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@ -43,8 +43,8 @@ For the proof we'll need a slight generalization of \autoref{thm2}:
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By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$
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By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$
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almost surely.
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almost surely.
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Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$.
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Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$.
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Since the first series $\sum_{n \ge 1} \bP(A_n) < \infty$,
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Since $\sum_{n \ge 1} \bP(A_n) < \infty$ by assumption,
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by Borel-Cantelli, $\bP[\text{infinitely many $A_n$ occur}] = 0$.
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Borel-Cantelli yields $\bP[\text{infinitely many $A_n$ occur}] = 0$.
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For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$
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For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$
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