In an exercise for class we were asked to prove that the graph of a continuous measurable function has measure zero.
Ok, so let us just look at some measurable function that is not necessarily continuous. For example the characteristic function on the set of irrationals. Then the graph is an uncountable union of points. I know that each point has measure zero, but how do we deal with the fact that we have uncountably many such points ?
I think the graph of such a function would then have measure equal to the measure of the irrationals which is not zero.
If $f:\mathbb{R}^{d}\to\mathbb{R}$, then in the referenced exercise, you most likely proved $\mathcal{G}(f)$ (read: graph of $f$) has $d+1$-dimensional measure $0$, which is an important distinction, because while it does not make sense in the framework of Lebesgue measure, $\mathcal{G}(f)$ has in general non-zero $d$-dimensional measure (think arc-length and surface area; this leads into the subject known as geometric measure theory).
So even for your discontinuous counter-example we have $\mu^{d=2}(\mathcal{G}(f))=0,$ which should be obvious.
Since the graph of your example is just a "countably punctured" horizontal line in $\mathbb{R}^{2}$, you could in principle compute $\mu^{d=1}(\mathcal{G}(f))$ without invoking geometric measure theory to see that it has non-zero measure, but you ought to refrain from going there since that line of thinking is somewhat misguided
I'm not actually familiar with any function where $\mu^{d+1}(\mathcal{G}(f))\neq0$. But see space filling curve to get a sense of what such a function (if one even exists) must exhibit.