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Let $\lambda$ the lebesgue measure on $\mathbb{R}^n$. If $N$ is a set with zero measure, then for all $\varepsilon>0$, exists a numerable collections of open balls ${B_n}$ such that $N\subset\bigcup_{n=1}^{\infty}B_n$ and $\sum_{n=1}^{\infty}\lambda(B_n)<\varepsilon$.

I think, in general we have $\lambda(A)=\inf\lbrace{\sum_{n=1}^{\infty}{\lambda(B_n):|A\subset\bigcup_{n=1}^{\infty}B_n} \rbrace}$

I easy show that $\lambda(A)\leq\inf\lbrace{\sum_{n=1}^{\infty}{\lambda(B_n):|A\subset\bigcup_{n=1}^{\infty}B_n} \rbrace}$

Because $A\subset\bigcup_{n=1}^{\infty}B_n$ then $\lambda(A)\leq\lambda(\bigcup_{n=1}^{\infty}B_n)\leq\sum_{n=1}^{\infty}{\lambda(B_n)}$ then taken the infimum we have the inequality.

I don't know how prove the other inequality. I think proving the result for a n-dimensional cube is a sufficient condition.

Note: I have the next definition for the lebesgue measure: $\lambda(A)=\inf\lbrace{ \sum_{i=1}^{\infty}{v(I_n)}:A\subset\bigcup_{i=1}^{\infty}{I_n}, I_n \textrm{ are open n-dimensional cubes} \rbrace}$ where $v(I_n)$ is the hiper-volume of the n-dimensional cube

I have this question because I read the first answer of this question: Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?

Gabriel Canedo
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Any set of diameter $d$ is contained in a closed ball of diameter $2d$. So a cube of diameter $d$ is covered by a ball of diameter $2d$. The volume of this ball is proportional to the volume of the cube since both volumes scale as $d^n$, so this proves the claim.

Mason
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  • what is not clear to me is that proportionality implies that the series with the measures of the balls converges to a quantity proportional to epsilon or that it even converges – Gabriel Canedo May 04 '22 at 13:09
  • Let $\varepsilon > 0$ be arbitrary. Cover the set $N$ with cubes whose volumes sum to $\leq \varepsilon$. Now cover each cube by a open ball of twice times the diameter. The volumes of the balls sum to $\leq C_n\varepsilon$. – Mason May 04 '22 at 18:28
  • Thank you :) This is a relation between Hausdorff Measure, right? – Gabriel Canedo May 04 '22 at 19:14
  • @GabrielCanedo The thing you are proving is easy; it is a special case of the result which says that in the definition of Lebesgue outer measure, one can replace "cubes" with "balls". The proof of this result uses covering lemmas. The result is related to equivalence of Hausdorff and Lebesgue measures. – Mason May 04 '22 at 23:16
  • Could you give me a reference book? For this covering lemmas... – Gabriel Canedo May 11 '22 at 05:03
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    Folland's book and "measure theory and fine properties of functions" by Evans and Gariepy have covering lemmas. I believe they are called Vitali covering lemma and Bescovitch covering lemma. – Mason May 11 '22 at 05:11