In Rosenthal's, "A First Look At Rigorous Probability Theory", $\exists$ this exercise:
Exercise 3.6.19.
Let $A_1,\ A_2,\ldots$ be independent events. Let $Y$ be a random variable which is measurable with respect to $\sigma(A_n, A_{n+1},\ldots)$ for each $n \in \mathbb{N}$. Prove that there is a real number a such that $P(Y = a) = 1.$
[Hint: Consider $P(Y \leq x)$ for $x \in \mathbb{R}$; what values can it take?]
I have a feeling that this has something to do with intersections?
Must we find x s.t. $P(Y \leq x \cap Y \geq x) = 1$ ?
What does measurable the given $\sigma$ -algebra have to do with this anyway? If $Y$ is measurable there, then $Y^{-1}(B) \in \sigma(A_n, A_{n+1},\ldots),\ \forall B \in \frak{B}$...? Choose $B = (-\infty,x]$?
I dunno. Help please. :(
Hint: Try to show that there exists $a \in \mathbb{R}$ such that $$P[Y \leq a] = 1 \text{ and } P[Y < a] = 0$$ by Kolmogorov $0$-$1$ law and $\lim_{x \to \infty} P[Y \leq x] = 1$ and $\lim_{x \to -\infty}P[Y \leq x] = 0$.
More explicitly, let $F(x) = P[Y \leq x]$ be the distribution function of $Y$, then $F$ is nondecreasing, right-continuous and takes only two values $0$ and $1$, what would happen?