Consider a sequence $\{p_n\}_{n \in \mathbb{N}}$ s.t. $p_n \in [0,1)$ and $\sum_{n=1}^{\infty} p_n < \infty$.
Prove $\prod_{n=1}^{\infty} (1-p_n) > 0$.
I think there's a way to do this without probability eg this or this. I mean there's no probability here originally. But I would like to try a probabilistic proof. Kinda like this or this
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.
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
$$= \lim_{N \to \infty}P(\bigcap_{n \ge N} (X_n=0))$$
By monotone convergence theorem
$$= \sup_{N \ge 1}P(\bigcap_{n \ge N} (X_n=0))$$
By independence,
$$ = \sup_{N \ge 1} \prod_{n=N}^{\infty} P(X_n=0)$$
$$= \sup_{N \ge 1} \prod_{n=N}^{\infty} (1-p_n)$$
$$\to \sup_{N \ge 1} \prod_{n=N}^{\infty} (1-p_n) = 1$$
$$\to \forall \epsilon > 0, \exists m \ge 1 \text{s.t.} \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1]$$
Since $p_n < 1 \ \forall n \in \mathbb{N}$,
$$\forall \epsilon \in (0,1), \exists m \ge 1 \text{s.t.} \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1]$$
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
'Verified by Landon Carter'