Krdytkyu

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.

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 $mathbb{R}_{>0} times (0, pi)^{d-2} times (0, 2pi)$ 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 $I < infty iff J < infty$.