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I am reading about the Lebesgue integral. My doubt is how to decide which integral to use Lebesgue or Riemann.

For instance, I was reading one example where we have a measurable set $E=[0,1]$ and sequence of function is defined as $f_{n}(x)= (n+1)x^n$
while calculating the integral $$\lim_{n\to\infty} \int_Ef_{n}(x)d\nu=1$$ where $\nu$ is a measure so here we have $\lim_{n\to\infty}\int_E (n+1)x^nd\nu=1$, so here they have calculated the integral directly using the Riemann so when to decide the we can use $dx$ instead of $d\nu$

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    The Lebesgue integral is more general than Riemann in every sense: if a function is Riemann integrable, then it is Lebesgue integrable and the two integrals agree. If a function is improperly Riemann integrable then it is 'improperly' Lebesgue integrable and again the two values agree. But if $\nu$ is some arbitrary measure then you can't think of it as a Riemann integral. – peek-a-boo Jul 31 '22 at 09:47
  • So here we clearly see that the function is Riemann integrable so it will be obviously Lebesgue integrable so they will agree here so, in this case, I can calculate my integral normally "I mean using $dx$ am right? – Exa Ready Jul 31 '22 at 09:49
  • Like I said in my last sentence, depends what the measure $\nu $ is. Is it the Lebesgue measure on the real line? – peek-a-boo Jul 31 '22 at 09:50
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    @peek-a-boo "If a function is improperly Riemann integrable then it is improperly Lebesgue integrable" is not always true. That's the case if the improper integral converges absolutely. However, Riemann improper integrals also include conditionally convergent integrals, like $\int\limits_{-\infty}^{\infty}\frac{\sin x}{x}dx$. However, the function $\frac{\sin x}{x}$ isn't Lebesgue integrable over $\mathbb{R}$. – Mark Jul 31 '22 at 09:52
  • @peek-a-boo yes it is a lebesgue measure – Exa Ready Jul 31 '22 at 09:54
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    @Mark you should re-read my comment I did say 'improper' Lebesgue integral. And frankly this 'counter example ' is rather silly. The limit of Lebesgue integrals over intervals as the intervals approach the whole real line gives the same answer. So once again, anything Riemann can do, so can Lebesgue. – peek-a-boo Jul 31 '22 at 09:57
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    @ExaReady then yes you can integrate this as usual. Note by the way that it is not Riemann vs Lebesgue which allows you to compute integrals easily. Rather it is the fundamental theorem of calculus which helps us compute integrals, and the Lebesgue integral has a more general version of that as well. So the Lebesgue integral with Lebesgue measure is better in every sense (except in technical difficulty). – peek-a-boo Jul 31 '22 at 09:59
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    In general, we do not defined "improper" Lebesgue integral. But, if we define it mimicking improper Riemann integral, we lose many of the nice properties of Lebesgue integral. For a more details on this topic, see, for instance :

    https://math.stackexchange.com/questions/2214482/is-there-a-general-theory-of-the-improper-lebesgue-integral

     – Ramiro Jul 31 '22 at 13:51

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