A question related to Kolmogorov's $0$-$1$ law

Question

Let $\{X_n\}_{n=1}^\infty$ be a sequence of mutually independent random variables such that

1. for each $n$ there exists a finite set $F_n$ such that $\mathbb{P}(X_n\in F_n)=1$;
2. $S=\lim_{n\rightarrow\infty}\sum_{i=1}^n X_i$ exists and is finite almost surely.

Prove that either there is a countable set $A$ such that $\mathbb{P}(S\in A)=1$ or there is no $\alpha\in\mathbb{R}$ such that $\mathbb{P}(S=\alpha)>0$.

It seems that we need to use Kolmogorov's $0$-$1$ law to solve above problem but I have no idea how to construct corresponding tail $\sigma$-algebra. Any help is appreciated.

Answer 1

Define $\mathcal{F}_n=\{a_1+a_2+\cdots+a_n|a_1\in F_1, a_2 \in F_2, \ldots,a_n\in F_n \}$. We will also use the notation $x+A=\{x +a| a\in A\}$, where $A\subset \mathbb{R}$ and $x\in \mathbb{R}$.

Suppose that there is $a$ such that $\mathbb{P}(S=a)>0$.

Define the event $A_k= \{ \sum_{i=k+1}^\infty X_i\in a- \mathcal{F}_k \} $. Then $A=\{\bigcap_{n} \bigcup_{m\geq n} A_m \}$ is a tail event, so by Kolmogorov's $0$-$1$ law, $\mathbb{P} (A)=0 \text{ or } 1$.

However, since $\{S=a\} \subset A$ we deduce that $\mathbb{P}(A)=1$. So, we conclude that $S$ is supported on a countable set, namely $\bigcup_{i} \left ((a-\mathcal{F_i})+\mathcal{F}_i\right) $ which a union of countable sets.