I am aware of the isoperimetric inequality, which states that if you fix the Lebesgue measure of a measurable set $A \subset \mathbb R^d. d \geq 2$ then the smallest possible value of the perimeter of $A$ is attained by an $n$-ball of appropriate radius. It also provides lower bounds on the volume of $A_{\epsilon} = \{x : \exists y\in A , \|x-y\|<\epsilon\}$ (usual norm) given the volume of $A$.

I wonder if it is possible to provide lower bounds on the following quantity for any measurable set $A\subset \mathbb R^d, d \geq 2$ with positive Lebesgue measure :$$ \int_{A_{\epsilon} \setminus A} \int_{A} f(\|x-y\|)dydx $$ where $f$ is a decreasing function. We may take $f(r) = r^{-d-\alpha}, \alpha \geq 0$ for convenience. This problem occurs when one is studying the small jumps of a Levy process (in the given case, the formula relates to a process jumping into the set $A$ from a "neighbourhood" of $A$). The estimate should only involve a function of $|A|$. We may assume $|A|$ to be fixed, but as close to $0$ as necessary.

The problem is that I haven't been able to get a grip on $A$ without having an idea of the geometry of $A$. The only thing I can assume about $A$ is that it's closed and has positive Lebesgue measure, which could make its geometry very wild (possibly fat Cantor set, lack of boundary regularity etc.). I wonder if it's possible to use isoperimetric ideas to parametrize this double integral and lower bound it, or to reduce this to the case of a ball which will be easier to handle.

For example, from evidence in papers, authors tend to neglect the term for $\alpha>0$ and achieve their results, because it is inferior to other terms in the Levy system formula, which look like $|A|$. On the other hand, precisely when $\alpha=0$, it seems that this double integral is the dominating term and furthermore the contribution involves a log factor ( I expect $|A| \log(1+|A|^{-2/d})$, but the contents of the $\log$ are still not fully clear) which I cannot quite understand. I am aware that in one-dimension the integral of $\frac 1x$ is $\log x$ and here we're dealing with $\frac 1{\|x-y\|^d}$ in $\mathbb R^d$ : that's a flag that a log could be on the cards, but it's still not clear why $|A|$ is inside the log.

Partial answers could attempt to explain the appearance of the log term in the expansion, or why the $\alpha>0$ situation has a contribution likely smaller than $|A|$.

(Note : even though the origins of this question are from Levy process theory, I've omitted the tag because it's peripheral to the discussion).

I am aware of the isoperimetric inequality, which states that if you fix the Lebesgue measure of a measurable set $A \subset \mathbb R^d. d \geq 2$ then the smallest possible value of the perimeter of $A$ is attained by an $n$-ball of appropriate radius. It also provides lower bounds on the volume of $A_{\epsilon} = \{x : \exists y\in A , \|x-y\|<\epsilon\}$ (usual norm) given the volume of $A$.

I wonder if it is possible to provide lower bounds on the following quantity for any measurable set $A\subset \mathbb R^d, d \geq 2$ with positive Lebesgue measure :$$ \int_{A_{\epsilon} \setminus A} \int_{A} f(\|x-y\|)dydx $$ where $f$ is a decreasing function. We may take $f(r) = r^{-d-\alpha}, \alpha \geq 0$ for convenience. This problem occurs when one is studying the small jumps of a Levy process (in the given case, the formula relates to a process jumping into the set $A$ from a "neighbourhood" of $A$). The estimate should only involve a function of $|A|$. We may assume $|A|$ to be fixed, but as close to $0$ as necessary.

The problem is that I haven't been able to get a grip on $A$ without having an idea of the geometry of $A$. The only thing I can assume about $A$ is that it's closed and has positive Lebesgue measure, which could make its geometry very wild (possibly fat Cantor set, lack of boundary regularity etc.). I wonder if it's possible to use isoperimetric ideas to parametrize this double integral and lower bound it, or to reduce this to the case of a ball which will be easier to handle.

For example, from evidence in papers, authors tend to neglect the term for $\alpha>0$ and achieve their results, because it is inferior to other terms in the Levy system formula, which look like $|A|$. On the other hand, precisely when $\alpha=0$, it seems that this double integral is the dominating term and furthermore the contribution involves a log factor ( I expect $|A| \log(1+|A|^{-1/d})$, but the contents of the $\log$ are still not fully clear) which I cannot quite understand. I am aware that in one-dimension the integral of $\frac 1x$ is $\log x$ and here we're dealing with $\frac 1{\|x-y\|^d}$ in $\mathbb R^d$ : that's a flag that a log could be on the cards, but it's still not clear why $|A|$ is inside the log.

Partial answers could attempt to explain the appearance of the log term in the expansion, or why the $\alpha>0$ situation has a contribution likely smaller than $|A|$.

(Note : even though the origins of this question are from Levy process theory, I've omitted the tag because it's peripheral to the discussion).

I am aware of the isoperimetric inequality, which states that if you fix the Lebesgue measure of a measurable set $A \subset \mathbb R^d$ then the smallest possible value of the perimeter of $A$ is attained by an $n$-ball of appropriate radius. It also provides lower bounds on the volume of $A_{\epsilon} = \{x : \exists y\in A , \|x-y\|<\epsilon\}$ (usual norm) given the volume of $A$.

I wonder if it is possible to provide lower bounds on the following quantity for any measurable set $A\subset \mathbb R^d$ with positive Lebesgue measure :$$ \int_{A_{\epsilon} \setminus A} \int_{A} f(\|x-y\|)dydx $$ where $f$ is a decreasing function. We may take $f(r) = r^{-d-\alpha}, \alpha \geq 0$ for convenience. This problem occurs when one is studying the small jumps of a Levy process (in the given case, the formula relates to a process jumping into the set $A$ from a "neighbourhood" of $A$). The estimate should only involve a function of $|A|$. We may assume $|A|$ to be fixed, but as close to $0$ as necessary.

The problem is that I haven't been able to get a grip on $A$ without having an idea of the geometry of $A$. The only thing I can assume about $A$ is that it's closed and has positive Lebesgue measure, which could make its geometry very wild (possibly fat Cantor set, lack of boundary regularity etc.). I wonder if it's possible to use isoperimetric ideas to parametrize this double integral and lower bound it, or to reduce this to the case of a ball which will be easier to handle.

For example, from evidence in papers, authors tend to neglect the term for $\alpha>1$ and achieve their results, because it is inferior to other terms in the Levy system formula, which look like $|A|$. On the other hand, precisely when $\alpha=1$, it seems that this double integral is the dominating term and furthermore the contribution involves a log factor ( I expect $|A| \log(1+|A|^{-2/d})$, but the contents of the $\log$ are still not fully clear) which I cannot quite understand. I am aware that in one-dimension the integral of $\frac 1x$ is $\log x$ and here we're dealing with $\frac 1{\|x-y\|^d}$ in $\mathbb R^d$ : that's a flag that a log could be on the cards, but it's still not clear why $|A|$ is inside the log.

Partial answers could attempt to explain the appearance of the log term in the expansion, or why the $\alpha>0$ situation has a contribution likely smaller than $|A|$.

(Note : even though the origins of this question are from Levy process theory, I've omitted the tag because it's peripheral to the discussion).

I am aware of the isoperimetric inequality, which states that if you fix the Lebesgue measure of a measurable set $A \subset \mathbb R^d. d \geq 2$ then the smallest possible value of the perimeter of $A$ is attained by an $n$-ball of appropriate radius. It also provides lower bounds on the volume of $A_{\epsilon} = \{x : \exists y\in A , \|x-y\|<\epsilon\}$ (usual norm) given the volume of $A$.

I wonder if it is possible to provide lower bounds on the following quantity for any measurable set $A\subset \mathbb R^d, d \geq 2$ with positive Lebesgue measure :$$ \int_{A_{\epsilon} \setminus A} \int_{A} f(\|x-y\|)dydx $$ where $f$ is a decreasing function. We may take $f(r) = r^{-d-\alpha}, \alpha \geq 0$ for convenience. This problem occurs when one is studying the small jumps of a Levy process (in the given case, the formula relates to a process jumping into the set $A$ from a "neighbourhood" of $A$). The estimate should only involve a function of $|A|$. We may assume $|A|$ to be fixed, but as close to $0$ as necessary.

The problem is that I haven't been able to get a grip on $A$ without having an idea of the geometry of $A$. The only thing I can assume about $A$ is that it's closed and has positive Lebesgue measure, which could make its geometry very wild (possibly fat Cantor set, lack of boundary regularity etc.). I wonder if it's possible to use isoperimetric ideas to parametrize this double integral and lower bound it, or to reduce this to the case of a ball which will be easier to handle.

For example, from evidence in papers, authors tend to neglect the term for $\alpha>0$ and achieve their results, because it is inferior to other terms in the Levy system formula, which look like $|A|$. On the other hand, precisely when $\alpha=0$, it seems that this double integral is the dominating term and furthermore the contribution involves a log factor ( I expect $|A| \log(1+|A|^{-2/d})$, but the contents of the $\log$ are still not fully clear) which I cannot quite understand. I am aware that in one-dimension the integral of $\frac 1x$ is $\log x$ and here we're dealing with $\frac 1{\|x-y\|^d}$ in $\mathbb R^d$ : that's a flag that a log could be on the cards, but it's still not clear why $|A|$ is inside the log.

Partial answers could attempt to explain the appearance of the log term in the expansion, or why the $\alpha>0$ situation has a contribution likely smaller than $|A|$.

(Note : even though the origins of this question are from Levy process theory, I've omitted the tag because it's peripheral to the discussion).