Sequence of independent events in a discrete probability space Let $(\Omega, \mathcal{A}, \Bbb{P})$ be a discrete probability space. Let $A_1, A_2,...\in \mathcal{A}$ be a sequence of independent events with $p_n = \Bbb{P}(A_n)$. Then $$\sum_{n\in \Bbb{N}} \min(p_n, 1-p_n)< \infty$$
My attempt: Suppose toward contradiction that the sum diverges. Then in particular $\sum_n p_n = \infty$. Applying Borel-Cantelli gives us $\Bbb{P}\left( \bigcap_{n\in \Bbb{N}} \bigcup_{k\geq n} A_n \right) =1$. Discreteness then implies that $\Omega = \bigcap_{n\in \Bbb{N}} \bigcup_{k\geq n} A_n$, so each $\omega \in \Omega$ is contained in infinitely many $A_n$'s.
On the other hand, since the sum diverges, we have that $0=\prod_{n\in \Bbb{N}} 1 - \min(p_n, 1-p_n) = \prod_{n\in \Bbb{N}} \max(1-p_n, p_n) \geq \prod_n p_n = \Bbb{P}(\bigcap_n A_n) \geq 0$, and thus $\emptyset = \bigcap_n A_n$.
That's not really a contradiction yet, but I feel like I'm almost there. What can I do?
 A: I believe the following the argument works, and it uses your proof by contradiction idea.
Since $(\Omega, \mathcal{A}, \mathbb{P})$ is a discrete probability space then clearly we can find an $\omega \in \Omega$ with $\mathbb{P}(\omega) > 0$.  Having chosen such an $\omega$ define the sequence $\{ A_n^\prime \}_{n=1}^{\infty}$ where $A_n^\prime = A_n$ if $\omega \in A_n^c$ and $A_n^\prime = A_n^c$ if $\omega \in A_n$ so that $\omega \notin A_n^\prime$ for every $n$.  Notice that $\sum_{n=1}^{\infty} p_n^\prime \geq \sum_{n=1}^{\infty} \min (p_n, 1 - p_n)$ so if the latter series diverges then so does $\sum_{n=1}^{\infty} p_n^\prime$.  Also, the $\{ A_n^\prime \}_{n=1}^{\infty}$ are independent and therefore this would imply $\mathbb{P}(A_n^\prime \,\, \text{i.o.}) = 1$.  But this last statement cannot be true by the construction of $A_n^\prime$, and hence $\sum_{n=1}^{\infty} \min (p_n, 1 - p_n) < \infty$.
A: Denote $\min(p_n, 1 - p_n)$ by $\alpha_n$.  For each $i$, let $B_i$ be either $A_i$ or $A_i^c$, then
\begin{align}
P(B_i) \leq \max(P(A_i), P(A_i^c)) = \max(p_i, 1 - p_i) = 1 - \alpha_i. \tag{1}
\end{align}
By the independence assumption and $(1)$, for every $n$:
\begin{align}
P(B_1\cap\cdots\cap B_n) = \prod_{i = 1}^nP(B_i) \leq \prod_{i = 1}^n(1 - \alpha_i) 
\leq \exp\left(-\sum_{i = 1}^n\alpha_i\right). \tag{2}
\end{align}
In the last inequality of $(2)$, we used the inequality $1 - x \leq e^{-x}$ for $x \in \mathbb{R}^1$.
Since $(\Omega, \mathcal{A}, P)$ is discrete, $\Omega = \{\omega_1, \omega_2, 
\ldots\}$ is countable and $\sum_k P(\{\omega_k\}) = 1$. For each $k$ and $n$, since $\omega_k$ lies in one of $2^n$ sets of form $B_1\cap\cdots\cap B_n$, it follows by $(2)$ that
\begin{align}
P(\{\omega_k\}) \leq P(B_1\cap\cdots\cap B_n) \leq \exp\left(-\sum_{i = 1}^n\alpha_i\right). \tag{3}
\end{align}
If $\sum_i \alpha_i = \infty$, letting $n \to \infty$ in $(3)$ yields
$P(\{\omega_k\}) = 0$, which would imply $\sum_k P(\{\omega_k\}) = 0$, contradiction.   Hence $\sum_i\alpha_i < \infty$.  This completes the proof.
