Consider Kolmogorov's Zero-One law:
Suppose that $(X_n)_{n \in \mathbb{N}}$ is a sequence of independent
random variables. Then any event belonging to the tail
$\sigma$-algebra $\bigcap_{n \in \mathbb{N}}\sigma(X_n,X_{n+1},\dots)$
has probability $0$ or $1$.
Now the book says:
Every random variable $X$ which is measurable with respect to the tail
$\sigma$-algebra is a.s. constant.
I do not quite see this, but I think it should be easy. So since $X$ is measurable with respect to the tail $\sigma$-algebra, we have that $$\{X \leq t\} \in \bigcap_{n \in \mathbb{N}}\sigma(X_n,X_{n+1},\dots)$$ for all $t \in \mathbb{R}$. Similarly also $\{X \geq t\}$. So by $$\{X = t\} = \{X \leq t\} \cap \{X \geq t\}$$ we get that $P(X = t)$ is either $0$ or $1$. Why should it be $1$ for some $t$? Or how does one see this?
Best Answer
You can also find that $F_X(t)=P(X\leq t)\in\{0,1\}$ for each $t$.
Combining this with $\lim_{t\to\infty}F_X(t)=1$ we will arrive at $F_X(t)=1$ for some $t$.
Now let $t_0:=\inf\{t\in\mathbb R\mid F(t)=1\}$.
Then $t_0\in\mathbb R$ and you can find that $P(X=t_0)=1$.