Examples of uncountable sets with zero Lebesgue measure

Let $(r_n)_{n\in\mathbb{N}}$ be a dense sequence in $\mathbb{R}$ (it could, for example, be an enumeration of the rationals). For $k \in \mathbb{N}$, let

$$U_k = \bigcup_{n\in\mathbb{N}} (r_n - 2^{-(n+k)},\, r_n + 2^{-(n+k)}).$$

$U_k$ is a dense open set with Lebesgue measure $\leqslant 2^{2-k}$, thus

$$N = \bigcap_{k\in\mathbb{N}} U_k$$

is a set of the second category (Baire), hence uncountable, and has Lebesgue measure $0$.

Your question is in some part historically related to the investigation of the Continuum Hypothesis CH, i.e. whether every uncountable subset of $\mathbb{R}$ is in bijection with $\mathbb{R}$. A perfect subset is closed set with no isolated points. It is well-known that the Cantor set continuous injects into all perfect subsets. (So perfect sets cardinality $2^{\aleph_0}$.) One way of solving CH was to show whether all uncountable set was the union of a perfect set and an countable set. This is called the perfect set property.

So one way of formalizing your idea of "not obtained from a Cantor Set" is to say that your set is not the Cantor set union a countable set. So taking this interpretation of question, a satisfactory answer would be to produce an uncountable measure zero set without a perfect subset.

Note that without the axiom of choice and with the axiom of determinacy, all uncountable sets have the perfect set property. So in such universe, you will not be able to produce any such example. In particular, if you are working $ZF$ set theory, you will not be able to produce such example (unless the axiom of determinacy is inconsistent, which no one knows.)

I am not sure if such a set exist in ZFC. However, by assuming some more axioms that are shown to be equiconsistent with ZFC, such a set does exist.

$Add(null)$ is the additivity of ideal of measure sets. That is if $\kappa < Add(null)$ is a cardinal, then the union of $\kappa$ many measure zero sets is still measure zero. $ZFC + \neg CH + Add(null) = 2^{\aleph_0}$ is consistent with $ZFC$.

So in such a universe, you can take any the Cantor set $C$ of measure zero. Use the axiom of choice to well order $C = \{a_\xi : \xi < 2^{\aleph_0}\}$. Then $\{a_\xi : \xi < \aleph_1\}$ is an uncountable measure zero set since $\aleph_1 < 2^{\aleph_0} = add(null)$.

You may not be too happy with this example I actually did use a Cantor set and carved out a uncountable measure zero subset. If you would like: Using the axiom of choice there exists disjoint uncountable set $A$ and $B$ such that $A$ and $B$ intersect every perfect subset of $\mathbb{R}$. In particular $A$ is does not contain any perfect subsets. (You can find this construction in $\textit{Descriptive Set Theory}$ 2C.4.) So this $A$ is very different from a Cantor set. In fact, if you look at the construction it is not clear if $A$ is even meaureable. Then apply the same construction above to carve out a uncountable measure zero subset of this set $A$.

So in conclusion, in the equiconsistent extension $ZFC + \neg CH + add(null) = 2^{\aleph_0}$ such a measure set exists. Using only $ZF$ and no choice, no such example can be found (if AD is actually consistent). I am still not sure if $ZFC$ can produce such an example.

A very simple example is the set of numbers not normal in base $10$. This set is not obtained from a Cantor set in any way (and is in fact dense in $[0,1)$) and has zero measure and is obviously uncountable. Additionally it has full Hausdorff dimension and is of the second category.

The set of badly approximable numbers works for this, too.

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.