Let $\mu $ be a positive Borel measure on $%
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^{d}$ such that $\mu \left( B\left( a,r\right) \right) \leq Cr^{n}$ for some
$n\in (0,d]$ and for any ball $B\left( a,r\right) $ in $%
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^{d}$. Could you help me to prove that $\int_{%
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^{d}}\frac{1}{\left\vert x\right\vert ^{n}}d\mu \left( x\right) =\infty $?
My effort: $\int_{%
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^{d}}\frac{1}{\left\vert x\right\vert ^{n}}d\mu \left( x\right) \geq \int_{%
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\mathbb{R}
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^{d}\backslash B\left( 0,1\right) }\frac{1}{\left\vert x\right\vert ^{n}}%
d\mu \left( x\right) =\sum_{k=0}^{\infty }\int_{B\left( 0,2^{k+1}\right)
\backslash B\left( 0,2^{k}\right) }\frac{1}{\left\vert x\right\vert ^{n}}%
d\mu \left( x\right) \geq \sum_{k=0}^{\infty }\frac{1}{\left(
2^{k+1}\right) ^{n}}\mu \left( B\left( 0,2^{k+1}\right) \backslash B\left(
0,2^{k}\right) \right) $.
Best Answer
For every positive $r$ and $n$, $$ \frac1{r^n}=\int_r^{+\infty}n\frac{\mathrm ds}{s^{1+n}}, $$ hence Tonelli theorem yields $$ \int_{\mathbb R^d}\frac1{\|x\|^n}\mathrm d\mu(x)=\int_{\mathbb R^d}\int_{\|x\|}^{+\infty}n\frac{\mathrm ds}{s^{1+n}}\,\mathrm d\mu(x)=n\int_0^{+\infty}m(s)\frac{\mathrm ds}{s^{1+n}}=(*), $$ where, for every nonnegative $s$, $$ m(s)=\mu(B(O,s)). $$ From this point, it is up to you to select some hypotheses on the function $m$ ensuring that, for the value of $n$ which interests you, the integral $(*)$ converges or that it diverges. As mentioned in the comments, the current hypothesis cannot work.
To make the integral $(*)$ converge at $0$, the control $m(s)\leqslant Cs^a$ when $s\to0$, for some $a\gt n$, is enough. To make the integral $(*)$ converge at infinity, the control $m(s)\leqslant Cs^b$ when $s\to\infty$, for some $b\lt n$, is enough.
On the other hand, if $m(s)\geqslant Cs^n$ when $s\to0$ or when $s\to\infty$, then $(*)$ diverges. This might be the result you had in mind. Note finally that the hypothesis $n\leqslant d$ is irrelevant but that $n$ must be positive.