moderately decreasing on $\mathbb R^n$ implies integrable In Stein's Fourier Analysis, he stated

if $f:\mathbb R^n\to \mathbb R$ is moderately decreasing, then $\int_{\mathbb R^n} f<\infty$.

where $f$ is cts and moderately decreasing if $\sup_{\mathbb R^n}|x|^{2}|f(x)|<\infty$ and $\int_{\mathbb R^n} f=\lim_N \int_{Q_N}f(x)=\lim_N\int_{-N}^{N}...\int_{-N}^{N}f(x_1,...,x_n)$.
I am trying to prove this by showing

$\sup_{\mathbb R^n}(1+x_1^2)(1+x_2^2)...(1+x_n^2)|f(x)|<\infty$ iff $f$ is moderate decreasing.

Since in the previous chapter Stein defined a function $f:\mathbb R^2\to \mathbb R$ is moderate decreasing iff $|f(x,y)|<\frac{A}{(1+x^2)(1+y^2)}$ for some $A>0$.
The necessary condition is clear as $x_1^2+...+x_n^2\le (1+x_1^2)(1+x_2^2)...(1+x_n^2)$. ButI I have no idea in showing the another side.
Edit: I took the wrong definition of moderate decrease. The definition should be $f$ is cts and moderately decreasing if $\sup_{\mathbb R^n}|x|^{n+1}|f(x)|<\infty$. So my attempt on the necessary condition also does not work.
 A: *

*Regarding the definition of moderate decrease, Stein says in Ch. 6 Section 1.2:


Two observations are in order. First, we may replace the square $Q_{N}$ by the ball $B_{N}=\left\{x \in \mathbb{R}^{d}:|x| \leq N\right\}$ without changing the definition. Second, we do not need the full force of rapid decrease to show that the limit exists. In fact it suffices to assume that $f$ is continuous and
$$
\sup _{x \in \mathbb{R}^{d}}|x|^{d+\epsilon}|f(x)|<\infty \quad \text { for some } \epsilon>0.
$$
For example, functions of moderate decrease on $\mathbb{R}$ correspond to $\epsilon=1$. In keeping with this we define functions of moderate decrease on $\mathbb{R}^{d}$ as those that are continuous and satisfy the above inequality with $\epsilon=1$.
The interaction of integration with the three important groups of symmetries is as follows: if $f$ is of moderate decrease, then
(i) $\int_{\mathbb{R}^{d}} f(x+h) d x=\int_{\mathbb{R}^{d}} f(x) d x$ for all $h \in \mathbb{R}^{d}$,
(ii) $\delta^{d} \int_{\mathbb{R}^{d}} f(\delta x) d x=\int_{\mathbb{R}^{d}} f(x) d x$ for all $\delta>0$,
(iii) $\int_{\mathbb{R}^{d}} f(R(x)) d x=\int_{\mathbb{R}^{d}} f(x) d x$ for every rotation $R$.



*That said, I did not verify Stein's definition in an answer to your previous question, and the reason was that (well, Stein doesn't always clearly mark where the definitions are...but also)  $\frac1{(1+x^2)(1+y^2)}$ is indeed integrable on $\mathbb R^2$.  Regardless, the method in that answer  gets you any power, (as the functions there are rapidly decreasing) which does finally answer the question with the corrected definition of moderate decrease when you use
$ (1+x^2+y^2) \le (1+x^2)(1+y^2)$ which gives $(1+|(x,y)|^3) \lesssim (1+x^2)^{3/2}(1+y^2)^{3/2}.$ I will eventually modify the answer there appropriately...


*For the task at hand, you will want to use polar/spherical coordinates (see the appendix of Stein for a statement of the change of variables theorem,  I believe Rudin has a proof. The radial case has a proof here but using the "coarea formula".) to show that $g(x)=\frac1{1+|x|^{d+1}}$ is integrable. In fact for any $h:\mathbb  R\to\mathbb R$,
$$\int_{ B_R(0) } h(|x|) dx = \omega_d \int_0^R h(r) r^{d-1} dr$$
where $\omega_d$ is some constant that depends only on $d$ and not on $h$ or $R$. ($\omega_d$ is the area of the unit sphere of $\mathbb R^d$.)
Then you can apply some sort of comparison theorem for integrals e.g. this one.
