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$f(x)=\displaystyle \frac{\sin x}{x}$ is Improper Riemann integrable in $[0,\infty)$ and the integral value $\displaystyle \int_0^{\infty}\frac{\sin x}{x}\,dx=\pi/2$, which is finite.

But , in sense of Lebesgue integration $\displaystyle (L)\int_0^{\infty}\left|\frac{\sin x}{x}\right|\,dx=\infty$ .

Now a function $f$ is said to be integrable over $E$ if $\displaystyle \int_E |f|$ exists and the value is finite. In sense of Riemann-improper integral , integration value is finite for this example. So why it is NOT Lebesgue integrable ?

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    $\begingroup$ A function $f$ is Lebesgue integrable if \begin{align} (L)\int^\infty_{-\infty} |f(x)|\ dx < \infty. \end{align} $\endgroup$ Oct 22, 2016 at 7:35
  • $\begingroup$ In the example $f(x) = \frac{\sin x}{x}$, it's not hard to check that \begin{align} \int^\infty_{-\infty}\left|\frac{\sin x}{x} \right|\ dx = \infty. \end{align} $\endgroup$ Oct 22, 2016 at 7:37
  • $\begingroup$ It's not Riemann integrable on $[0,\infty)$ either. $\endgroup$
    – zhw.
    Oct 22, 2016 at 16:37

2 Answers 2

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First off, this could be happen for the $f(x)=\sin(x^{2})$ with the same domain $(0,\infty).$ You can analyize it.

Second, the class of the improper Riemann integrable function is different with the class of Riemann integrable functions. Actually,

The class of Riemann integrable $\subset$ The class of Lebesgue integrable. But the class of the improper Riemann integrable is not comparable with Lebesgue one. You may interpret this is as an inappropriate property but, due to other properties of the Lebesgue integral this abstract integration is the most successful one. At this moment, I am not sure but maybe there is new abstract integration solving this deficiency.

Also, the key word like "compare Lebesgue with Riemann" and something like that can help you search more papers on this subject. As an instance, you can see this

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If $f:\mathbb{R} \to \mathbb{R}$ is a $\lambda$-measurable function, its integral is defined by $$ \int_{\mathbb{R}} f\,d\lambda = \int_{\mathbb{R}} f^+\,d\lambda - \int_{\mathbb{R}} f^-\,d\lambda $$ where $f^+ = \max(f,0),f^-=-\min(f,0)$. If $f = \sin x/x$, the integrals $\int_{\mathbb{R}} f^+\,d\lambda, \int_{\mathbb{R}} f^-\,d\lambda$ do not converge ($=\infty$), and $\infty - \infty$ is undefined.

You can still assign a meaningful value to this integral by taking limits like improper Riemann integration: $$ \int_{\mathbb{R}} f\,d\lambda = \lim_{r \to \infty} \int_{[-r,r]}f\,d\lambda $$

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