$\dfrac{1}{||x||}$ not Lebesgue integrable in $\mathbb {R^d}$ on set $E=\{x\in \mathbb {R^d}: ||x||≥1\}$

Question

I am trying to show the following: $\dfrac{1}{||x||}$ not Lebesgue integrable in $\mathbb {R^d}$ on set $E=\{x\in \mathbb {R^d}: ||x||≥1\}$

I tried to use Fubini's theorem and the fact that $\dfrac{1}{x}$ is not Lebesgue integrable in $\mathbb R$. My problem is in applying Fubini's theorem, i.e., I am not so sure how to apply it here and interchange bounds of integration in this case. I hope I can get some help here. Thanks

Answer 1

You can use Tonelli's theorem to change the order of integration - it does not require integrability, only measurability and positivity.

$$\int_E\frac{dx}{\|x\|} = \int_1^{\infty}r^{d-1}dr\int_{\mathbb S^d}\frac{1}{r}d\sigma =c_d\int_1^{\infty}r^{d-2}dr $$where $c_d$ is the measure of $\mathbb S^d$ - it is finite.

The last integral does not converge (the sufficient and necessary condition is $d-2<-1$, which does not hold).

Answer 2

Let $x_1,\ldots,x_d$ be the coordinates of $\mathbb{R}^d$, and let $F=\{x_1\geq1\}\subset\mathbb{R}^d$. On $F$ it is much easier to use Fubini and see that the function in question has a diverging integral. Since $F\subset E$, and the function is positive, it is enough.