Does an unconditional probability of 1 or 0 imply a conditional probability of 1 or 0 if the condition is possible? I can't seem to find any references saying that an event with probability 1 or 0 is independent of any event with positive probability, if that's even true and can't seem to be able to extend the argument here soooo......
Suppose we have a probability space $(\Omega, \mathscr F, \mathbb P)$ and let A and B be events s.t. $P(B) > 0$.
If $P(A) = 1 (/0)$, then $A = \Omega(/\emptyset)$ a.s.
$\to A \cap B = \Omega(/\emptyset) \cap B \ \text{a.s.} \ \tag{*}$
$$\to P(A \cap B) = P(\Omega(/\emptyset) \cap B)$$
$$\to P(A \cap B) = P(B(/\emptyset))$$
$$\to P(A|B)P(B) = P(B(/\emptyset))$$
$$\to P(A|B) = 1(/0) \ QED$$
Is that right? What is the justification or alternative for $(*)$?
 A: I almost surely do not know what is meant by a.s. in the equation
tagged with a $*$ in your question, but the proof of the independence
stuff is straightforward.
Given any event $B$, not necessarily of positive probability, we can
express it as the disjoint union of the events $A\cap B$ and
$A^c\cap B$, that is, $B =  (A\cap B) \cup (A^c\cap B)$.
Hence we have that
$$P(B) = P(A\cap B) + P(A^c\cap B).\tag{1}$$

*

*If $P(A) = 1$ (i.e. $P(A^c) = 0$), then, since $A^c \cap B \subset A^c$,
we have  $P(A^c \cap B) \leq P(A^c) = 0$, that is,
$P(A^c \cap B) = 0$. It follows from $(1)$ and the assumption that
$P(A) = 1$ that
$$P(B) = P(A\cap B) \Longrightarrow P(A)P(B) = P(A\cap B),$$
that is, $A$ and $B$ are independent events.


*If $P(A) = 0$ (i.e. $P(A^c) = 1$), then, since $A\cap B \subset A$,
we have that $P(A\cap B) \leq P(A) = 0$ and so
$$0 = P(A\cap B) = P(A)P(B),$$
that is, $A$ and $B$ are independent events.
Events of probability $1$ (or of probability $0$) have the
property that they are independent of all other events
including (somewhat surprisingly) themselves!
A: Prove $P(A|B) = 1$ if $P(A) = 1, P(B) > 0$:
$$P(A) = 1$$
$$\to 1_A = 1_\Omega \ \text{a.s.}$$
$$\to 1_A 1_B =1_\Omega 1_B \ \text{a.s.}$$
$$\to 1_{A \cap B} =1_B \ \text{a.s.}$$
$$\to P(A \cap B) = P(B)$$
$$\to P(A|B)P(B) = P(B)$$
$$\to P(A|B) = 1 \ QED$$
The last line assumes $P(B) > 0$.

Prove $P(A|B) = 0$ if $P(A) = 0, P(B) > 0$:
$$P(A) = 0$$
$$\to 1_A = 1_{\emptyset} \ \text{a.s.}$$
$$\to 1_A 1_B = 1_{\emptyset} 1_B \ \text{a.s.}$$
$$\to 1_{A \cap B} = 1_{\emptyset} \ \text{a.s.}$$
$$\to P(A \cap B) = P(\emptyset)$$
$$\to P(A|B)P(B) = 0$$
$$\to P(A|B) = 0 \ QED$$
Note: The last line assumes $P(B) > 0$.

Prove A and B are independent if $P(A) = 0$
$$A \cap B \subseteq A$$
$$\to 0 \le P(A \cap B) \le P(A) = 0$$
Also, $P(A)P(B) = 0$. Hence, we have
$$P(A \cap B) = P(A)P(B) \ QED$$
Note: This does not seem to assume that $P(B) > 0$

Prove A and B are independent if $P(A) = 1, P(B) > 0$
$$P(A \cap B) = P(A|B)P(B) = P(B) \tag{*}$$
$$P(A)P(B) = P(B)$$
$$\to P(A \cap B) = P(A)P(B) \ QED$$
Note: $(*)$ makes use of '$P(A|B) = 1$ if $P(A) = 1, P(B) > 0$'
