Examples of uncountable sets with zero Lebesgue measure I would like examples of uncountable subsets of $\mathbb{R}$ that have zero Lebesgue measure and are not obtained by the Cantor set.
Thanks.

In his answer, William defined "not obtained from the Cantor set" as "not the union of a [perfect] set and a countable set." In particular, any set not containing a perfect set is not obtained from the Cantor set. He then asked whether an uncountable, measure zero set of this type exists provably in ZFC.
The answer is yes. ZFC proves the existence of a Bernstein set - a set $B$ such that for every perfect set $P$, $B\cap P$ and $B^c\cap P$ are each nonempty. Now if $B$ is Bernstein and $P$ is perfect, we have in fact that $B\cap P$ is uncountable and does not contain a perfect set; clearly $B\cap P$ can't contain a perfect set (since then $B^c$ wouldn't meet it), and if $B\cap P$ were countable, then $P\setminus (B\cap P)$ would be an uncountable Borel set, hence (since Borel sets have the perfect set property) would have a perfect subset which would not meet $B$.
OK, so fix a Bernstein set $B$ and a measure zero perfect set $P$ (say, the usual Cantor set) and let $X=B\cap P$. Clearly $X$ has measure zero, and by the above $X$ is uncountable and does not contain a perfect subset, hence is not obtained from the Cantor set in the sense defined in the first paragraph.

Here's a sketch of the ZFC-proof that a Bernstein set exists. We use transfinite induction, and the observation that there are exactly as many reals as perfect sets. At stage $\alpha<2^{\aleph_0}$, look at the $\alpha$th perfect set; pick two reals not already used in this set, and put one of them in $B$, and the other in the complement of $B$. Since perfect sets have size continuum, and by induction we've only used $2\alpha$-many reals so far, we can always find such a pair of reals.
