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I am self-learning a bit of measure theory.

I want to show that some function $f:\mathbb{R}\rightarrow\mathbb{R}$ is not integrable. I have $$ \int_\mathbb{R}f=\int_\mathbb{R}f^+ - \int_\mathbb{R}f^- $$ and, as I understand, I must show that bot integrals on the RHS diverge to $\infty$. If at least one of them does not, then $f$ is integrable (the value of the integral being $+\infty$ or $-\infty$).

However, in Showing $\frac{\sin x}{x}$ is NOT Lebesgue Integrable on $\mathbb{R}_{\ge 0}$, the accepted answer only proves that one of them diverges. How come?

soap
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2 Answers2

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You are right, it would have been better if in that reply they had separated $n$ into even and odds. If $n=2k$ you get $$\int_{(2k-1)\pi}^{2k\pi}\frac{(\sin x)^-}{x}dx\ge \frac{1}{k\pi}$$ and so if you sum over all $k$ you get that $$\int_{0}^{\infty}\frac{(\sin x)^-}{x}dx=\infty.$$ Similarly, if you take $n=2k-1$ you get $$\int_{(2k-2)\pi}^{(2k-1)\pi}\frac{(\sin x)^+}{x}dx\ge \frac{2}{(2k-1)\pi}$$ and so if you sum over all $k$ you get that $$\int_{0}^{\infty}\frac{(\sin x)^+}{x}dx=\infty.$$ Edit: Alternatively, you need to prove that the improper Riemann integral of $\frac{\sin x}{x}$ exists. This can be done by integrating by parts $$\int_{\pi/2}^t \frac{\sin x}{x}dx=-\frac{\cos t}{t}-\int_{\pi/2}^t \frac{\cos x}{x^2}dx.$$ Then $-\frac{\cos t}{t}\to 0$ as $t\to\infty$ and $\left\vert \frac{\cos x}{x^2}\right\vert \le \frac1{x^2}$ which is integrable in $[\pi/2,\infty)$. Hence, the improper Riemann integrable exists. This together with $$\int_0^\infty \frac{|\sin x|}{x}dx=\infty$$ implies that $$\int_{0}^{\infty}\frac{(\sin x)^+}{x}dx=\int_{0}^{\infty}\frac{(\sin x)^-}{x}dx=\infty.$$ PS: Jason is right. To prove that a function is not Lebesgue integrable it suffices to prove that $\int|f|=\infty$ (but if you want to give an example for that, just use $f(x)=\frac1{x}$ in $(1,\infty)$. To prove that the Lebesgue integral of a function does not exist, you need to prove that $\int (f)^+=\int (f)^-=\infty$. This follows either by direct computation, or, if appropriate, as in this case, if the proper Riemann integrable is finite and $\int|f|=\infty$

Gio67
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A function is integrable if both positive and negative integrals are finite. A function can integrate to $\infty$ or $-\infty$, but we do not call such a function integrable.

Jason
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  • Agree but if you are not Lebesgue integrable, it could still be that the Lebesgue integral is defined. For this particular function it is not, that's why you should compute positive and negative parts separately. – Gio67 May 07 '17 at 16:48
  • No, I disagree. If you want to show that the function is not Lebesgue integrals it is necessary and sufficient to show one integral diverges. If the improper Riemann integral exists and is finite then it is automatic that the other integral must diverge too. – Jason May 07 '17 at 17:49
  • I edited my answer. Hope we agree now. – Gio67 May 07 '17 at 18:23
  • Yes, I would say we do. – Jason May 07 '17 at 18:32