Total variation

Question

I cannot decide if the next function has bounded variation: in the segment $(0,1)$

$$f(x)=\begin{cases} \frac{1}{m^{2} n^{2}},&\text{if $x$ is rational}\\\\ 0,&\text{otherwise}. \end{cases}$$ the rational $x$ is $m/n$ and it is the reduced form.

so far showed it is not absolute continuous

thanks

Answer 1

More generally, assume that $f(x)=0$ for every $x$ with the exception that $f(x_n)=a_n$ for some countable set $\{x_n\mid n\in\mathbb N\}$ and some nonnegative sequence $(a_n)_n$ such that $\sum\limits_na_n$ converges.

Then $f=g-h$ where $$ g(x)=\sum_{n\in\mathbb N}a_n\mathbb 1_{x\geqslant x_n},\qquad h(x)=\sum_{n\in\mathbb N}a_n\mathbb 1_{x\gt x_n}. $$ Since $g$ and $h$ are both finite valued and nondecreasing, $f$ has finite variation.

Can you prove that the function $f$ in your post fits this setting, that is, that $f$ corresponds to some $(a_n)_n$ such that $\sum\limits_na_n$ converges?