How do I prove that if $g'$ and $fg'$ are integrable then $f$ is integrable along $g$?

Question

Let $g:[a,b]\rightarrow \mathbb{R}$ be a differentiable function of bounded variation such that $g'$ is Riemann-integrable.

Let $f:[a,b]\rightarrow \mathbb{R}$ be a bounded function.

If $fg'$ is Riemann-integrable, then is $f$ Riemann-Stieltjes integrable along $g$?

I have proven that "If $f|g'|$ is Riemann-integrable then $f$ is Riemann-Stieltjes integrable along $g$"

However, I think this does not help the original problem and I'm stuck for hours.

Thank you in advance

EDIT:

Define $g_1(x)=1/2(V_a^x(g) + g(x) )$ and $g_2(x)=1/2(V_a^x (g) - g(x))$, hence $g=g_1 - g_2$.

I have proven that, for a fixed $\epsilon$, there exists a partition $P=\{x_0,...,x_n\}$ of $[a,b]$ such that for every $s_i,t_i\in [x_{i-1},x_i]$ such that $|\sum_{i=1}^n (f(s_i) -f(t_i)) (g(x_i) - g(x_{i-1})) | < \epsilon$.

Note that the value in the absolute value is the "difference of two sums", namely, $\sum (f(s_i) - f(t_i) ) (g_1(x_i)- g_1(x_{i-1}))$ and $\sum (f(s_i) - f(t_i) ) (g_2(x_i)- g_2(x_{i-1}))$.

Hence we get $|(U(P,f,g_1) - L(P,f,g_1) ) - (U(P,f,g_2) - L(P,f,g_2))| < \epsilon$.

The thing we must show to prove the problem is to show that $|U(P,f,g_1)-L(P,f,g_1) | < \epsilon$ and $|U(P,f,g_2) - L(P,f,g_2) | <\epsilon$. Triangle inequality does not help this

Gaps between lower sums and upper sums $g_1$ and $g_2$ possibly delete each other mutually.

Answer 1

Okay. It is true for the Lebesgue integral. But it is very technical.

Reference: Giovanni Leoni: A First Course in Sobolev Spaces

1. A differentiable function has the Lusin N property (Corollary 3.14 cf the reference above).

2. A continuous function with bounded variation and Lusin N property is absolutely continuous (Corollary 3.27).

3. For a absolutely continuous integrator, the statement holds. In fact you have $\int f dg = \int fg'$ (in Lebesgue sense!).