Kolmogorov 0-1 law and terminal sigma algebra I was wondering about the following: Assume $X : \Omega \rightarrow S$ is measurable w.r.t. a terminal sigma algebra on $\Omega$ and some sigma algebra on $S$. What are the requirements we have to put on $S$ such that $X$ is constant a.s. 
I know that it holds for $S= \mathbb{R}$ but what about more general sets $S$?
 A: So let's suppose that $(\Omega,\mathcal T,\Bbb P)$ is a probability space, $(S,\mathcal S)$ is a measurable space, and $X:\Omega\to S$ is $\mathcal T/\mathcal S$ measurable. Let's also suppose that $\mathcal T$ (like the tail $\sigma$-algebra in the context of Kolmogorov's 0-1 Law) is trivial, in the sense that $\Bbb P(A)=0$ or $1$ for all $A\in\mathcal T$.
We'd like to deduce that $X$ is constant a.s.; namely  that there is an element $s\in\mathcal S$ such that $\Bbb P(X=s)=1$. One thing we need to know is that $\mathcal S$ has enough elements to distinguish between points in $S$, so let's assume that the atoms of $\mathcal S$ are the points of $S$: If $y,y'$ are points of $S$ such that $1_B(y)=1_B(y')$ for all $B\in\mathcal S$, then $y=y'$. Let's also assume that $\mathcal S$ is countably generated: there is a sequence $\mathcal B=\{B_n\}$ of elements of $\mathcal S$ such that $\mathcal S=\sigma(B_n:n\in\Bbb N)$. This ensures that the atoms of $\mathcal S$ are $\mathcal S$-measurable: for each $y\in S$, $\{y\}=\cap\{B: B\in\mathcal B: y\in B\}\cap\{B^c: B\in\mathcal B, y\notin B_n\}\in\mathcal S$. 
Because $X$ is $\mathcal T/\mathcal S$-measurable, and $\mathcal T$ is trivial,
$$
\Bbb P(X\in B)=0\hbox{ or }1,\qquad\forall B\in\mathcal B.
$$
Let $\mathcal B_0:=\{B\in\mathcal B:\Bbb P(X\in B)=0\}$ and $\mathcal B_1:=\{B\in\mathcal B:\Bbb P(X\in B)=1\}$. These are countable collections, so the intersection 
$$
C:=\left(\cap\{B^c:B\in\mathcal B_0\}\right) \cap \left(\cap\{B:B\in\mathcal B_1\}\right)
$$
 is an element of $\mathcal S$ with $\Bbb P(C)=1$. Also, $\mathcal C$ is a singleton, by the preceding discussion.
Thus, $C=\{s\}$ for some $s\in S$, and $\Bbb P(X=s)=1$.
