Let $\{X_i:i\in\{1,\cdots,n\}\}$ be a set of random variables.
I want to show that $\sigma\{X_i:i\in\{1,\cdots,n\}\}=\sigma\{\prod_{k=1}^iX_k:i\in\{1,\cdots,n\}\}$.
It is easy to see that $\sigma\{X_i:i\in\{1,\cdots,n\}\}\supseteq\sigma\{\prod_{k=1}^iX_k:i\in\{1,\cdots,n\}\}$ since $\prod_{k=1}^iX_k$ is measurable with respect to $\sigma\{X_i:i\in\{1,\cdots,n\}\}$.
The other inclusion causes me problems:
$X_i=\frac{\prod_{k=1}^iX_k}{\prod_{k=1}^{i-1}X_k}$
I would like to say that both the numerator and the denominator are measurable with respect to $\sigma\{\prod_{k=1}^iX_k:i\in\{1,\cdots,n\}\}$ and therefore the entire fraction is measurable with respect to $\sigma\{\prod_{k=1}^iX_k:i\in\{1,\cdots,n\}\}$.
Is this reasoning correct? Mainly I have doubts about divide-by-zero errors.