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Let $f(x) = \sin(x)/x$. Consider the following:

$$\int_{\mathbb{R}} f(x)dx = \lim_{n\to\infty} \int_{\mathbb{R}} f(x)\chi_{[-n,n]}dx$$

so that each $\int_{\mathbb{R}} f(x)\chi_{[-n,n]}dx$ is Riemann integrable and so it is Lebesgue integrable (integrating over compact space). Since $\int_{\mathbb{R}} f(x)\chi_{[-n,n]}dx \to \int_{\mathbb{R}} f(x)dx$ and so by completeness of $L^1(\mathbb{R})$, we have that $f\in L^1(\mathbb{R})$ so in particular, $\int_{\mathbb{R}} f(x)dx < \infty$ in the Lebesgue integral sense.

I know I have made a mistake here somewhere but I am having trouble seeing exactly where. Any help is appreciated.

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  • $\begingroup$ This question is related... $\endgroup$ – Fabian Apr 9 '18 at 21:05
  • $\begingroup$ I know but I arrived at the conclusion that this function IS Lebesgue integrable using (what I assume is) faulty logic. Therefore, I thought this was worth asking in a different thread (not sure if this is the right word). I already know it is a well established fact that this function is not Lebesgue integrable. $\endgroup$ – Anmol Bhullar Apr 9 '18 at 21:07
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    $\begingroup$ $L^1(\mathbb R)$ is complete with respect to the $L^1$ norm. You would need to show that $f\chi_{[-n,n]}$ converges to $f$ in the $L^1$ norm and not just pointwisely. In fact, $\| f \chi_{[-n,n]} \|_1$ is not even bounded, so the sequence does not converge in $L^1$. It is easier to see the error if you just consider the sequence $\chi_{[-n,n]}$ on $\mathbb R$. Each function is $L^1$, but it does not converge with respect to the norm and the pointwise limit of the functions is clearly not in $L^1$. $\endgroup$ – Trevor Norton Apr 9 '18 at 21:13
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    $\begingroup$ @TrevorNorton That should be an answer. $\endgroup$ – Noah Schweber Apr 9 '18 at 21:14
  • $\begingroup$ thank you for that comment Trevor, I see where I went wrong now. I thought it was a general fact that $f\chi_{[-n,n]}\to f$. I see that I am wrong now. $\endgroup$ – Anmol Bhullar Apr 9 '18 at 21:15
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$L^1(\mathbb R)$ is complete with respect to the $L^1(\mathbb R)$ norm. You would need to show that $f\cdot\chi_{[−n,n]}$ converges to $f$ in the $L^1$ norm and not just pointwisely. In fact, $\|f\cdot\chi_{[−n,n]}\|_1$ is not even bounded, so the sequence does not converge in $L^1$. It is easier to see the error if you just consider the sequence $\chi_{[-n,n]}$. Each function is $L^1$, but it does not converge with respect to the norm and the pointwise limit of the functions is clearly not in $L^1$.

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    $\begingroup$ +1 for being super clear and complete $\endgroup$ – Severin Schraven Apr 9 '18 at 21:27
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    $\begingroup$ @SeverinSchraven So is this a Banach answer? $\endgroup$ – TheGeekGreek Apr 9 '18 at 21:36
  • $\begingroup$ @TheGeekGreek what do you mean by Banach answer ? $\endgroup$ – Anmol Bhullar Apr 10 '18 at 2:47
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    $\begingroup$ @AnmolBhullar This is just a joke, because Severin said that your answer is complete. A Banach space is a complete normed space, complete in the sense of Cauchy sequences. You can safely ignore this comment. $\endgroup$ – TheGeekGreek Apr 10 '18 at 6:33
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    $\begingroup$ @TheGreekGeek Indeed, pun intended ^^ $\endgroup$ – Severin Schraven Apr 10 '18 at 7:04
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The problem is, that your sequence is not a Cauchy sequence and thus you cannot use the completeness of $L^1$.

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