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.
Let $(\Omega,\mathcal B,\mu)$ be a probability space and $\mathcal A$ a subsigmaalgebra of $\mathcal B$. The following statements are equivalent:
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 sigmaalgebra, for example open intervals with rational endpoints. The (countable) collection $(X_k^{1}(S_j),(k,j)\in\Bbb N^2)$ generated $\mathcal A$. 


Yes, you can define separable by saying that it means "generated by a countable collection of (real) random variables", and you can always assume that all these variables take values in $[0,1]$. Then you can find a single random variable generating the same $\sigma$algebra by intertwining the digits of all your initial random variables (the countable collection of all the digits of all random variables is itself countable, so that you can encode them in a single sequence of digits). 

