Motivation: In a software project I know well, for testing purposes, the evolution of a particular price is simulated by repeatedly multiplying by a random number picked between $0.995$ and $1.005$.
Let us formalize this.
Let $(X_i)_{i=1}^\infty$ be a sequence of independent random variables, each with a (continuous) uniform distribution in $[0.995,1.005]$, $$X_i \sim U([0.995,1.005])$$ For each $n\in\mathbb{N}$ we define the random variable $$P_n = \prod_{i=1}^n X_i$$ which represents the price at day $n$.
I call for a proof of the fact that the price tends to zero as $n$ grows without bound. One way of formalizing this is to define a map $P_\infty$ by: $$P_\infty : \omega\mapsto\begin{cases} \lim_{n\to\infty}P_n(\omega),&\text{if the limit exists} \\ \text{undefined},&\text{otherwise} \end{cases}$$ where $\omega$ is an abstract outcome of the probability space. Then can you show that $P_\infty$ is defined almost everywhere (and is measurable) so that we can see $P_\infty$ as a random variable, and prove that the distribution of $P_\infty$ is that of a constant, i.e. $$\mathrm{Pr}(P_\infty = 0) = 1$$ where $\mathrm{Pr}$ is the probability measure?
Bonus question: One way to "fix" the issue with the price going towards zero, would be to change the distribution of the $X_i$ into $X_i=\exp(D_i)$ where the $D_i$ are independent and all distributed $D_i \sim U([-0.005,0.005])$.
However, suppose for the purpose of the exercise, that we were required to keep the $X_i$ uniformly distributed, and could only change the lower endpoint, into, say, $\alpha$. So now $$X_i \sim U([\alpha,1.005])$$ and otherwise everything is as before. For too low values of $\alpha$, $P_\infty$ will be $0$ as before. But for higher values of $\alpha$, it appears that $P_n$ would (almost surely) go to $+\infty$ instead.
So the bonus question is, what "Goldilocks" value of $\alpha$ should be chosen so that $P_\infty$ is neither $0$, nor $+\infty$?