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My question is pretty much what it is in the title. My intuition tells me that it should be true because in other case we could always find intervals in which the function is greater than 0 and this should lead to a contradiction based on the fact that $\int |f| < \infty$. But on the other hand I feel that it may happen that on a zero measure set there are points on this doesn't hold or something like that.

In a more particular sense, I'm trying to prove that if $f$ is a complex valued function such that $f$ is lebesgue integrable over the real line then $\int_0^\pi |f(re^{i\theta})|re^{-r\sin \theta} d\theta \rightarrow 0$ as $r \rightarrow +\infty$ and I feel that the previous result should help on proving this, but I'm a bit stuck, so any help would be appreciated.

H4z3
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1 Answers1

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Let $f(x):\mathbb{R} \to \mathbb{R}$, $f(x) = 0$ if $x \notin \mathbb{N}$, and $f(x) = 1$ if $x \in \mathbb{N}$. We have that $\int_{\mathbb{R}}|f(x)|dx = 0$, and $\lim_{x\to \infty}|f(x)| \not= 0.$

I believe that the statement may be: Let $f \in L^{1}(\mathbb{R})$, if $\exists$ $\lim_{x\to \infty}|f(x)| = \ell$, then $\ell = 0$.

I hope this answer helps you.

ZAF
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  • Is the limit taken everywhere? What if you take a function that has spikes as you go to infinity but with smaller and smaller width? – user57 Sep 05 '24 at 15:41
  • Your function is $0$ in $L^1(\mathbb{R})$ because $\mathbb{N}$ is a null set. – psl2Z Sep 05 '24 at 16:59