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Let $$f(x)=\frac{\sin\left(\frac{1}{t}\right)}{t}.$$

Show that $\int^1_0 f(x)\,dx$ exists but $f\notin L^1$, i.e., on $(0,1)$ $f$ is improperly Riemann integrable but not Lebesgue integrable.

Attempt:

Affirmation: $\int^1_0 f(x)\,dx$ exists (I am not very sure how to prove this either).

Now, using the following

Theorem: $f\in L^1 \iff |f| = f^+ + f^- \in L^1$

I can reduce the proof that $f\notin L^1$ by showing that either $\int^1 _0 f^+\,d\lambda = \infty$ or $\int^1 _0 f^-\,d\lambda = \infty$, but I haven't been able to show this either.

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Alfredo Lozano
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    Important distinction: you want to say that this $f$ is improperly Riemann integrable and not Lebesgue integrable. To see that, try taking the substitution $u=1/x$ to make an integral over $[1,\infty)$. – Ian Apr 15 '17 at 19:31
  • @Ian thanks, I'll correct that, this is left for excercise in (Grabinsky, "Teoría de la Medida") and used later as an example, so I asume it can be shown – Alfredo Lozano Apr 15 '17 at 19:35
  • You should probably change the title as well. – Michael McGovern Apr 15 '17 at 19:38
  • @MichaelMcGovern is this right? – Alfredo Lozano Apr 15 '17 at 19:40
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    It is improperly Lebesgue integrable (for the same reason that it is improperly Riemann integrable). – Angina Seng Apr 15 '17 at 19:43
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    @JackD'Aurizio Please read carefully what I wrote. – Angina Seng Apr 15 '17 at 19:48
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    @JackD'Aurizio I've never seen a definition of "improperly Lebesgue integrable", but I expect it's like the definition of "improperly Riemann integrable", that is, the function is integrable over proper subintervals, and the limit of these integrals as the bounds of the subintervals tend to the bounds of the interval exists. – Daniel Fischer Apr 15 '17 at 19:54
  • @JackD'Aurizio Why do you assert that this improperly Lebesgue integrable function is not improperly Lebesgue integrable? – Angina Seng Apr 15 '17 at 19:58
  • These comments start to bother me. If you claim that such a function is improperly Lebesgue integrable, please tell me, what is $$\int_{0}^{+\infty}\left|\frac{\sin x}{x}\right|,dx $$ or $$\lim_{\varepsilon\to 0^+}\lim_{M\to +\infty}\int_{\varepsilon}^{M}\left|\frac{\sin x}{x}\right|,dx $$ if not $+\infty$ ? – Jack D'Aurizio Apr 15 '17 at 20:02
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    It's not clear what "improperly Lebesgue integrable" should even mean, but it is true that for every $\delta>0$, $\int_{\delta}^1 \sin(1/x)/x dx$ exists as a Lebesgue integral and that the limit as $\delta \to 0^+$ of that exists. – Ian Apr 15 '17 at 20:06
  • Ian is on the money: $\int_\delta^1 \sin(1/t),dt/t$ exists as a Lebesgue integral and it tends to a limit, as $\delta\to0^+$. Therefore the improper Lebesgue integral $\int_0^1\sin(1/t),dt/t$ exists. Is this clear enough. Mr D'Aurizo, is "improper Lebesgue integrability" simply a synonym for "Lebesgue integrability" to you? If so, then why bother with the concept, or the name? – Angina Seng Apr 15 '17 at 20:12
  • @Alfredo Lozano Does this help? https://math.stackexchange.com/questions/67198/does-int-0-infty-frac-sin-xxdx-have-an-improper-riemann-integral-or – Michael McGovern Apr 15 '17 at 20:18
  • Gentlemen, you can't fight in here! This is the War Room. – Mark Viola Apr 15 '17 at 21:17
  • Also see this post – hbghlyj Jun 12 '23 at 19:41

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