Let $g\in L_1(\mathbb{R},m)$ bounded function where $m$ is the Lebesgue measure in $\mathbb{R}$. If
$$\lim_{x\to\pm\infty} g(x)=0,$$
show that for all functions $f\in L_1(\mathbb{R},m)$ we have:
$$\lim_{n\to\infty}\frac{1}{n}\int_{\mathbb{R}}g(x)\cdot f\left(\frac{x}{n}\right)\,dm=0.$$
Attempt: I tried exactly $2$ hours, I followed distinct ways but I have not idea how to do, I don't know how to apply Fatou's Lemma or LMCT or LDCT if is possible.
Let $x=ny$. Then $$\frac{1}{n}\int_{\mathbb{R}}g(x)f(\frac{x}{n})dx=\int_{\mathbb{R}}g(ny)f(y)dy.\tag{1}$$ Let $n\to\infty$ and apply dominated convergence theorem to the right hand of side $(1)$. The conclusion follows.