I am trying to answer the following question: Let $g$ be a bounded lebesgue measurable function on $\mathbb{R}$ such that $\lim_{n} \int_I g(nx) dx = 0$ for all intervals $I \subset [0,1]$. Suppose $f \in L^1([0,1])$. Show $\lim_{n} \int_0^1 f(x) g(nx) dx = 0$.
I have the idea that I can treat $g(nx)$ as a radon nikodym derivative, use the jordan decomposition to split $g(nx)dx$ into two nonnegative measures, and show that those assign $0$ mass to the unit interval as $n$ goes to infinity. In this way the desired integral would also tend to $0$. But, I have not been able to properly flesh out the details and would like some help.
The reason I want to use the jordan decomposition to split the measure is so that I can use theorems requiring nonnegative functions like Fatou's lemma to show convergence. I am aware that there are proofs of this problem using other methods but I specifically want to use the idea above for practicing usage of radon nikodym.
Since $\lim_{n\to\infty} \int_I g(nx)\, dx = 0$ for all intervals $I \subset [0,1]$,we get $$\lim_{n\to\infty}\int_0^1f(x)g(nx)\,dx=0$$ for any step function $f(x)=\sum_{k=1}^n a_k\chi_{I_k}$ where $I_k\subset[0,1]$ is an interval.
Now suppose $f\in L^1$. Since step functions are dense in $L^1$, for any $\epsilon>0$ we can write $f=f_1+f_2$ where $f_1$ is a step function and $\int_0^1|f_2|\,dx<\epsilon$. Using what we've proven for step functions and the boundedness of $g$ gives that $$\lim_{n\to\infty}\int_0^1f(x)g(nx)\,dx=0.$$