I believe that the answer is "in some sense" no, and that this follows from the definition of a set with measure zero. See: https://en.wikipedia.org/wiki/Null_set
The intuitive idea is that any subset of the real line with measure zero can be covered by open intervals whose "total length" can be arbitrarily small.
If we take $\epsilon = \frac{1}{n}$, then any measure zero set is either the limit of a decreasing nested sequence of open interval covers (whose "total length" is $\frac{1}{n}$ for each $n$) in which case the set is Borel measurable,
or it is a proper subset of a measure-zero Borel set, so not the limit of the decreasing nested sequence, but a subset of a limit of a decreasing nested sequence.
If any set of measure zero were the limit of an increasing nested sequence of sets, all of the sets in the sequence would themselves have to be measure zero.
However, it may be possible to find such an increasing sequence of uncountable sets, all of which have measure zero, such that their limit is again an uncountable set with measure zero.
So the answer depends on which nested sequences you are willing to consider.