Is it true that all subsets of the Cantor set have Jordan content zero?
What is the definition of countably generated Boolean algebra?
Does the Boolean algebra of subsets $[0,1]$ which possess Jordan content is countably generated?
Sorry I updated the question.
The middle-thirds Cantor set has Jordan measure zero -- it is the intersection of a sequence of finite unions of intervals whose total length converge to zero.
Therefore every subset of it also has Jordan measure zero, since you can find a simple superset of it with arbitrary small content.
If the Boolean algebra of Jordan measurable sets were countably generated, it would itself be countable because then by definition every set in the algebra is generated from some finite sequence of joins, meets and complements from the generator elements. And these sequences of operations can be coded as integers using routine constructions.
However, every interval $[0,\alpha]$ is Jordan measurable, and there are uncountably many such intervals (namely one for each $\alpha\in [0,1]$). Therefore the algebra cannot be countably generated.