If the given function is of bounded variation. Let  $f:[0,1]\rightarrow R$   be a  function  defined  by
$ f(x)= 0 $ if  x  is  irrational
$f(x)$=${1}\over {q^{2}}$ for $x$=${p}\over{ q} $  where  $p$  and  $q$  are  relatively  prime
$f(0)=0$
 $f(1)=1$ 
Then  is  $f$  of  bounded  variation$?$ 
If  I  take  partitions of  the  interval $[0,1]$ as  $\{x_{0},x_{1},.......x_{n}\}$  such  that  $x_{i}$  is  irrational for  odd  $i$ and  rational  for  even  $i$. Then $\sum_{1}^{n}|\Delta f_{k} |$=$\sum_{i=1}^{[n/2]}$ ${1}\over {q^{2}} $.Now  taking  limit $n\rightarrow \infty$  this  series  is $\sum_{i=1}^{\infty}$${1}\over {n^{2}}$ which  is  convergent  hence  bounded  by  some $M$$\in R$. So  by  definition  of  bounded  variation  this  function  is  of  bounded variation. Am  I  correct? But I  did not use  the  definition  of  $f$  at  $0$  and  $1$  .

You can't bound the variation by $\sum \frac{1}{n^2}$, since $f$ attains the value $\frac{1}{n^2}$ multiple times (for $n > 2$), e.g. we have $f\bigl(\frac{1}{6}\bigr) = f\bigl(\frac{5}{6}\bigr) = \frac{1}{6^2}$. Generally, the value $\frac{1}{n^2}$ is attained at each reduced fraction with denominator $n$. A fraction $\frac{k}{n}$ is reduced if and only if $\gcd(k,n) = 1$, and the number of such $k$ not exceeding $n$ is given by Euler's totient function: $f$ attains the value $\frac{1}{n^2}$ exactly $\varphi(n)$ times.
To see that $f$ is of unbounded variation, it suffices to consider the primes. For any finite set $F$ of primes, there are partitions of $[0,1]$ such that all points $\frac{k}{p},\, 1 \leqslant k < p$ with $p\in F$ occur in the partition, and between any two successive such points there occurs an irrational partition point. Thus, for any $x\in (0,+\infty)$, there is a partition with
$$\sum \lvert \Delta f_k\rvert = 2\sum_{p \leqslant x} \frac{p-1}{p^2} \geqslant 2\sum_{p \leqslant x} \frac{1}{p} - 2\sum_{n = 1}^\infty \frac{1}{n^2}.$$
Since
$$\sum_{p \leqslant x} \frac{1}{p} \sim \log \log x$$
(Mertens' second theorem), we see that the variation of $f$ is unbounded.
The values $f(0)$ and $f(1)$ are irrelevant, except that they need to be defined so that we have a function defined on the whole interval $[0,1]$. If a function of bounded variation is changed at finitely many points, the resulting function still has bounded variation - the value of the total variation usually changes, of course.
