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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.

Anmol Bhullar
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  • This question is related... – Fabian Apr 09 '18 at 21:05
  • 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. – Anmol Bhullar Apr 09 '18 at 21:07
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    $L^1(\mathbb R)$ 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 $L^1$. – Trevor Norton Apr 09 '18 at 21:13
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    @TrevorNorton That should be an answer. – Noah Schweber Apr 09 '18 at 21:14
  • 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. – Anmol Bhullar Apr 09 '18 at 21:15
  • @TrevorNorton You should make your comment an answer. – Severin Schraven Apr 09 '18 at 21:19
  • @SeverinSchraven I have submitted my comment as an answer now. – Trevor Norton Apr 09 '18 at 21:25

2 Answers2

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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 $L^1$.

Trevor Norton
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The problem is, that your sequence is not a Cauchy sequence and thus you cannot use the completeness of $L^1$.

Severin Schraven
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