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I was reading a proof that submanifolds $M^d \subset \mathbb R^n$ with $d<n$ has measure zero. In the proof they use the fact:

Let $f:U\longrightarrow\mathbb{R}^n$ $\mathcal{C}^1$ where $U$ is an open susbet of $\mathbb{R}^n$, then if $E$ is a borelian of null measure, then $f(E)$ is of null measure too.

If I have this result, I know how to conclude. Can you help me showing this lemma? Thank you!

Let $B$ be a closed ball of $U$ and $E$ a borelian of null measure, then since $f$ is $\mathcal{C}^1$, there exists a $K>0$, such that $f$ is $K$-lipschitz continuous (by compacity of $B$ and by mean value inequality). We conclude that the image by $f$ of a cube $C$ of measure $\delta>0$ is a set of measure at most $K^n\delta$.

For all $\varepsilon>0$, we can cover $E\cap B$ by an union of cube $C$ such that $\lambda_n(C)\leq\varepsilon$. I don't understand this one since we don't think we have compacity of $E\cap B$... I just understood that we can obtain a $f(C)$ arbitrarily small if we take a cube $C$ arbitrarily small.

Arctic Char
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BlueCharlie
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    Does this answer your question? Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets? – Arctic Char Aug 06 '20 at 19:35
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    Suppose you have compacity, then we are done, right? Now assume that for every compact set K, $f(E\cap K) = 0$. Since $\mathbb{R}^n = \cup_{k=1}^\infty (K_k)$ where $K_k = { x\in\mathbb{R}^n : ||x||\leq k }$ we have that $m( f(E) ) \leq \sum m ( f(E\cap K_k) ) = 0$. – Lucas Resende Aug 06 '20 at 19:35
  • @ArcticChar Thanks for your answer. The main answer in this article doesn't really answer my issue. It says that "if N has measure zero then, $N\subset\bigcup_{k=1}^\infty B_k$, $B_k$ being a ball of radius $r_k$". Since I don't have any hypothesis on $N$ such has compacity, I don't really see where that fact comes from. – BlueCharlie Aug 07 '20 at 16:42
  • @Lucas Resende Thanks for you answer, but there is one thing I don't really understand. If $E$ is compact, I understand that we are done. But if $E$ isn't, my goal is to prove that $\lambda(E\cap K)=0$. I have to cover it by a finite union of small cubes or small balls and I don't know how to do that. – BlueCharlie Aug 07 '20 at 16:49
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    @BlueCharliebl they are using the outer regularity of lebesgue measure. – Arctic Char Aug 08 '20 at 12:24

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