My original problem is given $X_i\sim^{iid}U[0,1]$, find $$\lim_{n \rightarrow \infty} (X_1X_2 \cdots X_n)^{1/n} = \lim_{n \rightarrow \infty} (\prod_{i=1}^{n} X_i)^{1/n}$$
Well, $$\lim_{n \rightarrow \infty} (X_1X_2 \cdots X_n)^{1/n} = \lim_{n \rightarrow \infty} (\prod_{i=1}^{n} X_i)^{1/n} = \lim_{n \rightarrow \infty} \exp\{\frac{1}{n}\sum_{i=1}^{n} \log(X_i)\}$$ I'm thinking of using the weak law of large numbers.
So, I would like to compute $$E(\log(X_i))=\int^1_0 \log(x)\ dx$$
However, I'm supposed to be using the Lebesgue integral, and the only theorem that I know of that ensures me the Lebesgue integral will be equal to the improper Riemann integral is when the latter ($\int f \, dx$) exists and $f$ is non-negative, which is not my case $\because \log(x) < 0 \ \forall x \in (0,1)$
Therefore, what other theorems could ensure me that improper Riemann integral is the same as the Lebesgue integral?
Any help would be appreciated.
$−\log{x}$ is non-negative, and the integral is linear.
In general, a function $f$ is Lebesgue-integrable on $X$ when it has an improper Riemann integral on $X$, and the (Lebesgue-)integrals of the positive and negative parts of $f$ are finite; this can be expressed as "$f$ is measurable and $\int_X \lvert f \rvert<\infty$", if I'm not forgetting some stupid special cases.