Is this a bounded variation function?
$$ f(x) = \begin{cases} 0, & \text{if $x$ is irrational} \\ \frac{1}{q}, & \text{if $x = \frac {p}{q}$, with $\frac {p}{q}$ irreducible} \end{cases}$$
I would say it IS a bounded variation function because it is a constant function (and constant functions are of bounded variation) except for a countable set of discontinuities.
Consider a sequence of partitions $\Pi_i$ of $[0,1]$, indexed by $i$, such that $x_0(i)=0$, $x_{2k}(i)=1/k$, for $k=1,2,\cdots,i$ and $x_{k}$ is any (finite) set of irrationals that interlace with $x_{2k}$ (technically, let $y_k$ be the $x_k$ rewritten in increasing order to make a bonafide partition). Then
$$\sum_{x_k \in \Pi_i}|f(x_{k+1}(i)-f(x_k(i))|=\sum_{k=1}^{i}\frac{1}{k},$$
which goes to infinity as $i\rightarrow\infty$. so $f$ is not of bounded variation.
