There are these: 1, 2, 3, but I wanna try different ways.
Let $(\Omega, \mathfrak{F}, \mathbb P)$ be such probability space with each (event) $E \in \mathfrak F$ having trivial probability. Consider a random variable $X$.
An attempt:
Show that $X$ is almost surely equal to $c = \inf\{x:F_{X}(x)=1\}$, with $F_X$ as cdf (cumulative distribution function or, well, just distribution function) of $X$.
Actually $c$ is finite. If $c=\pm \infty$, then $(x:F_{X}(x)=1) = \emptyset$ or $\mathbb R$. Then $F_X(x) < 1 \ \forall x$ or $F_X(x) = 1 \ \forall x$.
By definition of such finite c, $$P(X \le c) = F_X(c) = 1 = F_X(d) \forall \ d \ge c$$
Now either $P(X \ge c) = 1$, in which case we're done or $P(X \ge c) = 0$.
Now $P(X \ge c) = 0 \to P(X < c) = 1 \to \exists \ d < c$ s.t. $P(X \le d) = 1$
But $\forall \ d < c$,
$$P(X \le d) = F_X(d) < 1 \to P(X \le d) = 0$$
Question 1: Is this correct?
Question 2: Is there perhaps an even simpler way to go about this namely without guessing what the constant $c$ is?
Actually, what I might try to do is show the $c$ exists and then later (not sure if I have to 1st show $c$ is finite for this next part) say it's the infimum (minimum actually? idk) of when the cdf reaches 1.
I think to consider any extended real $a$ and then the event $\{X=a\}$. Well, we can't have $P(X=a)=1$ for more than 1 extended real $a$. But from this, all we can say is that $P(X=a)=1$ for at most one possible $a$.
Of course it's possible we don't have $P(X=a)=1$ for any $a$, i.e. we have $P(X=a)=0$ for all $a$ such as when $X$ is a continuous(/an absolutely continuous or whatever) random variable. (But I guess this is easy to show for every discrete random variable [even if we don't assume finite] ?)
But if $X$ were continuous then $P(X \in (g,h)) = 1$ means...yeah you could probably argue a contradiction here to show continuous random variables don't exist in such a probability space (and apparently neither do binomial distributions) but in higher probability of course not all random variables are either discrete or continuous.
Well, maybe this simpler approach works for only half the battle namely the at most 1 part, but the at least 1 part really requires more.
- Note: Simpler need not mean easier/less machinery. Could be more difficult/more machinery. Like perhaps Lebesgue's decomposition theorem can be used? What I understand of this theorem is that every random variable basically has a 'continuous part' and a 'discrete part' (Eg this and this). Maybe I can somehow show that in such a probability space every random variable is discrete and then (immediately?) conclude every random variable is a.s. constant?
