If $u \in L^1(0,\infty)$, then $|u(x)| \to 0$ as $x \to \infty$?

Question

Let $u \in L^1(0,\infty)$. Does this mean necessarily that $|u(x)| \to 0$ as $x \to \infty$?

I think it has to decay otherwise the integral will be infinite. Can I get a hint on how to prove this? Thank you.

Answer 1

Actually, it is not true. A counterexample is the function that is equal to $1$ only for intervals of the type $[2^n, 2^n + 2^{-n}]$ for every integer $n$.

Because the intervals on which it is positive are getting thinner and thinner, the integral of the function is positive, even though it is equal to $1$ even for some very large values of $x$.