Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $n$ be a positive integer. $\; $ Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be everywhere Frechet differentiable.
Let $S$ be a subset of $\mathbb{R}^n$ with Lebesgue measure zero.

Does it follow that $\{f(s) : s\in S\}$ has Lebesgue measure zero?

(I know it would if $D(f)$ is continuous.)

share|improve this question
    
Yes. In fact, if $f$ has a Frechet derivative $Df$ on a measurable set $S\subseteq{R}^n$ then $$\mu(f(S))\le\int_S\Vert Df\Vert^n\,d\mu$$ where $\mu$ is the Lebesgue measure on $\mathbb{R}^n$. In particular, if $\mu(S)=0$ then the right hand side is zero. You can prove this using the Vitali covering lemma. –  George Lowther Aug 23 '11 at 2:08
    
... and how could I prove that using the Vitali covering lemma? –  Ricky Demer Aug 23 '11 at 2:26
    
I just posted the proof of your question. I was going to wait until I had time to give a fuller answer but, as you asked, I posted the partial answer now. –  George Lowther Aug 23 '11 at 2:40
    
In fact, you can show that the full change of variables formula holds in this generality, involving the Jacobian determinant of $f$ (which is bounded by $\Vert Df\Vert^n$). Then, you actually have equality rather than just the inequality I gave in my answer. –  George Lowther Aug 23 '11 at 4:06

2 Answers 2

up vote 7 down vote accepted

Yes. In fact, if $f$ has a Frechet derivative $Df$ on a measurable set $S\subseteq{R}^n$ then $$ \begin{align} \mu(f(S))\le\int_S\left\vert{\rm det} Df\right\vert\,d\mu&&(1) \end{align} $$ where $\mu$ is the Lebesgue measure on $\mathbb{R}^n$. In particular, if $\mu(S)=0$ then the right hand side is zero regardless of the value of $Df$. Actually, (1) is an immediate consequence of the change of variables formula for multidimensional integration, which can be shown to apply in the generality required here. Note that a general form of Sard's theorem also follows from (1), and says that the image of the set of points where $Df$ is singular has zero measure.

The Vitali covering lemma gives a quick proof that $f(S)$ has zero Lebesgue measure. If $f$ is Frechet differentiable at a point $x$ then, for any $K > \Vert Df\Vert$ we have $f(B_r(x))\subseteq B_{Kr}(f(x))$ for all small enough $r > 0$. So, $\mu(f(B_r(x))\le K^n\mu(B_r(x))$. Now, if $\Vert Df\Vert < K$ on a measurable set $S$ then you can choose open $U\supseteq S$ with $\mu(U)$ as close to $\mu(S)$ as you wish. For each $x\in S$ choose $r_x\in(0,1)$ such that $B_{r_x}(x)\subseteq U$ and $f(B_{r}(x))\subseteq B_{Kr}(f(x))$ for all $r\le 5r_x$. The Vitali covering lemma implies that there is a countable subset $S^\prime\subseteq S$ such that $\{B_{r_x}(x)\colon x\in S^\prime\}$ are disjoint and $$ \bigcup_{x\in S^\prime}B_{5r_x}(x)\supseteq\bigcup_{x\in S}B_{r_x}(x)\supseteq S. $$ So, $$ \begin{align} \mu(f(S))&\le\sum_{x\in S^\prime}\mu(f(B_{5r_x}(x)))\\ &\le 5^nK^n\sum_{x\in S^{\prime}}\mu(B_{r_x}(x))\\ &=5^nK^n\mu\left(\bigcup_{x\in S^\prime}B_{r_x}(S)\right)\\ &\le5^nK^n\mu(U). \end{align} $$ Letting $\mu(U)$ decrease to $\mu(S)$ gives $\mu(f(S))\le5^nK^n\mu(S)$. In particular, if $\mu(S)=0$ then $\mu(f(S))=0$. Even if the Frechet derivative just exists on $S$, but is not bounded, we can let $S_n$ be the subset of $S$ on which $\Vert Df\Vert\le n$. Then, using monotone convergence, $\mu(f(S))=\lim_{n\to\infty}\mu(f(S_n))=0$.

This gives a positive answer to the question asked, where $\mu(S)=0$. We can go further and use the Vitali covering theorem to prove (1). In fact, the full change of variables formula $$ \int_S\vert{\rm det} Df\vert\,d\mu=\int_{f(S)}\#\left(f^{-1}(\{y\})\cap S\right)\,d\mu(y) $$ can be shown to hold in the generality asked here. This can be proven with a similar argument as above involving the Vitali covering theorem, and also using the limits $\lim_{r\to0}\mu(f(B_r(x))/\mu(B_r(x))=\vert{\rm det}Df(x)\vert$ on $S$.

share|improve this answer
    
Dear George Lowther, I am sorry for bothering you in such an very old post. I learned your answer recently from here. In your equation $(1)$, it is clear to me if I understand $\mu(f(S))$ as the Lebesgue outer measure of $f(S)$, but I cannot see why $f(S)$ is Lebesgue measurable. When $Df$ is locally integrable, it is easy to see that $f$ is absolutely continuous and hence $f(S)$ is always Lebesgue measurable when $S$ is, but what is the general story? Moreover, I am also concerned about that whether $f(S)$ is Borel when $S$ is Borel. –  23rd Jun 8 '13 at 11:09
1  
@Landscape: While I think that using the Lebesgue outer measure is enough for this question, it is true that $f(S)$ will be Lebesgue measurable whenever $S$ is. Note that, being continuous, $f$ takes compact sets to compact sets and, using the argument in this answer, Lebesgue-null sets to Lebesgue-null sets (hence, Lebesgue measurable). As every Lebesgue measurable set $S$ can be written as a countable union of compact sets and a Lebesgue-null set, it follows that $f(S)$ is Lebesgue measurable. –  George Lowther Jun 8 '13 at 12:24
    
George Lowther, thank you for your reply. I was only focusing on equation $(1)$ and ignoring that $f$ maps Lebesgue-null sets to Lebesgue-null sets had already been proved by you. Sorry for my carelessness. I still want to know if $f$ necessarily maps Borel sets to Borel sets. A similar question is my post here, i.e. does $f$ necessarily map sets of first Baire category to sets of first Baire category? –  23rd Jun 8 '13 at 12:37
1  
@Landscape: No, $f$ will not in general map Borel sets to Borel sets. Even the standard projection from $\mathbb{R}^2$ to $\mathbb{R}$ doesn't take Borel sets to Borel sets (a standard result), but is clearly a smooth map. Just embed $\mathbb{R}$ back into $\mathbb{R}^2$ to see that smooth $f:\mathbb{R}^2\to\mathbb{R}^2$ need not take Borel sets to Borel sets. –  George Lowther Jun 8 '13 at 12:54
    
Dear George Lowther, thank you for your excellent counter-example for the $n>1$ case! Just in case that you are interested, I will add the $n=1$ case to my question mentioned above. By the way, I just viewed your profile page and I really appreciate your taste in selecting questions to answer: as far as I see, the questions you asked are always interesting and definitely non-trivial. What I can only do to express my thanks and appreciation is to upvote your answers and share them with others. :) –  23rd Jun 8 '13 at 13:09

More generally, if $f:\ X \to Y$ is a locally Lipschitz function from one $\sigma$-compact metric space to another and $S \subseteq X$ has $d$-dimensional Hausdorff measure 0, then $f(S)$ also has $d$-dimensional Hausdorff measure 0. This is pretty much immediate from the definition of Hausdorff measures. In your case, Lebesgue measure on ${\mathbb R}^n$ coincides with $n$-dimensional Hausdorff measure.

share|improve this answer
1  
That's exactly how I got "(I know it would if $D(f)$ is continuous.)". –  Ricky Demer Aug 25 '11 at 4:37
    
Could you sketch the proof? It's relevant to my question: when is the image of a null set also null? –  Ricardo Buring Sep 14 '12 at 13:39
1  
If $f$ has Lipschitz constant $c$, i.e. $\text{dist}(f(x) , f(y)) \le c \text{dist}(x ,y)$, and $S$ is covered by open sets $U_j$ with diameter $< \delta$ and $\sum_j (\text{diam}\ U_j)^d < \epsilon$, then $f(S)$ is covered by open sets $V_j$ of diameter $< c \delta$ and $\sum_j (\text{diam}\ V_j)^d < c^d \epsilon$. If $f$ is only locally Lipschitz, break $X$ up into countably many pieces on which $f$ is Lipschitz. –  Robert Israel Sep 14 '12 at 18:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.