We have that for $f$ $\in$ $L^{1}$$(\mathbb{R}^n)$, $g$ measurable and bounded on $\mathbb{R}^n$ and, for any rectangle $R$,
$$ \lim_{m(R)\rightarrow\infty} \frac{1}{m(R)}\int_Rg(x)dx = 0 $$
Then:
$$ \lim_{k\rightarrow\infty} \int_\mathbb{R^n}g(kx)f(x)dx = 0 $$
This can be proved by solving the problem for step functions and using their density on $L^1$. What I'm trying to do is to replace the assumption. More precisely, if I have that the integral of $g$ on $\mathbb{R}^n$ is limited or, let's say, for any rectangle $R$ and a constant $M$
$$ \left|{\int_Rg(x)dx}\right| \leq \frac{M.m(R)}{1+m(R)} $$
Does the conclusion still hold?