Kolmogorov's $0-1$ Law:
For any terminal event $A$ we have that either $\mathbb{P}(A)=1$ or $\mathbb{P}(A)=0$. Alternatively any $F_{\infty}$ measurable random variable (so basically a terminal random variable) is almost surely constant.
My question:
I understand very well the first part but can't really grasp the intuition about the second one. Could someone help me?
An almost surely constant random variable should be constant everywhere but in sets of measure $0$, so can such a random variable have three of four different values? Because I have seen a proof of the second part using the first part and it chooses $ c= \sup \{x:\mathbb{P}(Y \leq x)=0\} $ and shows that $\mathbb{P}(Y = c)=1$. As far as I understand this $c$ is the switching point from events with probability zero to events with probability one but I still don't get why one would choose this $c$ or how do you come up with it?
(by the way my question might be a bit confusing, sorry for that)
Not only can such a random variable have $3$ or $4$ values, it can have infinitely many values. Simply consider the random variable $X$ on the probability space $([0,1],\mathcal{B}([0,1]),P)$ where $P[(a,b]] = b- a$ (i.e. $P$ is the Lebesgue measure) given by $$ X(u) = \begin{cases} 0 & u \notin \mathbb{Q} \\ u & u \in \mathbb{Q} \\ \end{cases}. $$ Clearly $X = 0$ $P$-a.s., but $X$ takes on infinitely many values. This can be made more egregious if you like (uncountably infinitely many values, etc) but I think the point is made.
As for the choice of $c$, defining it in the way you've described guarantees that $Y \leq c$ holds $P$-a.s., and so only one inequality is left to verify. I'm not sure of the proof you've seen (I would have proven this by writing any r.v. as a sequence of simple functions and shown it that way), but presumably another way to go would be to take $c = \inf\{x \mid P[Y \geq x] = 0 \}$ and then work to verify the converse inequality.