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Let $f(x) \sim g(x)$ and $f(x) \in L_1(\mathbb{R})$. Then it is stated that $g(x) \in L_1(\mathbb{R})$, is that true?

I feel like the answer is positive, but I can not bound $g(x)$ from above with something..

Updated: I have this question because of the following topic

I do not understand implication: "if $M\neq 0$ then $|f(x)|\sim |M|$ (in $+\infty$) this is impossible because a constant function $x\to |M|$ is not integrable on $[0,+\infty[$ but $f$ is integrable on this interval".

How did we get that stated equality is impossible? if functions are equivalent, we can not say something about their integrals?

John Doe
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  • What do you mean by "equivalent"? – Zima Jun 11 '24 at 15:52
  • @Zima, $f(x) \sim g(x)$ means that $\lim_{x \to \infty} \frac{f(x)}{g(x} = 1$ – John Doe Jun 11 '24 at 16:00
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    Under that assumption, definitely not. Take any $L^1$ function for $f$ and let $g(x)=f(x)$ for $x\geq 0$ and $g(x)=1$ when $x<0$. This is not in $L^1(\mathbb{R})$. You need stronger constraints on $f$ and $g$. – Josh B. Jun 11 '24 at 16:03
  • @JoshB., my question is related to this topic. it seems like this statement about equal functions both in $L_1$ is used there (in Hamou's answer) : $f \in L_1$ and $f \sim M$ but $M$ is not in $L_1$ and somehow he gets contradiction – John Doe Jun 11 '24 at 16:19
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    Can you please modify your original question to reflect what you are trying to ask? Your question has been answered as stated, but the link you provided discusses a different, albeit tangentially similar, problem. – Josh B. Jun 11 '24 at 17:34
  • @JoshB., done. I do not understand why $|f(x)| \sim |M|$ is impossible for $|M| \neq 0$. Initially I've created this question to figure this out. – John Doe Jun 11 '24 at 17:45
  • If $M$ is a nonzero constant, ask yourself, what is $\int_{-\infty}^\infty M,dx$? It's not defined, so a constant function is not in $L^1(\mathbb{R})$. If $f\sim M$, then that means that as $x$ gets large $f(x)$ approaches $M$ (assuming $M$ is positive), and so $f$ is at least larger than $M/2$. Thus, by a similar argument, $f$ does not belong to $L^1(\mathbb{R})$. – Josh B. Jun 12 '24 at 13:32

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