TITLE: If $\mathcal{F}_t$ is separable why is $\mathcal{F}_\infty$ generated by a random variable?
QUESTION [5 upvotes]: I am reading this introduction to enlargement of filtration and at the beginning of section 2.4 there is a claim that I cannot justify but seems like it should be well known.  The author claims that if $\mathcal{F}_t$ is the natural filtration of a Brownian Motion, then since $\mathcal{F}_t$ is separable it follows that $\mathcal{F}_\infty$ is generated by a real random variable.  I assume that separable means that there is a countable collection of random variables which generates $\mathcal{F}_t$,though I have not been able to find a definition for the term 'separable' in this context.  In any case, I wonder if someone could help me with a proof (or a reference for one) of this result.

REPLY [5 votes]: Let $(\Omega,\mathcal B,\mu)$ be a probability space and $\mathcal A$ a sub-sigma-algebra of $\mathcal B$. The following statements are equivalent: 

$\mathcal A$ admits a countable set of generators. 
There is a bounded random variable $X\colon \Omega\to \Bbb R$ such that $\sigma(X)=\mathcal A$. 
There are countably many bounded random variables $X_k\colon \Omega\to \Bbb R$ such that $\sigma(X_k,k\in\Bbb N)=\mathcal A$. 

Indeed, if $1.$ is satisfied, let $\(A_k,k\in\Bbb N\)\subset 2^\Omega$ generating $X$. Then define $X:=\sum_{k=1}^{+\infty}3^{-k}\chi_{A_k}$. Then $X^{-1}(3^{-k})=A_k$. 
$2.\Rightarrow 3.$ is obvious taking $X_k=X$. 
$3.\Rightarrow 1.$ Let $S_j$ be countably many sets generating Borel sigma-algebra, for example open intervals with rational endpoints. The (countable) collection $(X_k^{-1}(S_j),(k,j)\in\Bbb N^2)$ generated $\mathcal A$.