If $\Omega\subseteq\mathbb{R}^n$ is bounded, then $\int_\Omega|x-y|^{1-n}\,d\lambda < \infty$

Question

Let

• $\Omega\subseteq\mathbb{R}^n$ be bounded with $n\ge 2$
• $\left|\;\cdot\;\right|$ be the euclidean norm
• $\lambda$ be the Lebesgue measure on the Borelian $\sigma$-algebra of $\mathbb{R}^n$

I want to show, that $$I:=\int_\Omega|x-y|^{1-n}\,d\lambda(y)<\infty\;\;\;\text{for all }x\in\mathbb{R}^n\;.\tag{1}$$

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I've tried to consider $$\varepsilon(x):=\sup_{y\in\Omega}|x-y|\;\;\;\text{for }\mathbb{R}^n\;.\tag{2}$$ I think we need to achieve something like $$I\le\omega_n\int_0^{\varepsilon(x)}1\;dr<\infty\tag{3}$$ where $\omega_n$ is the $n$-dimensional volume of the unit sphere in $\mathbb{R}^n$. But how can we obtain $(3)$?

Answer 1

Let $R$ be so large that $\Omega\subset B(x,R)$ (where $B(x,R)$ is the ball centered at $x$ and with radius $R$). Then, since the function is positive, you have, by domain-monotonicity, $$ \int_\Omega|x-y|^{1-n}\lambda^n(y)\leq \int_{B(x,R)}|x-y|^{1-n}\lambda^n(y). $$ Now, let $u=x-y$, and you get that the integral to the right equals $$ \int_{B(0,R)}|u|^{1-n}\lambda^n(u). $$ Can you proceed from here?