 Solution

Let $C_n$ be the union of the $2^n$ closed intervals remaining after $n$ steps of the Cantor set construction. Their total length is $(2/3{)}^n$. Any point outside $C_n$ has a ternary expansion with a $1$ among its first $n$ digits, so $A^c\subseteq C_n$. Endpoints of deleted intervals also belong to $A$ because they have an alternative ternary expansion containing a $1$, as in $2/3=0.1222\dots {}_3$.

A partition using all endpoints of $C_n$ has lower Darboux sum at leastSince $A$ is dense, every upper Darboux sum is one. Letting $n\to \infty$ gives equal upper and lower integrals, and therefore