$Y(\omega)=P(A\mid \mathscr B)(\omega)=P(A),\ \, \forall \omega \in \Omega.$$
Since, $\not \exists \ \omega \in \emptyset$ $Y(\omega)=P(A\mid \mathscr B)(\omega)$ is not defined over $\emptyset.$
That is,
$$P(A\mid \{\Omega,\emptyset\})=P(A)=P(A\mid \Omega)$$
is a constant. 

$P(A\mid \emptyset)$ was not defined. But this was not because of $P(\emptyset)=0$. In the theory of conditional expectation events of zero probability may appear as conditions. This and that $P(A)=P(A\mid \Omega)$ may be tempting to try to say something about $P(A\mid \emptyset)$. It seems that $P(A\mid \emptyset)$ has to remain undefined.
