Let $u$: $(0,1)\to \mathbb R$ be defined as $$ u(x):=\sum_{n=1}^\infty \frac{1}{2^n} \cos(2^n\pi x)$$ I am trying to prove that $\operatorname{Var}(u)<\infty$
It is clear that $\operatorname{Var}(\frac{1}{2^n} \cos(2^n\pi x))=2$ but then I will have $\operatorname{Var} (u)\leq \sum_{n=1}^\infty 2 =\infty$... dead end.....
I then try to define $$u_N:=\sum_{n=1}^N \frac{1}{2^n} \cos(2^n\pi x)$$ and show that $\operatorname{Var}(u_N)$ is uniformly bounded, but I got nothing from this path too.
This is a modified form of the Weierstrass function with $a=1/2$ and $b=2$. As the classical Weierstrass functions (those with $ab>1+\frac32\pi$), this one is nowhere differentiable, but this fact is much harder. It was proved less than 100 years ago, by Hardy: the Wikipedia article has a reference.
Since the functions of bounded variation are differentiable a.e., this function is not of bounded variation.