Does $x \ f(x) \in L^1(\mathbb{R})$ (lebesgue integrable) imply $f(x)$ is in $L^1(\mathbb{R})$

Question

I am wondering what conditions must be true so that $$x \ f(x)\in L^1(\mathbb{R}) \Longrightarrow f(x) \in L^1(\mathbb{R})$$

I have been trying to find a counter-example but I have not found any yet.

Any help? Is there a general theorem/rule regarding my confusion?

EDIT: I am interested in the domain being all of $\mathbb{R}$.

Answer 1

Consider the set $[0,1]$ with Lebesgue measure and the function $f(x)=\frac{1}{x}$.

Then $\int_{[0,1]}{xf(x) dm(x)}=1$ but $\int_{[0,1]}{f(x)}=\infty$.

For your edit just consider $f(x)=\chi_{[0,1]}\frac{1}{x}$.