Prove $S \doteq \sum_{n=1}^{\infty} p_n < \infty \to \prod_{n=1}^{\infty} (1-p_n) > 0$ assuming $0 \leq p_n < 1$.
Hint: Show that $S < 1 \to \prod_{n=1}^{\infty} (1-p_n) \geq 1 - S$.
Assuming the hint is true:
If $S \geq 1$ but $S < \infty$, then $\exists k \in \mathbb{N}$ s.t. $\sum_{n=k}^{\infty} p_n < 1$.
Define $T \doteq \sum_{m=1}^{\infty} q_m = \sum_{n=k}^{\infty} p_n < 1$ where $q_1 = p_k, q_2 = p_{k+1}, ...$
By hint, $T \doteq \sum_{m=1}^{\infty} q_m < 1 \to \prod_{m=1}^{\infty} (1-q_m) > 0 \to \prod_{n=k}^{\infty} (1 - p_n) > 0 \to \prod_{n=1}^{\infty} (1 - p_n) > 0$ (a reason why $p_n < 1$, I guess)
QED
Is that right?
There is an answer here, but I would like to try a different approach using continuity of measure, monotone convergence theorem and definition of supremum.
What I tried (I'm not sure if my lim, inf and sup statements are right):
Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space, and let $X_n$ be iid Bernoulli random variables with parameter $p_n$ and in $(\Omega, \mathscr{F}, \mathbb{P})$. Then,
$$\sum_{n=1}^\infty p_n<\infty \iff \sum_{n=1}^\infty P(X_n = 1) <\infty$$
By Borel-Cantelli Lemma 1 we have, $$P(\limsup(X_n = 1))=0$$
which is equivalent to $$P(\liminf(X_n = 0)) = 1$$
$$ \ $$
$$\to 1 =P(\bigcup_{N \ge 1} \bigcap_{n \ge N} (X_n=0))$$
By continuity of measure $\color{red}{\text{(I think?)}}$
$$= \lim_{N \to \infty}P(\bigcap_{n \ge N} (X_n=0))$$
By monotone convergence theorem $\color{red}{\text{(I think?)}}$
$$= \sup_{N \ge 1}P(\bigcap_{n \ge N} (X_n=0))$$
By independence,
$$ = \sup_{N \ge 1} \prod_{n=N}^{\infty} P(X_n=0)$$
$$= \sup_{N \ge 1} \prod_{n=N}^{\infty} (1-p_n)$$
$$\to \sup_{N \ge 1} \prod_{n=N}^{\infty} (1-p_n) = 1$$
$$\to \forall \epsilon > 0, \exists m \ge 1 \text{s.t.} \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1] \ \color{red}{\text{(I think?)}}$$
Since $p_n < 1 \ \forall n \in \mathbb{N}$,
$$\forall \epsilon \in (0,1), \exists m \ge 1 \text{s.t.} \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1] \ \color{red}{\text{(I think?)}}$$
Hence $\forall \epsilon \in (0,1), \exists m \ge 1 \text{s.t.}$
$$\prod_{n=1}^\infty(1-p_n)=\prod_{n=1}^{m-1}(1-p_n)\prod_{n=m}^{\infty}(1-p_n) > 0 \ \because$$
$\prod_{n=1}^{m-1}(1-p_n) > 0 \because p_n < 1 \ \forall n \ge 1$
$\prod_{n=m}^{\infty}(1-p_n) > 0 \because \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1]$ QED
Any mistakes? Again, I'm not sure if my lim, inf and sup statements are right.