Prove that there is a continuous $f$ not Riemann-Stieltjes integrable with respect to $\phi$

Question

Let $V[\phi;a,b]=\infty$, show there is a continuous function $f$ such that $$\int_a^b f\ d\phi$$ not exists.

Info: This result seems easy to follow from the results already proved. But I couldn't answered yet. $V$ represents the total variation of $\phi$, $V=\sup_{\Gamma}\sum_{i}|\phi(x_i)-\phi(x_{i-1})|=\infty$, $\Gamma$ is a partition of $[a,b]$. The hint says use the following two results:

If $V[\phi; a,b]=\infty$. Show there exists a point $z\in[a,b]$ and a monotone sequence $(z_n)_{n\geq 1}$ such that $z_n\to z$ and such that $\sum_{i=1}^{\infty} |\phi(z_i)-\phi(z_{i-1})|=\infty$

Which I have already proved here: There exists $(z_n)_{n\geq 1}$ monotone $z_n\to x_0$ such that $\sum |\phi(z_i)-\phi(z_{i-1})|=\infty$

If $\{a_k\}$ is a sequence of positive terms with $\sum a_k=\infty$, then there is another sequence of positive terms $\{\xi_k\}$ such that $\xi_k\to 0$ and $\sum\xi_k a_k=\infty$.

Any help is appreciated. The complicated hint should be there for a reason, I can't see yet. I tried using for example $f\equiv 1$ but then $\int_a^b fd\phi= \lim_{m\to\infty}\sum_{i=1}^m (\phi(x_i)-\phi(x_{i-1}))$ this is not precisely $\infty$, at least not immediatly, because here $\phi$ could be any function (including discontinuous).

Answer 1

One solution would be to use Riesz representation theorem, i.e. if the integral is finite for any continuous function then since it is linear it is represented by a function of a bounded variation say $\psi$. Then one would have to prove that if $\int fd\phi = \int f d\psi$ for every continuous function $f$ then $\phi = \psi + constant$. (Which probably is also part of Riesz's theorem i.e. the function we integrate against is unique).
But going back to the hints in the problem. You have to construct continuous function $f$ such that $\sum f(y_i)(\phi(x_{i+1})-\phi(x_i))$ tends to $\infty$. Ideally, $f(y_i) = sign(\phi(x_{i+1}) - \phi(x_i)$ but such function is not continuous of course.
Here come your hints, take $\xi_k\to 0$ such that $\sum_{i=1}^{\infty} \xi_k|\phi(z_i)-\phi(z_{i-1})|=\infty$ and define $f$ to be $f(z_i) = \xi_k * sign (\phi(x_{i+1}) - \phi(x_i))$. That is, at this moment, $f$ is defined on a convergent sequence only but extend it linearly to continuous function on $[0,1]$.