A series of independent random variables is given, $ X_n $, which is uniformly distributed in an interval $ [0,1]$. We are supposed to determine a limit:
$$ \lim_{n\to \infty} (\prod_{i=1}^n X_i )^{1/n} $$
Since we're doing measure theory, and we can use the law of large numbers (Kolmogorov), I was trying to transform the product into a sum and then determine the limit. I am not sure how to do that, this product looks like a geometric mean and I wanted to transform it to an average mean with this law with logarithms. It looks way too complicated, so I wanted to ask if I'm missing out on a certain property of uniformly distributed random variables so it can be done in an easier way.
Let's take the logarithm of the limit and then undo that by taking the exponential, and we can nice $\ln$ inside $\lim$ so the desired result is $\exp\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\ln X_i$. In other words, it's $e^\mu$ with $\mu:=\mathbb{E}(\ln X_1) =\int_0^1 \ln x dx$. I'll leave pricing $\mu=-1$ as an exercise.