David Williams "Probability with Martingales" 4.12: Tail sigma algebras warning I am struggling to understand this passage in David Williams "Probability with Martingales"  4.12:

I do not even understand the notation highlighted : The 2 comma-separated items inside the $\sigma()$ are already sigma algebras individually , and while i understand sigma-algebras generated by collections of sets, r.v, $\pi$-systems etc . I have seen no definition of $\sigma($collection of sigma-algebras) ... 


*

*Is it the sigma-algebra generated by all the sets in the underlying sigma-algebras ?

*even so, I am not sure why $Y_0$ is as stated in the hint ... $Y_0$ is not in $\cap \tau_n$ because there is a > in the definition of $\tau_n$ , so that even $\tau_0$ does not have $Y_0$, so I am not sure why $Y_0$ is measurable in $\mathcal{L}$

 A: $\sigma(\mathcal{A},\mathcal{B})$ denotes the smallest $\sigma$-algebra containing both $\mathcal{A}$, $\mathcal{B}$.
Hints to solve the exercise:


*

*Deduce from $$Y_0 =  \frac{X_n}{Y_1 \cdots Y_n}$$ that $Y_0$ is $\sigma(\mathcal{Y},\mathcal{T}_{n-1})$-measurable. Since this holds for any $n \in \mathbb{N}$, this already proves that $Y_0$ is measurable with respect to $\mathcal{L} := \bigcap_n \sigma(\mathcal{Y},\mathcal{T}_n)$.

*Recall the following statement: If a $\sigma$-algebra $\mathcal{A}$ is generated by some family $\mathcal{G}$ (i.e. $\mathcal{A} = \sigma(\mathcal{G})$) and $\mathcal{G}$ is $\cap$-stable (i.e. $G,H \in \mathcal{G} \implies G \cap H \in \mathcal{G}$), then an event $B$ is independent from $\mathcal{A}$ if and only if $B$ is independent from $\mathcal{G}$.

*Show that $\mathcal{G} := \{G_1 \cap G_2 : G_1 \in \mathcal{Y}, G_2 \in \bigcap_n \mathcal{T}_n\}$ is $\cap$-stable, and that $\sigma(\mathcal{G}) = \sigma(\mathcal{Y}, \bigcap_n \mathcal{T}_n)$.

*By step 2, $Y_0$ is independent from $\mathcal{R} = \sigma(\mathcal{G})$ if  $$\mathbb{P}(\{Y_0 \in A\} \cap G) = \mathbb{P}(Y_0 \in A) \cdot \mathbb{P}(G) \tag{1}$$ for all $G \in \mathcal{G}$ and any Borel set $A$. To prove this, note that any set $G \in \mathcal{G}$ is of the form $$G = G_1 \cap G_2$$ where $G_1 \in \mathcal{Y}$ and $G_2 \in \bigcap_n \mathcal{T}_n$. Use the fact that $Y_0$ and $\mathcal{Y}$ are independent and that $\mathbb{P}(G_2) \in \{0,1\}$ for any $G_2 \in \bigcap_n \mathcal{T}_n$ (by Kolmogorov's 0-1 law) to prove $(1)$.

