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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.)

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  • $\begingroup$ 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. $\endgroup$ Commented Aug 23, 2011 at 2:08
  • $\begingroup$ ... and how could I prove that using the Vitali covering lemma? $\endgroup$
    – user57159
    Commented Aug 23, 2011 at 2:26
  • $\begingroup$ 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. $\endgroup$ Commented Aug 23, 2011 at 2:40
  • $\begingroup$ 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. $\endgroup$ Commented Aug 23, 2011 at 4:06

2 Answers 2

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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 $S$.

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  • $\begingroup$ 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. $\endgroup$
    – 23rd
    Commented Jun 8, 2013 at 11:09
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    $\begingroup$ @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. $\endgroup$ Commented Jun 8, 2013 at 12:24
  • $\begingroup$ 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? $\endgroup$
    – 23rd
    Commented Jun 8, 2013 at 12:37
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    $\begingroup$ @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. $\endgroup$ Commented Jun 8, 2013 at 12:54
  • $\begingroup$ 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. :) $\endgroup$
    – 23rd
    Commented Jun 8, 2013 at 13:09
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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.

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    $\begingroup$ That's exactly how I got "(I know it would if $D(f)$ is continuous.)". $\endgroup$
    – user57159
    Commented Aug 25, 2011 at 4:37
  • $\begingroup$ Could you sketch the proof? It's relevant to my question: when is the image of a null set also null? $\endgroup$ Commented Sep 14, 2012 at 13:39
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    $\begingroup$ 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. $\endgroup$ Commented Sep 14, 2012 at 18:00
  • $\begingroup$ Why do we need to deal with Hausdorff measure, not Lebesgue measure directly? It seems to me that Hausdorff measure does not play any role? $\endgroup$
    – Bach
    Commented Aug 2, 2019 at 5:58

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