Of course.
Restrict Cantor's staircase to the Cantor set.
[Here is the graph of Cantor's staircase. Now remove the (open) horizontal line-segments.]
The domain consists of the Cantor set set $C$, that is reals of the form $$ x = \sum_{k=1}^\infty 2\epsilon_k 3^{-k} $$ where $(\epsilon_k) \in \{0,1\}^{\mathbb N}$. That is, $x \in [0,1]$ has an expansion in base $3$ using only the digits $0$ and $2$. This set $C$ has Lebesgue measure zero.
The image of such a point $x$ is $$ f(x) = \sum_{k=1}^\infty \epsilon_k 2^{-k} $$ The range of the function is $[0,1]$, which has Hausdorff dimension $1$, so the graph of $f$ has Hausdorff dimension at least $1$.