Let $B(x,r)$ be an open ball in $\mathbb R^n$ , then is the Lebesgue measure of the boundary of $B(x,r)$ zero ?
Is the Lebesgue measure of any open ball in Euclidean space zero?
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measure-theory
metric-spaces
lebesgue-measure
2 Answers
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Your question and title don't match.
If you know that $|B(x,r)| = c_n r^n$ for all $r > 0$ and you know that $\partial B(x,r) \subset B(x,r+\epsilon) \setminus B(x,r - \epsilon)$ for all $\epsilon > 0$ and you know that $B(x,r-\epsilon) \subset B(x ,r + \epsilon)$ for all $\epsilon > 0$ then you know that $$|\partial B(x,r)| \le c_n \left[ (r+\epsilon)^n - (r - \epsilon)^n \right]$$ for all $\epsilon > 0$.
What happens as $\epsilon \to 0^+$?
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Title question: No; the proof is that, as in Euclidean spaces all metrics are equivalent, for each ball you can put inside a $||\dot{}||_{\infty}$-ball, call it $B$, and we know lebesgue measure is product measure in Euclidean space, so $B$ has positive measure, therefore $B(x,r)$ also has positive measure (monotony).