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Let $\mu $ be a positive Borel measure on $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{d}$ 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 $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{d}$. Could you help me to prove that $\int_{% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{d}}\frac{1}{\left\vert x\right\vert ^{n}}d\mu \left( x\right) =\infty $?

My effort: $\int_{% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{d}}\frac{1}{\left\vert x\right\vert ^{n}}d\mu \left( x\right) \geq \int_{% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{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) $.

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What about $n$-dimensional Hausdorff measure restricted to the $n$-sphere (embedded in the obvious way in $\mathbb{R}^d$)? –  Jose27 Dec 31 '12 at 5:18
    
Which page of Stein's book is this from? as @Jose27 remarked, it can't be true as stated. Here's a one-dimensional counterexample ($d=n=1$): $d\mu(x) = \frac{|x|}{x^2+1}|dx|$. –  user53153 Dec 31 '12 at 6:38
    
@PavelM : I wanna learn that Hardy-Littlewood-Sobolev inequality (for p=1) is wrong just like lebesgue measure case in Stein book. But now,for a measure that satisfy the above condition. –  beginner Dec 31 '12 at 8:10
    
@PavelM: How to check that that your measure satisfying $\mu (B(a,r))\leq Cr^n$? Since $d\mu(x)=\frac{|x|}{x^2+1} dx$, then $\mu(x)=\int \frac{|x|}{x^2+1} dx$, right? –  beginner Dec 31 '12 at 8:21
    
The function $|x|/(x^2+1)$ is bounded (say, by $M$), which implies $\mu(B(a,r))\le 2Mr$. (Integral estimated by supremum * size of the region of integration). –  user53153 Dec 31 '12 at 8:28
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1 Answer 1

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.

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I think he needs the condition to be reverse inequality on the measure of the ball, that is $\mu(B(a,r))\geq Cr^n$. Then using your answer he gets a lower bound of an integral of the form $\int_1^\infty s^{-1}ds$, which explodes. –  Eric Naslund Jan 1 '13 at 1:28
    
@EricNaslund I agree that everything points to this--but this is the OP's job to clear the hypothesis up. –  Did Jan 1 '13 at 9:47
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