Why $\sum_n E(\xi_n^2\wedge1)<\infty \Rightarrow \sum_n 1_{\{\xi_n>1\}}<\infty$ for independent, symmetric random variables.

Question

Let $\xi_1, \xi_2,...$ be independent, symmetric random variables. Then $\sum_n E(\xi_n^2\wedge1)<\infty \Rightarrow \sum_n 1_{\{\xi_n>1\}}<\infty$ a.s..

This is a step in Theorem 4.17 from Foundations of Modern Probability edition 2 by Olav Kallenberg.

It says it can be derived from Fubini's theorem but I don't know how.

Answer 1

$E\sum_n I_{\xi_n >1}=\sum P(\xi_n >1) \leq \sum E(\xi_n^{2} \wedge 1)<\infty$ because $E(\xi_n^{2} \wedge 1) \geq E(\xi_n^{2} \wedge 1)I_{\xi_n >1} \geq (1\wedge 1) P(\xi_n >1)$. This implies that $\sum_n I_{\xi_n >1}<\infty$ almost surely.