Showing that $\int\int |x-y|^{-t}d\mu(x)d\mu(y)<\infty$ provided that $\mu(\mathbb{R}^n)<\infty,\mu(B(x,r))<cr^s, s>t>0$ with $\mu$ a Radon measure

Question

Let $t > 0$ and $\mu$ be a generic Radon measure on $\mathbb{R}^n$ and write

$$I := \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|x-y|^{-t}d\mu(x)d\mu(y)$$

Mattilas's Geometry of Sets and Measures in Euclidean Spaceas claims in chapte r8 that if the it holds true that

1.) $\mu(\mathbb{R}^n)<\infty$

2.) for some $s > t$ we have $\mu(B(x,r))\leq cr^s$ for some $c>0$ independent of $x,r$

then necessarily $I < \infty$ without any further comments on the matter. Mattila provides the following integral equality prior the claiming that $I < \infty$: For $x\in\mathbb{R}^n$, we have

$$\int_{\mathbb{R}^n}|x-y|^{-t}d\mu(y) = t\int_0^\infty r^{-t-1}\mu(B(x,r))dr$$

Well, I naive person, like myself, just wants to insert that equality brutely just to $I$ and estimate:

$$I = \int_{\mathbb{R}^n}t\int_0^\infty r^{-t-1}\mu(B(x,r))drd\mu(x)\leq \int_{\mathbb{R}^n}t\int_0^\infty r^{-t-1}cr^sdrd\mu(x)$$

whence

$$I\leq ct\int_{\mathbb{R}^n}\int_0^\infty r^{s-t-1}drd\mu(x) \leq ct\mu(\mathbb{R}^n)\left[\frac{1}{s-t}r^{s-t}\right]_0^\infty$$

where

$$\left[\frac{1}{s-t}r^{s-t}\right]_0^\infty = \infty$$

So clearly this is not the way. But then, what is? What should the sequence of equalities/inequalities be?

Answer 1

Combining 1) and 2), we have a better bound on $\mu(B(x,r))$:

$$ \mu(B(x, r)) \leq \min\{ cr^s, \mu(\mathbb{R}^n) \} $$

Plugging this to $I$,

\begin{align*} I &= \int_{\mathbb{R}^n} t \int_{0}^{\infty} r^{-t-1} \mu(B(x,r)) \, \mathrm{d}r\mathrm{d}\mu(x) \\ &\leq \int_{\mathbb{R}^n} t \int_{0}^{\infty} r^{-t-1} \min\{ cr^s, \mu(\mathbb{R}^n) \} \, \mathrm{d}r\mathrm{d}\mu(x) \\ &= \mu(\mathbb{R}^n) t \int_{0}^{\infty} \min\{ cr^{s-t-1}, \mu(\mathbb{R}^n) r^{-t-1} \} \, \mathrm{d}r. \end{align*}

Now denoting the integrand of the last integral by $f(r) := \min\{ cr^{s-t-1}, \mu(\mathbb{R}^n) r^{-t-1} \}$, we know:

1. $ f(r) = cr^{s-t-1} $ whenever $r$ is sufficiently close to $0$, and

2. $ f(r) = \mu(\mathbb{R}^n) r^{-t-1} $ whenever $r$ is sufficiently large.

This is enough to show that $I < \infty$.

Answer 2

Just to provide a justification for the identity in the reference of the OP:

\begin{align} \int_{\mathbb{R}^n}|x-y|^{-t}\mu(dy)&=t\int_{\mathbb{R}^n}\Big(\int^\infty_{|x-y|}r^{-(t+1)}\,dr\Big)\,\mu(dy)\\ &=t\int_{\mathbb{R}^n}\Big(\int^\infty_0r^{-(t+1)}\mathbb{1}_{(|x-y|,\infty)}(r)\,dr\Big)\,\mu(dy)\\ &=t\int^\infty_0r^{-(t+1)}\Big(\int_{\mathbb{R}^n}\mathbb{1}_{[0,r)}(|x-y|)\mu(dy)\Big)\,dr\\ &=t\int^\infty_0r^{-(t+1)}\mu(B(x;r))\,dr\\ &\leq t\int_{0}^{\infty} r^{-t-1} \min\{ cr^s, \mu(\mathbb{R}^n) \} \, dr\ \end{align}

The estimate then follows by assumptions (1) and (2) in the OP's as described by @SangchulLee:

\begin{align} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|x-y|^{-t}\mu(dy)\mu(dx)&\leq t\int_{\mathbb{R}^n}\int_{0}^{\infty} r^{-t-1} \min\{ cr^s, \mu(\mathbb{R}^n) \} \, dr\,\mu(dx)\\ &=ct\int_{\mathbb{R}^n}\int^{(\mu(\mathbb{R}^n)/c)^{1/s}}_0r^{s-(t+1)}\,dr\,\mu(dx)\\ &\qquad+t\int_{\mathbb{R}^n}\int^\infty_{(\mu(\mathbb{R}^n)/c)^{1/s}}r^{-(t+1)}\mu(\mathbb{R}^n)\,dr\,\mu(dx)\\ &=\frac{c^{\frac{t}{s}}t}{s-t}\big(\mu(\mathbb{R}^n)\big)^{2-\frac{t}{s}}+ c^{\frac{t}{s}}\big(\mu(\mathbb{R}^n)\big)^{2-\frac{s}{t}}<\infty \end{align}