Here is a question from a past homework which I was never able to solve.

Suppose $X_1,X_2,\dots$ are independent with the following properties.

1) For each $n$, there is a finite set $F_n$ with $P(X_n \in F_n) = 1$.

2) $S \doteq \lim_{n\to\infty}\sum_{i=1}^n X_i$ exists and is finite almost surely.

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

My attempt was to suppose that there existed an $\alpha \in \mathbb{R}$ with $P(S = \alpha) > 0$, and then construct a countable set $A$ containing the support of $S$ using this $\alpha$. We have \begin{align*} P(S \text{ exists and is finite}) = 1, \end{align*} so that \begin{align*} 0 < P(S = \alpha) &\leq P\left(\sum_{i=2}^{\infty}X_i \in \bigcup_{x_1 \in F_1}\{\alpha- x_1\} \right) \\ &\leq P\left(\sum_{i=3}^{\infty}X_i \in \bigcup_{x_1 \in F_1, x_2 \in F_2}\{\alpha- x_1-x_2\} \right), \end{align*} and so on. Letting $B_n$ denote the set \begin{align*} B_n \doteq \left\{\sum_{i=n}^{\infty}X_i \in \bigcup_{i < n} \bigcup_{x_i \in F_i} \{\alpha - \sum_i x_i\}\right\}, \end{align*} taking limits and pushing the limsup inside $P$, the above would imply that \begin{align*} 0 < P(B_n \text{ i.o.}) \end{align*} I think that this might also be a tail event, in which case the above would imply that $B_n$ happens infinitely often with probability $1$. But all this says is that $S - S_n$ falls into a countable set infinitely often. Is this somehow enough to conclude that $S$ lives in a countable set?

  • 1
    $\begingroup$ There is no reason to believe $\alpha$ is unique in having $P(S = \alpha) > 0$, so there cannot be a method of constructing $A$ from just one such $\alpha$. $\endgroup$ – Paul Sinclair Jul 14 '18 at 20:26
  • $\begingroup$ Hint: if $\mathbb P(S=\alpha)>0$, then there exists a sequence $a_n$ such that $$\mathbb P\left(\bigcap_{n=1}^\infty \{X_n=a_n\}\right) > 0$$ and $$\sum_{n=1}^\infty a_n = \alpha. $$ Since the $F_n$ are finite, there are only countably many such sequences, and therefore $A$ is at most countable. $\endgroup$ – Math1000 Jul 24 '18 at 4:01

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