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I am working through Achim Klenke's text entitled "Probability Theory", and I came across the following interesting exercise:

Let $(X_i)_{i\in\mathbb{N}}$ be independent, square-integrable random variables with $\mathbb{E}(X_i)=0$ for all $i$. Suppose that $\sum_{i=1}^\infty \mathbb{E}(X_i^2)<\infty$. Conclude that there exists a real random variable $X$ with $\sum_{i=1}^n X_i \xrightarrow{n\to\infty} X$ almost surely.

I attempted to prove this via Borel-Cantelli, namely, I tried to show that $\mathbb{P}(\{\omega:\sum_{i=n}^\infty X_i(\omega)\xrightarrow{n\to\infty}0 \})=1$, since the sequence will be summable if and only if the remainders are going to zero. In the details of B-C, though, for a fixed $\epsilon>0$ this requires showing that $\mathbb{P}(|\sum_{i=n}^\infty X_i| > \epsilon \;\;\;i.o.) =0$. An application of Chebyshev's inequality and using independence then gives

$$\mathbb{P}\left(\left|\sum_{i=n}^\infty X_i\right|>\epsilon\right) \leq \frac{1}{\epsilon^2}\sum_{i=n}^\infty \mathbb{E}(X_i^2)<\infty.$$ But now we certainly need not have that this is summable over all $n$ (take $X_i$ to be Bernoulli with possible values $\pm 1/i$).

I imagine my choice of Chebyshev's wasn't strong enough, or the entire approach is off. Suggestions?

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Another neat proof of this fact, though maybe not the one you want right now, is that $M_n = \sum_{i=1}^n X_i$ is a martingale which is $L^2$-bounded and hence uniformly integrable. –  Nate Eldredge Oct 22 '12 at 13:01

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We can use this answer to see that we just have to check convergence in probability, what it's done in the OP.

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I still wonder whether there is a Borel-Cantelli approach to this problem. –  Matt Spencerman Oct 23 '12 at 12:30

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