Is there a null-set whose translations generate the set of all null-sets? Under the Lebesgue measure, is there a null-set $N$ whose translations generate the set of all null-sets when closed under countable unions and countable intersections?
I know that such an $N$ cannot be countable, since the set of all countable sets produce nothing new under countable unions and countable intersections.
 A: No, there isn't. This follows from pure cardinality considerations: Since every subset of the Cantor set is a nullset, and the Cantor set is of cardinality $\frak c$, there are at least $2^{\frak c}$ nullsets (and of course at most $2^{\frak c}$ nullsets). But given a nullset $N$ there are only $\frak c$ translations of $N$ and closing under countable union and intersection won't raise this (for details see the proof that there are $\frak c$ Borel sets).

In the comments, bof has asked if we replace countable intersection with taking subsets, if such a magical nullset exists, and I believe the answer is still no, although there are still some issues with the proof. I'll condition on the result that for "most" $a\in[0,1]$ (it suffices to be positive measure, but I think there are countably many exceptions), $aC+b$ has measure zero in the cantor measure on $C$ (where $C$ is the $[0,1]$-Cantor set).
The trick is that these scaled Cantor sets are "unreachable" from each other; no amount of countable unions will make one big in the measure on the other. Given $N$, there are at most a nullset of $a\in[0,1]$ such that there is a $b$ with $N+b$ having measure $>0$ in the cantor measure on $aC$, because otherwise $N$ would have positive measure (not sure about justification here), and so there is an $a\in[0,1]$ such that $N+b$ has measure zero in the cantor measure on $aC$. Then any countable union of such sets and any subset of this is also measure zero in $aC$, and so $aC$ is a nullset which is missed by $N$.
