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.
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 $0$.
Denote $E_2^c =$ the set of real numbers $x$ on the interval $[0, 1]$ such that in the decimal form of $x = 0.i_1 i_2 i_3 \dots$ there is no digits $2$.

Write $x \in E_2^c$ in decimal form
    $$
 x = 0. i_1 i_2 i_3 \dots, \ \ 
 i_k = 0, 1, 3, 4, 5, 6, 7, 8, 9, \ \ 
 x = \sum_{k=1}^{\infty} \frac{i_k}{10^k}, 
 $$
    and assume that the first digit $2$ appeals at $k$th decimal of $x$.

Denote $E_{2,1}$ as the set of real numbers $x$ on the interval $[0, 1]$ such that the first digit $2$ appeals at $1$th decimal of $x$:
    $E_{2,1} 
 = \{x | x = 0. 2 \dots\} 
 = [0.2, 0.3)$, 
    then $m(E_{2,1}) = 10^{-1}$;

Denote $E_{2,2}$ as the set of real numbers $x$ on the interval $[0, 1]$ such that the first digit $2$ appeals at $2$th decimal of $x$:
    $E_{2,2} 
 = \{x | x = 0. i_1 2 \dots\} 
 = \cup_{i_1} [0.i_1 2, 0.i_1 3)$, 
    $i_1 \in \{0, 1, 3, 4, 5, 6, 7, 8, 9\}$, 
    then $m(E_{2,2}) = 9 \times 10^{-2}$;

Inductively, 
    \begin{align}
 E_{2,k} 
 &amp;= \{x | x = 0. i_1 \dots i_{k-1} 2 \dots\} \\ 
 &amp;= \cup_{i_1, \dots, i_{k-1}} [0. i_1 \dots i_{k-1} 2, 0. i_1 \dots i_{k-1} 3),
 \end{align}
    where $i_1, \dots, i_{k-1} \in \{0, 1, 3, 4, 5, 6, 7, 8, 9\}$, 
    then $m(E_{2,k}) = 9^{k-1} \times 10^{-k}$;

Define $E_2 =$ the set of real numbers $x$ on the interval $[0, 1]$ such that in the decimal form of $x = 0.i_1 i_2 i_3 \dots$ there is digits $2$.

Then the measure of $E_2$ is 
    $$
 m(E_2) 
 = m(\cup_{k=1}^{\infty} E_{2,k})
 = \sum_{k=1}^{\infty} m(E_{2,k})
 = \sum_{k=1}^{\infty}  9^{k-1} \times 10^{-k} 
 = 1.
 $$

Thus, the measure of $E_2^c$ is 
    $$
 m(E_2^c) 
 = m([0, 1] \setminus E_2) 
 = m([0, 1]) - m(E_2) 
 = 1 - 1 
 = 0. 
 $$