Let $A \subset \mathbb{R}^n$ be open and bounded. Is it true that $$\int_A \int_A |x-y| \, dx \, dy \leq R|A|^2$$ where $R$ is some number (eg. the radius of a ball containing $A$) and $|A|$ is measure of $A$?
How do I show this rigourously?
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3$\lvert x-y\rvert \leqslant \operatorname{diam}(A)$ for $x,y \in A$. – Daniel Fischer Dec 12 '13 at 23:09
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2Hint: try bounding $|x-y|$ – PVAL-inactive Dec 12 '13 at 23:09
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What are your ideas? – LASV Dec 12 '13 at 23:30
1 Answers
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Since $A$ is bounded, the measure $|A|$ is finite.
If you choose two points $X$ and $Y$ at random within $A$, independently of each other, the probability of either falling within any subset $B$ of $A$ being $|B|/|A|$, then the expected distance $\mathbb E|X-Y|$ is $$ \frac{1}{|A|^2}\int_A \int_A |x-y|\,dx\,dy. $$ Try to show that if a random variable is always within certain bounds, then so is its expected value.