Lebesgue integrable function at $[0,+\infty)$

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Let $f:[0,+\infty) \to \mathbb R$ be a Lebesgue integrable function. Prove that there is a sequence $(x_n)_{n\geq 1}$ such that $\lim_{n \to \infty}x_n=+\infty$ and $\lim_{n \to \infty}f(x_n)=0$.

What I thought of was trying to show it by the absurd: suppose that for every sequence $\lim_{n \to \infty}x_n=+\infty$, $\lim_{n \to \infty}f(x_n) \neq 0$ or diverges. Then $\lim_{n \to \infty} |f(x_n)|>0$ or it diverges ($\lim_{n \to \infty} |f(x_n)|=+\infty$).

On the other hand, if $f$ is integrable, then $|f|$ is integrable, i.e., $$\int_{[0,+\infty)}|f|dx<\infty$$

So, to arrive to an absurd I would like to show that this integral is not finite. How could I prove this?

I would appreciate any suggestions or hints.

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Let us prove that for every $n$ there is $x_n > n$ such as $$ |f(x_n)| < \frac 1n $$

Otherwise, you have $$ \exists n \ \ \ \forall x > n \ \ \ |f(x)| \ge \frac 1n $$ in which case $\int |f| = \infty$.