# $f \circ | \cdot |$ Lebesgue Integrable $\iff$ g is Lebesgue Integrable

Define $$f: [0, \infty[ \to \bar{\mathbb R}$$ measurable

Show that:

$$f \circ | \cdot |: \mathbb R^d \to \bar{\mathbb R}$$ as a lebesgue integrable function $$\iff$$ $$g: \mathbb R_{\geq 0}\to \mathbb R, g(r):=r^{d-1}f(r)$$ is lebesgue integrable

Ideas:

"$$\Rightarrow$$" $$g$$ is simply $$f$$ scaled by a constant $$r^{d-1}$$ and therefore measurable and

$$\int_{[0,\infty[}g(r)dr=\int_{[0,\infty[}r^{d-1}f(r)dr$$

I would like to say that find a constant $$c \in \mathbb R$$ so that $$|r^{d-1}|\leq c$$ but obviously $$r \in [0,\infty[$$ so this approach would not make sense. The only other solution that comes to mind would be substitution, so

set $$|x| = r \Rightarrow dx =dr$$ (Is this even correct?) but here I do not get any further.

"$$\Leftarrow$$"

$$f \circ |\cdot|$$ is by definition measurable

All I notice is that on $$r \in [1,\infty[$$: $$f(r) \leq g(r)$$ and then using the monotonicity of integrals I could use $$\int_{[1,\infty[}f(r)dr \leq \int_{[1,\infty[}g(r)dr < \infty$$ but this still does not help me on $$]0,1[$$

Any help would be greatly appreciated.

• No, $f \circ |\cdot|: \mathbb R^d \to \bar{\mathbb R}$, while $f: [0, \infty[ \to \bar{\mathbb R}$ – SABOY Jan 4 '19 at 20:02
• Do you know the "polar coordinates" formula? The case $d=2$ should be familiar. – John Dawkins Jan 4 '19 at 23:05
• See Rudin's RCA for polar coordinates in $\mathbb R^{n}$. – Kavi Rama Murthy Jan 5 '19 at 0:03

As you have already seen measurability of $$f$$ and $$g$$ is not a problem. It is enough to show that for $$f \geq 0$$ we have $$\newcommand{\RR}{\mathbb{R}}$$
$$I := \int_{\mathbb{R}^n} f(|x|) \,d x < \infty \iff J := \int_\mathbb{R_{\geq0}} r^{d-1} f(r) \, d r <\infty$$
This is integration using $$d$$-dimensional spherical coordinates. They provide a diffeomorphism $$\Psi$$ from $$\mathbb{R}_{>0} \times (0, \pi)^{d-2} \times (0, 2\pi)$$ to $$\mathbb{R}^n \setminus N$$ where $$N$$ is some set of measure zero ($$N$$ is something like $$\mathbb{R}_{>0} \times \{0\} \times \mathbb{R}^{d-2}$$, but it depends on the details of the parametrization and doesn't really matter). The special thing about spherical coordinates is that $$|\Psi(r, \phi_1, \dots, \phi_{d-1})| = r$$
The Jacobi determinant of this map is $$\det(\mathrm{D} \Psi(x)) = r^{d-1} \sin^{n-2} \phi_1 \sin^{n-3} \phi_2 \dots \sin \phi_{d-1}$$. Therefore, by the change of variables formula
\begin{align} I &= \int_{\RR_{>0}} \int_{(0, \pi)^{d-2}} \int_{(0, 2 \pi)} f(|\Psi(r, \phi_1, \dots, \phi_{d-1})|)\det(\mathrm{D} \Psi(x)) \,d \phi_{d-1} \,d \phi_{d-2} \dots \,d \phi_1 \,d r \\ &= \int_{\RR_{>0}} \int_{(0, \pi)^{d-2}} \int_{(0, 2 \pi)} f(r) r^{d-1} \sin^{n-2} \phi_1 \sin^{n-3} \phi_2 \dots \sin \phi_{d-1} \,d \phi_{d-1} \,d \phi_{d-2} \dots \,d \phi_1 \,d r \\ &= \int_{\RR_{>0}} f(r) r^{d-1} \, dr \cdot \int_{(0, \pi)^{d-2}} \int_{(0, 2 \pi)} \sin^{n-2} \phi_1 \sin^{n-3} \phi_2 \dots \sin \phi_{d-1} \,d \phi_{d-1} \,d \phi_{d-2} \dots \,d \phi_1 \\ &= \int_{\RR_{>0}} f(r) r^{d-1} \, dr \cdot K = J \cdot K \end{align} where $$K$$ is the second integral in the line before the last. One clearly sees that the absolute value of the integrand in $$K$$ is bounded by $$1$$. Therefore $$K \leq 2 \pi^{d-1}$$. So $$I$$ and $$J$$ are the same up to a finite constant and therefore $$I < \infty \iff J < \infty$$.