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I'm working on the following exercise from Achim Klenke's "Probability Theory: A Comprehensive Course" (exercise 6.1.4):

Let $X_1, X_2, \ldots$ be independent, square integrable, centered random variables with $\sum_{i=1}^\infty \mathbf{Var}[X_i] < \infty$. Show that there exists a square integrable $X$ with $X = \lim_{n \to \infty} \sum_{i=1}^n X_i$ almost surely.

Chebyshev's inequality gives us $$ \mathbf P\left[|S_m - S_n| > \epsilon\right] \leq \epsilon^{-2} \mathbf{Var}\left[ \sum_{i=m+1}^n X_i\right] = \epsilon^{-2} \sum_{i=m+1}^n \mathbf{Var}\left[X_i\right] \xrightarrow{m,n \to \infty} 0. $$ whence $(S_n)_{n \in \mathbb N}$ is a Cauchy sequence in probability. Thus $S_n \xrightarrow{\mathbf P} X$. Using a similar strategy, we can in fact show that $S_n \to X$ in $L^2$.

Now, to prove almost sure convergence, I'd like to use the following result (Corollary 6.13 in Klenke):

Let $(E,d)$ be a separable metric space. Let $f, f_1, f_2, \ldots$ be measurable maps $\Omega \to E$. Then the following statements are equivalent.

(i)$\quad f_n \to f$ in measure as $n \to \infty$.

(ii)$\quad$For any subsequence of $(f_n)_{n \in \mathbb N}$, there exists a sub-subsequence that converges to $f$ almost everywhere.

and somehow use the fact that we're working with a sum of centered random variables to show that in fact every subsequence converges a.s. But I'm not sure how to do this since our $X_i$ are not nonnegative. I tried reconstructing the proof of this theorem, but I've only been able to show once again that there are a.e. convergent subsequences.

My other thought was to apply the Borel-Cantelli lemma to the events $B_n(\epsilon) := \left\{ |X - S_n| > \epsilon\right\}$ and prove that $\limsup_{n \to \infty} B_n(\epsilon) =: B(\epsilon)$ has probability $0$, but in the latter case I don't know how to approximate the probability of $B_n(\epsilon)$. Chebyshev doesn't seem available to us since strictly speaking we don't know what $X$ looks like, only that $S_n$ converges in $L^2$ to it. Even if we could say $X - S_n = \sum_{i=n+1}^\infty X_i$, the above approximation using Chebyshev with $|X - S_n|$ instead of $|S_m - S_n|$ would work out to $$ \mathbf P\left[|X - S_n| > \epsilon\right] \leq \epsilon^{-2} \sum_{i=n+1}^\infty \mathbf{Var}[X_i] $$ which would sum to $\epsilon^{-2} \sum_{n=1}^\infty n\mathbf{Var}[X_n]$, but I don't see why this series converges.

Any thoughts on how to prove $S_n \to X$ almost surely?

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Since $$ \lim_{n\to\infty}\mathsf{P}\left(\sup_{k\ge n}|S_n-S_k|> \epsilon\right)=0, $$ the set on which the sequence $\{S_n\}$ is not Cauchy, $$ N=\bigcup_{\epsilon>0}\bigcap_{n\ge 1}\left\{\sup_{j,k\ge n}|S_j-S_k|>\epsilon\right\} $$ is a null set ($\because \sup_{j,k\ge n}|S_j-S_k|\le 2\sup_{k\ge n}|S_n-S_k|$). So you define $X:=\lim_{n\to\infty} S_n1_{N^c}$.

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  • $\begingroup$ This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n \sim \mathrm{Ber}_{1/n}$.) $\endgroup$ – D Ford Dec 16 '18 at 14:27
  • $\begingroup$ @DFord "Why does this not imply..."? Why should it? $\endgroup$ – d.k.o. Dec 16 '18 at 18:20
  • $\begingroup$ Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which $\{X_n\}$ is not Cauchy is a null set. $\endgroup$ – D Ford Dec 16 '18 at 18:26
  • $\begingroup$ @DFord Does $\mathsf{P}(\sup_{k\ge n}|X_n-X_k|>\epsilon)$ converge to $0$ in that case? $\endgroup$ – d.k.o. Dec 16 '18 at 18:28
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    $\begingroup$ Cauchy in prob. means that $\mathsf{P}(|X_j-X_k|>\epsilon)\to 0$ as $j,k\to \infty$. For the a.s. convergence you need a stronger condition. $\endgroup$ – d.k.o. Dec 16 '18 at 19:00
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$var (\sum _{i=n}^{m} X_i) =\sum _{i=n}^{m} var(X_i) \to 0$ as $n,m \to \infty$so the partial sums of $\sum X_i$ form a Cauchy sequence in $L^{2}$. Hence there is a square integrable random variable $X$ such that $\sum _1^{n} X_i \to X$ in $L^{2}$. Now convergence in mean square implies convergence in probability and for series of independent random variables convergence in probability implies almost sure convergence.

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  • $\begingroup$ ``For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this. $\endgroup$ – D Ford Dec 16 '18 at 14:28
  • $\begingroup$ It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord $\endgroup$ – Kavi Rama Murthy Dec 16 '18 at 23:22
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Your try is not bad, but I doubt that those methods will give you the result. We should show that the sum of independent random variables $$S_n = \sum_{i=1}^n X_i \to S_\infty $$ almost surely. But we know that in general $$ P(\sum_{i=1}^\infty |X_i|<\infty) =1 $$ fails. This means that the series is conditionally convergent in most cases, and it is known that some kind of maximal inequality such as $$ P(\max_{n\in \mathbb{N}} |S_n|>\lambda) \leq \frac{C}{\lambda^2}\sum_{n=1}^\infty \operatorname{Var}(X_n) $$ provides sufficient and necessary condition for the almost sure convergence. It is necessary since we need to control the oscillation of the sequence $$ n\mapsto S_n(\omega) $$ for almost all $\omega\in \Omega$. Fortunately there are several known maximal inequalities such as Kolmogorov's maximal inequality, Etemadi's inequality or martingale maximal inequalities. In particular, Kolmogorov's inequality can establish that $$ S_n \text{ converges a.s.} \iff S_n \text{ converges in probability.} $$ (Or you can see this: https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem.) If you are allowed to use more powerful tools such as martingale convergence theorem, then $$S_n = \sum_{i=1}^n X_i \to S_\infty $$ almost surely follows immediately.

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  • $\begingroup$ Is there a clear reason why $S_n$ converges a.s. $\iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it. $\endgroup$ – D Ford Dec 16 '18 at 18:48
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    $\begingroup$ It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = \max_{j\leq n} P(|S_n-S_j|>a).$ Then $P(\max_{j\leq n}|S_j|>x) \leq \frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+\cdots +X_n$ and $X_i$ are independent (need not be identical). $\endgroup$ – Song Dec 16 '18 at 18:53

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