# Levy's theorem? [duplicate]

Possible Duplicate:
how to show convergence in probability imply convergence a.s. in this case?

Good evenig! I stumbled upon this theorem:

For independent random variables $(X_n)_{n\in\mathbb{N}}$ and $S_n := \sum_{i=1}^n X_i$ the following two statements are equivalent:

1. $\exists S_\infty \forall \varepsilon >0 : \lim_{n\to\infty} P(\vert S_\infty - S_n\vert>\varepsilon)=0$
2. $\exists S_\infty : P(\lim_{n\to\infty} S_n = S_\infty) = 1$

In the notation above $S_\infty$ denotes a random variable. Unfortunately no proof of this theorem is given and i can not find any in my books or via google. The only hint : The theorem is named " Levy's theorem".

Does anyone know where I can find a proof ? I am very thankful for any suggestions. With best regards!

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## marked as duplicate by Davide Giraudo, Norbert, Beni Bogosel, Chris Eagle, SashaNov 1 '12 at 16:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

See Theorem 5.3.4 in Kai Lai Chung, A course in probability theory, 3rd ed., Academic Press, 2001.

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Thank you for this reference. The book of Kai Lai Chung was not known to me. It seems to be a nice introductory course on probability! – Mat Nov 1 '12 at 10:20
Introductory is debatable but this is certainly an excellent book, well worth the time you might choose to spend working on it. – Did Nov 1 '12 at 11:00