% TODO \begin{goal}
% TODO     We want to drop our assumptions on finite mean or variance
% TODO     and say something about the behaviour of $ \sum_{n \ge 1}  X_n$
% TODO     when the $X_n$ are independent.
% TODO \end{goal}
\begin{theorem}[Theorem 3, Kolmogorov's three-series theorem] % Theorem 3
    \label{thm3}
    Let $X_n$ be a family of independent random variables.
    \begin{enumerate}[(a)]
        \item Suppose for some $C \ge 0$, the following three series
            of numbers converge:
            \begin{itemize}
                \item  $\sum_{n \ge 1} \bP(|X_n| > C)$,
                \item $\sum_{n \ge 1}  \underbrace{\int_{|X_n| \le C} X_n d\bP}_{\text{\vocab{truncated mean}}}$,
                \item $\sum_{n \ge 1}  \underbrace{\int_{|X_n| \le  C} X_n^2 d\bP - \left( \int_{|X_n| \le  C} X_n d\bP \right)^2}_{\text{\vocab{truncated variance} }}$.
            \end{itemize}
            Then $\sum_{n \ge  1}  X_n$ converges almost surely.
        \item Suppose $\sum_{n \ge 1}  X_n$ converges almost surely.
            Then all three series above converge for every $C > 0$.
    \end{enumerate}
\end{theorem}
For the proof we'll need a slight generalization of \autoref{thm2}:
\begin{theorem}[Theorem 4] % Theorem 4
    \label{thm4}
    Let $\{X_n\}_n$ be independent and \vocab{uniformly bounded}
    (i.e. $\exists M < \infty : \sup_n \sup_\omega |X_n(\omega)| \le M$).
    Then $\sum_{n \ge 1} X_n$ converges almost surely
    $\iff$ $\sum_{n \ge 1} \bE(X_n)$ and $\sum_{n \ge 1}  \Var(X_n)$
    converge.
\end{theorem}
\begin{refproof}{thm3}
    Assume, that we have already proved \autoref{thm4}.
    We prove part (a) first.
    Put $Y_n = X_n \cdot \One_{\{|X_n| \le  C\}}$.
    Since the $X_n$ are independent, the $Y_n$ are independent as well.
    Furthermore, the $Y_n$ are uniformly bounded.
    By our assumption, the series
    $\sum_{n \ge  1}  \int_{|X_n| \le  C} X_n d\bP = \sum_{n \ge 1}  \bE[Y_n]$
    and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le  C} X_n d\bP \right)^2 = \sum_{n \ge 1}  \Var(Y_n)$
    converges.
    By \autoref{thm4} it follows that $\sum_{n \ge 1}  Y_n < \infty$
    almost surely.
    Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$.
    Since the first series $\sum_{n \ge 1}  \bP(A_n) < \infty$,
    by Borel-Cantelli, $\bP[\text{infinitely many $A_n$ occcur}] = 0$.


    For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$
    for almost every $\omega$.
    Fix an arbitrary $C > 0$.
    Define
    \[
    Y_n(\omega) \coloneqq  \begin{cases}
        X_n(\omega) & \text{if} |X_n(\omega)| \le  C,\\
        C &\text{if } |X_n(\omega)| > C.
    \end{cases}
    \] 
    Then the $Y_n$ are independent and $\sum_{n \ge 1} Y_n(\omega) < \infty$ 
    almost surely and the $Y_n$ are uniformly bounded.
    By \autoref{thm4} $\sum_{n \ge 1} \bE[Y_n]$ and $\sum_{n \ge 1} \Var(Y_n)$
    converge.
    Define
    \[
    Z_n(\omega) \coloneqq \begin{cases}
        X_n(\omega) &\text{if } |X_n| \le  C,\\
        -C &\text{if } |X_n| > C.
    \end{cases}
    \] 
    Then the $Z_n$ are independent, uniformly bounded and  $\sum_{n \ge 1}  Z_n(\omega) < \infty$ 
    almost surely.
    By \autoref{thm4} we have
    $\sums_{n \ge 1} \bE(Z_n) < \infty$
    and $\sums_{n \ge 1} \Var(Z_n) < \infty$.
    
    We have
    \[
        \bE(Y_n) &=& \int_{|X_n| \le  C} X_n d \bP + C \bP(|X_n| \ge C)\\
        \bE(Z_n) &=& \int_{|X_n| \le  C} X_n d \bP - C \bP(|X_n| \ge C)\\
    \] 
    Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n d\bP$
    the second series converges,
    and since
    $\bE(Y_n) - \bE(Z_n)$ converges, the first series converges.
    For the third series, we look at
    $\sum_{n \ge 1} \Var(Y_n)$ and
    $\sum_{n \ge 1} \Var(Z_n)$ to conclude that this series converges
    as well.
\end{refproof}