Sets of convergence of independent random variables

Question

I'm reading a claim on p. 30 of Kallenberg's Foundations of Modern Probability (1st ed.) that implies the following claim:

Claim: Suppose $\xi_1, \xi_2, \dotsc:\Omega \to \mathbb{R}^{\geq 0}$ are independent random variables. Suppose $\mathcal{F}_1, \mathcal{F}_2, \dotsc$ are the sigma algebras generated by these random variables, form the join of the tail $\mathcal{T}_k = \vee_{i=k}^\infty \mathcal{F}_i$, i.e. the sigma algebra generated by $\mathcal{F}_k, \mathcal{F}_{k+1} \dotsc$. Now take the tail algebra $\mathcal{T} = \cap_{j=1}^\infty \mathcal{T}_j$. Form the sum of the $\xi_i$, $S_n = \xi_1 + \dotsb + \xi_n$. Then the sets of convergence of $(S_n)$ and $(S_n/n)$ are $\mathcal{T}$-measurable.

He refers to lemma 1.9, which says that various limits of measurable functions to $\mathbb{R} \cup \{\infty\}$ are themselves measurable. I don't see his logic...it would work if $S_n$ were $\mathcal{T}$-measurable, and so their limit would be $\mathcal{T}$-measurable. but $S_n$ are not $\mathcal{T}$-measurable: for instance $\xi_1 + \xi_2$ isn't measurable with respect to $\vee_{i=3}^\infty \mathcal{F}_i \supset \mathcal{T}$ in general.

Answer 1

I think the logic is:

$$\text{Set of convergence of }\xi_1 + \xi_2 + \dotsb = \text{Set of convergence of } \xi_k + \xi_{k+1} + \dotsb$$

and the latter is $\mathcal{T}_k$ measurable. So just let $k$ increase and this shows that the set of convergence is $\mathcal{T}_k$ measurable for all $k$ and so $\mathcal{T}$ measurable.

Answer 2

I don't have the book so I can't comment on Kallenberg's argument, but here's how I would prove it:

If we use the Cauchy criterion for convergence and translate this criterion into events, we can write $$ \{\lim_{n\to\infty}S_n\;\mathrm{exists}\}=\bigcap_{k=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{m\geq n\geq N}\{|S_m-S_n|<\frac{1}{k}\}$$ (using $\frac{1}{k}$ instead of a general $\varepsilon>0$ to make the outer intersection a countable one). If we define $$ E_{k,N}=\bigcap_{m\geq n\geq N}\{|S_m-S_n|<\frac{1}{k}\}$$ then $E_{k,N}\in\mathcal{T}_N$, and for each $k$ we have $$E_{k,1}\subset E_{k,2}\subset E_{k,3}\subset\cdots$$ so the union $$ \bigcup_{N=1}^{\infty}E_{k,N}$$ is a tail event for each $k$, which proves that $\{\lim_{n\to\infty}S_n\;\mathrm{exists}\}$ is a tail event. Something similar should work for $\frac{S_n}{n}$.