real analysis - For lebesgue integrable function $f$, show that there exists a sequence ${x_n} rightarrow infty$ such that $x_n| f(x_n) | rightarrow 0$

As in the title, $f$ is any integrable function, and we want to show that there exists $x_n \rightarrow \infty$ as $n \rightarrow \infty$ with the desired property. So far I imagine that it's easy to find $x_n$ increasing slowly enough that the tails of $|f(x_n)|$ will go to zero faster than $x_n$ diverges as $n \rightarrow \infty$. However, it's not true that all integrable functions decay to zero, as I've come across some counterexamples of that. I want to use some consequences of the integrability of $f$ but haven't found the right way to apply them.

I also thought of choosing some slowly diverging $x_n$ like $\log(n)$ and supposing that $x_n |f(x_n)|$ does not converge to zero in order to obtain some contradiction of $f$ being integrable.

Any hints or directions are greatly appreciated!

Answer

The assertion is equivalent to

$$\liminf_{x \to \infty} \big( x |f(x)|\big) =0. \tag{1}$$

We prove by contradiction that $(1)$ holds for any Lebesgue integrable function $f$. Suppose that $(1)$ does not hold true, then we can find $c>0$ and $R>0$ such that

$$x |f(x)| \geq c \quad \text{for all $x \geq R$}.$$

Thus,

$$\int_{x \geq R} |f(x)| \, \lambda(dx) \geq c \int_{x \geq R} \frac{1}{x} \, \lambda(dx) = \infty$$

in contradiction to our assumption that $f$ is integrable.

Remark: The proof actually shows that

$$\liminf_{x \to \infty}\big( h(x) |f(x)| \big) = 0$$

for any Lebesgue integrable function $f$ and any function $h: \mathbb{R} \to (0,\infty)$ such that

$$\int_{x \geq 1} \frac{1}{h(x)} \, \lambda(dx)=\infty,$$

e.g. $h(x) := x |\log x|$.