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If a real-valued $f(t)$ is absolute continuous on a domain $[a,\,b]$, Does it imply it is also absolute integrable $\int_a^b |f(t)|dt < \infty$?

If is not true in general, please give some counter-example "easy to check" (closed-form function).

Also, if possible, since could be relevant, I want to check the following two scenarios:

  1. Unbounded domain $\mathbb{R}$: $\quad [a,\,b] \equiv (-\infty,\,\infty)$
  2. Bounded domain $\in\mathbb{R}$: $\quad [a,\,b],\,\, -\infty<a<b<\infty$

I am trying to understand the consequences of having defined a function in a bounded domain, and for the absolute continuity definitions on Wiki I get a bit lost, since in general it is says is the condition required to made the Fundamental Theorem of Calculus to work (so I expect this to be true), but I don´t know if I missing something more abstract (since is also defined through measures), or If there exists some counterexamples that could defy this intuition.

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If your function is absolutely continuous, it is continuous in particular. Thus it is absolutely integrable if the domain is bounded. If the domain is unbounded, that is not necessarily the case. Just take a non-zero constant on $[0,\infty)$.

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  • $\begingroup$ I know that from the Extreme Value Theorem, if a function $f(t)$ is continuous and have compact support $[a,\,b]$ then is bounded $\|f\|_\infty < \infty$, and I am trying to understand if it also implies that $\|f\|_1 < \infty$, is that right? Why is not included in the first proposition/theorem if is already granted? $\endgroup$
    – Joako
    Commented Apr 25, 2022 at 18:26
  • $\begingroup$ @Joako That is correct. If your space has a finite measure (in this case, a bounded interval with the Lebesgue measure), then $L^p\subset L^q$ if $p>q$. In particular, $L^\infty$ is contained in all of the $L^p$. If the measure is infinite, then such inclusion is false. $\endgroup$
    – GReyes
    Commented Apr 25, 2022 at 22:31
  • $\begingroup$ But if is false in general (given $\|f\|_\infty < \|f\|_1$ for a bounded continuous real-valued function)... Why been "absolute continuous" implies then $\|f\|_1<\infty$? (as I understand is true for your first answer to the main question) $\endgroup$
    – Joako
    Commented Apr 25, 2022 at 22:44
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    $\begingroup$ The correct inequality on a bounded interval is $\|f\|_1\le (b-a)\|f\|_\infty$ (which implies $L^\infty\subset L^1$. $\endgroup$
    – GReyes
    Commented Apr 26, 2022 at 4:56
  • $\begingroup$ Thanks you very much!, I understand it now.... hope you can add it to the main answer. I am going to accept it anyway. Best regards. $\endgroup$
    – Joako
    Commented Apr 26, 2022 at 13:49

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