TITLE: A continuous function on $[0,1]$ not of bounded variation
QUESTION [25 upvotes]: I'm looking for a continuous function $f$ defined on the compact interval  $[0,1]$ which 
is not of bounded variation.
I think such function might exist. Any idea?
Of course the function $f$ such that
$$
f(x) =
\begin{cases}
1  & \text{if $x \in [0,1] \cap \mathbb{Q}$} \\\\
0 & \text{if $x \notin [0,1] \cap \mathbb{Q}$}
\end{cases}
$$
is not of bounded variation on $[0,1]$, but it is not continuous  on $[0,1]$.

REPLY [7 votes]: Here is an general idea. A function of bounded variation on $[0,1]$ is necessary differentiable a.e.. In another word, any continuous function which is not differentiable on a positive measure set is not of bounded variation. So, Weierstrass Function is not of bounded variation for sure.
Also, some function which oscillate too much can not be bounded variation as well, for example
$$ u(x)=x^a\sin(\frac{1}{x^b}) $$
is not of bounded variation as long as $1 \le a\leq b$.