P-trivial $\sigma-$algebra I want to proof the following:
Consider an probability space with a P-trivial $\sigma-$Algebra $G$. Show that for any real-valued $G$-measurable random variable $X$, there exists $c \in \mathbb{R}$ such that$X=c$ almost surely.
My idea:
Considering cdf implies: $F_X(t)=P(X\leq t) \in \{0,1\}$
Hence $G$ is trivial, there is $x_0 := \inf\{x: F_X(x)=1\}$. Thus $P(X=x_0)= F_X(x_0)- \lim_{x<x_0} F_X(x)=1$.
Therefore $X$ is constant. But why does this hold almost surely?
 A: Your argument is essentially correct. It needs just a small adjustments. Here is a detailed proof, using your argument.
Consider $F_X(t)$  for $t$ in the extended real line, $\overline{\Bbb R}= \Bbb R \cup \{-\infty , +\infty\}$.
Then $F_X(+\infty)=P(X\leq +\infty) =1$, and so let $c := \inf \{x\in \overline{\Bbb R} : F_X(x)=1\}$.
Claim: $F_X(c) = 1$.
In fact, since $c := \inf \{x\in \overline{\Bbb R} : F_X(x)=1\}$, there is a decreasing sequence $\{x_n\}_n$ in  $\{x\in \overline{\Bbb R} : F_X(x)=1\}$, such that $x_n \to c$. So
$$F_X(c) = P(X\leq c) = \lim_{n \to \infty} P(X\leq x_n) = \lim_{n \to \infty} F_X(x_n) = \lim_{n \to \infty} 1=1$$
(You can also see the claim is true, directly from the fact that $F_X$ is continuous from the right.)
Now, $P(X=c)= F_X(c)- \lim_{x \to a^{-}} F_X(x)=1$.
So, $P(X \neq c)=0$, which means that $X=c$ almost surely.
Remark: Note that we can not conclude that $\{x: X(x) \neq c\} =\emptyset$. So we can not conclude that $X=c$. We can only conclude that $P(X \neq c)=0$, that is why we have $X=c$ almost surely.
