Prove that if $\sum_{t=1} ^{\infty} x_t = \infty$ where $x_t \in [0,1], \forall t$, then $\prod_{t = 1}^{\infty} (1-x_{t}) = 0$ [closed]

Question

Prove that if $\sum_{t=1} ^{\infty} x_t = \infty$ where $x_t \in [0,1], \forall t$, then $\prod_{t = 1}^{\infty} (1-x_{t}) = 0$.

Many thanks.

P/S:

• Not homework but it is a part of the proof in a research paper I am reading.
• It could be trivial but I may have missed something and could not figure it out yet.
• I checked that this result is true for $p$-series.
• My initial idea is that let $a_t := \prod_{\tau=1}^t (1-x_{\tau}), \forall t$ and then try to prove that the series $\sum_{t}a_t$ converges but I have not been able to prove it yet.

Answer 1

The proof can be found in this reference.

Answer 2

Sketch proof (relating the infinite product to the infinite sum via the inequality $e^x \geq 1 + x, \forall x$):

Since $x_t \leq -\log(1 - x_t), \forall t$ and $\sum_{t=1}^{\infty} x_t = \infty$, we have $\sum_{t=1}^{\infty} (-1)\log(1- x_t) = \infty$. Thus, $\prod_{t=1}^{\infty} (1-x_t)$ must either does not exist or equals $0$ (if exists). Since $a_t := \prod_{\tau=1}^{t} (1-x_{\tau}), \forall t$ is a non-increasing sequence and bounded below (by $0$), thus $\lim_{t \rightarrow \infty} a_t$ exists which is then $0$.