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A comment by Aloizio Macedo in this post claims claims

Every "improperly Riemann integrable function" is also "improperly Lebesgue integrable". That Lebesgue doesn't know how to handle this as much as Riemann (not counting H-K) is a commonly wide-spread myth: for instance, you just take $\lim_{x \to \infty} \int_{[c,x]} f$ in the same way as Riemann. The difference is that Lebesgue has a meaning on itself for the value $\int_{[c,\infty]} f$, whereas Riemann doesn't, and this can differ from the "improper".

I'm skeptical of the claim "and this can differ from the 'improper'." Perhaps he means "differ" in that the Lebesgue integral $\int_{[c,\infty)}f$ might not exist, whereas the improper Riemann or Lebesgue integral defined by $\lim_{x \to \infty} \int_{[c,x]} f$ might exist. This I agree with, as there are classic examples like $f(x)=\sin(x)/x.$

But perhaps he is making a stronger claim, that there exists a function for which the ordinary Lebesgue integral and improper Lebesgue integral exist, but have different values. This I find hard to imagine.

Hence: Is there a Lebesgue-integrable function $f:[c,\infty)\to\mathbb R$ such that $\lim_{x\to\infty}\int_{[c,x]}f\neq\int_{[c,\infty)}f$ ?

Both integrals above are Lebesgue integrals, and the limit is assumed to exist.

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  • $\begingroup$ No. There does not exist "a function for which the ordinary Lebesgue integral and improper Lebesgue integral exist, but have different values." But this requires proof since the two definitions are very different. $\endgroup$ May 5, 2022 at 22:01
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    $\begingroup$ You're assuming $f$ is Lebesgue integrable, so you can apply dominated convergence ($|f|$ is a dominating function) to prove the equality. $\endgroup$
    – peek-a-boo
    May 5, 2022 at 22:14
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    $\begingroup$ @peek-a-boo So something like: $(f\cdot\chi_{[c,n]})_{n\in\mathbb N}\to f$ pointwise, with each $|f_n|\leqslant|f|$, hence $\lim_{n\to\infty}\int_{[c,n]}f=\lim_{n\to\infty}\int_{[c,\infty)}f\cdot\chi_{[c,n]}=\int_{[c,\infty)}f$? $\endgroup$
    – WillG
    May 5, 2022 at 22:22
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    $\begingroup$ Yes, though more accurately, since you're talking about the limit $x\to\infty$, you should take an arbitrary sequence $\{x_n\}_{n\in\Bbb{N}}$ with $x_n\to \infty$ (the basic equivalence between function limits and sequential convergence). Then, $f\cdot\chi_{[c,x_n]}\to f$ pointwise and is dominated by $|f|$. $\endgroup$
    – peek-a-boo
    May 5, 2022 at 22:24
  • $\begingroup$ @peek-a-boo Feel free to copy those comments into an answer, and I will accept it. $\endgroup$
    – WillG
    May 5, 2022 at 22:28

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Just typing up my comments. The equality here follows from the dominated convergence theorem because you're assuming $f$ is Lebesgue integrable. Take any sequence $\{x_n\}_{n\in\Bbb{N}}$ with $x_n\to \infty$, and apply dominated convergence to $\{f\cdot \chi_{[c,x_n]}\}_{n\in\Bbb{N}}$ with $|f|$ as the dominating function. And to conclude, just keep in mind the equivalence between function limits $x\to \infty$ and sequential convergence of arbitrary sequences $x_n\to\infty$.

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