So that the reader is not under the impression that there are only few measure zero sets and that these sets are uncomplicated, here is an uncountable, compact, measure zero subset of \([0,1]\text{,}\) which contains no intervals. Any \(x \in [0,1]\) can be expanded in ternary:

\begin{equation*}
x = \sum_{n=1}^\infty d_n 3^{-n} ,
\qquad \text{where } d_n=0, 1, \text{ or } 2.
\end{equation*}

\begin{equation*}
C \coloneqq \Bigl\{ x \in [0,1] : x = \sum_{n=1}^\infty d_n 3^{-n},
\text{ where } d_n = 0 \text{ or } d_n = 2 \text{ for all } n \Bigr\} .
\end{equation*}

That is, \(x\) is in \(C\) if it has a ternary expansion in only \(0\)s and \(2\)s. If \(x\) has two expansions, as long as one of them does not have any \(1\)s, then \(x\) is in \(C\text{.}\) Define \(C_0 \coloneqq [0,1]\) and

\begin{equation*}
C_k \coloneqq \Bigl\{ x \in [0,1] : x = \sum_{n=1}^\infty d_n 3^{-n},
\text{ where } d_n = 0 \text{ or } d_n = 2 \text{ for all } n=1,2,\ldots,k \Bigr\} .
\end{equation*}

Clearly,

\begin{equation*}
C = \bigcap_{k=1}^\infty C_k .
\end{equation*}