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Define for $f\in C_{c}^{\infty}(\mathbb{R}^n)$, the function, $$ I^{\epsilon}f(x)=\int_{|x-y|<\epsilon}\frac{|\nabla\phi(y)|}{|x-y|^{n-1}}\,dy. $$ For a fixed $\epsilon>0$ and for any $j\geq 1$, define $$ J^{j,\epsilon}(x)=\int_{\frac{\epsilon}{2^{j}}\leq|x-y|\leq\frac{\epsilon}{2^{j-1}}}\frac{|\nabla\phi(y)|}{|x-y|^{n-1}}\,dy. $$ Then I want to prove $$ \sum_{j=1}^{\infty} J^{j,\epsilon}(x)=I^{\epsilon}(x). $$ each $J^{j,\epsilon}$ is finite because $|x-y|$ lies outside zero, and $J^{j,\epsilon}$ is dominated by $c(n)\epsilon M{\nabla\phi}(x)$ for every $x\in\mathbb{R}^n$ and $\epsilon>0$, where $M$ is the maximal function which is finite, since $\nabla\phi$ is infinitely differentiable with compact support. But from here I cannot prove the above equality.

I came around this problem while proving theorem 15.23. in the book of Juha Heinonen titled as "Nonlinear potential theory of degenerate elliptic equations".
Please help me in this regard. Thanks.

Mathlover
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  • don't you have a sum of non-overlapping (annular) regions in $J^{j,\varepsilon} $ ? The ranges of integration in $J^{j,\varepsilon}$ do not intersect for different $j$ and cover the set $|x-y| < \varepsilon$ as $j=1,2,...$. – Hayk Oct 09 '18 at 19:39
  • Yes, I have nonoverlaping regions. But I have to integrate over infinite union of such regions. Can you please tell how to proceed from here. Thanks. – Mathlover Oct 09 '18 at 20:29
  • the infinite count does not matter. You need to use countable additivity of the Lebesgue integral (googling the term will provide a sufficient base for the topic) which states that if you partition the domain of integration into (at most) countable union of disjoint measurable sets, then the (possibly infinite) sum of integrals over these sets equals to the integral of the function over the original domain. – Hayk Oct 09 '18 at 20:38
  • I think you are giving the argument in the link: – Mathlover Oct 09 '18 at 20:39
  • https://math.stackexchange.com/questions/668958/prove-the-countable-additivity-of-lebesgue-integral – Mathlover Oct 09 '18 at 20:39
  • But here the assumption that $f$ is integrable over the countable union of sets is assumed which I am not able to prove in my case. – Mathlover Oct 09 '18 at 20:40
  • additivity is one of the most basic properties of the Lebesgue integral, you need just that. Your function $f$ is smooth and compactly supported, while the $|x-y|$ terms comes with an exponent which makes it integrable on compact sets of $\mathbb{R}^n$, namely, use the fact that if $\alpha < n$ then $$\int\limits_{\mathbb{R}^n } \frac{dx}{|x|^{\alpha}} < \infty$$. – Hayk Oct 09 '18 at 20:41
  • Thank you very very much, I should have pointed out this fact. Thanks a lot Sir. – Mathlover Oct 09 '18 at 21:08
  • you're welcome! – Hayk Oct 09 '18 at 21:12

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