Let $\varphi : \mathbb{R}^d \rightarrow \mathbb{R}$ be a measurable function that verifies the following property: for every rectangle $Q$ one has:
$$\left| \int_Q \varphi(x) dx \right| \le \frac{M \cdot m(Q)}{1+m(Q)}$$
For some constant $M,$ independent of $Q$. Show that for any $f \in L^1(\mathbb{R}^d)$ we have:
$$ \lim_{k\rightarrow \infty} \int \varphi(kx) f(x) dx = 0.$$
My try:
It is easy to see that if $f$ is a characteristic of an interval, then the limit holds, by the density of these functions, there should be some way to pass the limit, but I don't know how to do it. Any hints are appreciated.
Hint: From the condition $$ \frac1{m(Q)}\left| \int_Q \varphi(x) dx \right| \le \frac{M}{1+m(Q)}, $$ deduce that $$ |\varphi(x)|\le M \ \ \text{ for almost every }x\in\Bbb R^d. $$ Then, define $$ J =\{f\in L^1 : \lim_{k\to\infty}l_k(f) = 0\}, $$where $$ l_k : L^1\ni f\mapsto \int \varphi(kx)f(x) \mathrm dx,\quad k\ge 1. $$ Show that $J$ is a dense, closed subspace of $L^1$. This easily implies that $J=L^1$.