If $\sum_{i=1}^\infty i^2\mathbb{P}(i\leq X_n<i+1)\leq C\leq \infty$, prove $\mathbb{P}(X_n \geq n\ i.o.) = 0$

This question comes from Rosenthal's 3.6.13

Let $X_1, X_2,\dots$ be defined jointly on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$, with $\sum_{i=1}^\infty i^2\mathbb{P}(i\leq X_n<i+1)\leq C\leq \infty$ for all $n$. Prove that $\mathbb{P}(X_n \geq n\ i.o.) = 0$

I am thinking if I can prove $\sum_{n=1}^\infty \mathbb{P}(X_n \geq n) < \infty$, then Borel-Cantelli Lemma can apply, but got no luck. What I got are:

\begin{align} \sum_{i=1}^\infty i^2\mathbb{P}(i\leq X_n<i+1) &= \mathbb{P}(X_n \geq 1) - \mathbb{P}(X_n\geq 2) + 2^2 \mathbb{P}(X_n \geq 2) - \mathbb{P}(X_n\geq 3) + \dots \\ &= \sum_{i=1}^\infty (2i-1)\mathbb{P}(X_n \geq i) \\ \mathbb{P}(X_n \geq n) & = \sum_{i=n}^\infty \mathbb{P}(i\leq X_n<i+1) \end{align}

However, none of these lead me to an answer. If it requires the Kolmogorov Zero-One Law, please explain me a little. I am confused about the definition of "tail field".

• Is there are typo in the statement of the problem? Should it say, $\cdots \le C < \infty,$ and not $\cdots \le C \le \infty$? – Dfrtbx Jan 5 '18 at 17:45
• @Dfrtbx It must be $C<\infty$ such that the tail sum $\sum_{i=k}^{k+m} i^2 P(X_n \geq i) \to 0$ for large enough $k$ as $m\to \infty$, which is what we prefer for the proof. Yet, only state close to $0$ is not small enough. It has to go to $0$ fast, which requires some algebra I'm missing. – Jango Jan 5 '18 at 18:06

The event $X_n \ge n$ i.o. is indeed a tail event. You can tell because if you change any finite number of the $X_n,$ it won't change the truth value of $X_n \ge n$ i.o. By the Kolmogorov 0-1 law, $\mathbb{P}(X_n \ge n \text{ i.o})$ is either 0 or 1. Therefore, it suffices to show that it is not equal to 1.
So assume that $\mathbb{P}(X_n \ge n \text{ i.o}) = 1.$ For infinitely many $n,$ we have that $$\sum_{k=1}^\infty k^2\mathbb{P}(k \le X_n < k+1) = \sum_{k=n}^\infty k^2\mathbb{P}(k \le X_n < k+1)\ge n^2,$$ contrary to hypothesis. $\square$