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@ -63,9 +63,9 @@ In order to prove \autoref{thm2}, we need the following:
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|X_1(\omega) + X_2(\omega)| > \epsilon \},\\
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|X_1(\omega) + X_2(\omega)| > \epsilon \},\\
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\ldots\\
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\ldots\\
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A_i &\coloneqq& \{\omega: |X_1(\omega)| \le \epsilon,
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A_i &\coloneqq& \{\omega: |X_1(\omega)| \le \epsilon,
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|X_1(\omega) + X_2(\omega)| \le \epsilon, \ldots,
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|X_1(\omega) + X_2(\omega)| \le \epsilon, \ldots, %
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|X_1(\omega) + \ldots + X_{i-1}(\omega)| \le \epsilon,
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|X_1(\omega) + \ldots + X_{i-1}(\omega)| \le \epsilon,\\
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|X_1(\omega) + \ldots + X_i(\omega)| > \epsilon\}.
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&& ~ ~|X_1(\omega) + \ldots + X_i(\omega)| > \epsilon\}.
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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It is clear, that the $A_i$ are disjoint.
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It is clear, that the $A_i$ are disjoint.
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We are interested in $\bigcup_{1 \le i \le n} A_i$.
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We are interested in $\bigcup_{1 \le i \le n} A_i$.
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