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\centerline{\bf HAUSDORFF MEASURES AND THE MORSE--SARD THEOREM}

\vskip .5in

\centerline{\sc Carlos Gustavo T. de A. Moreira}

\centerline{IMPA}

\centerline{Estrada Dona Castorina, 110}
\centerline{22460-320, Rio de Janeiro, RJ}
\centerline{e-mail: gugu@impa.br}

\vskip .5in

\begin{abstract}
 Let $F\colon U\subset \re^n\to\re^m$ be a differentiable function and $p<m$ an integer. If $k\ge1$ is an integer, $\alpha\in [0,1]$ and $F\in C^{k+(\alpha)}$, if we set $C_p(F)=\{x\in U \mid \text{ rank}(Df(x))\le p\}$ then the Hausdorff measure of dimension $(p+\frac{n-p}{k+\alpha})$ of $F(C_p(F))$ is zero.
\end{abstract}

\vskip .5in

\section{Introduction}

The Morse-Sard theorem is a fundamental theorem in analysis that is in the basis of transversality theory and differential topology. The classical Morse-Sard theorem states that the image of the set of critical points of a function $F\colon\re^n\to\re^m$ of class $C^{n-m+1}$ has zero Lebesgue measure in $\re^m$. It was proved by Morse ([M]) in the case $m=1$ and by Sard ([S1]) in the general case.

Due to its theoretical importance, the Morse-Sard theorem was generalized in many directions.
Many of these generalizations are related with Hausdorff measures and
Hausdorff dimensions.

Given a metric space $X$ and a positive real number $\al$, we define the
Hausdorff measure of dimension
$\al$ associated to a covering $\mathcal{U} = (U_\la)_{\la\in L}$ of $X$ by
bounded sets $U_\la$ by 
$m_\al(\mathcal{U}) = \sum_{\la\in L}(\text{diam } U_\la)^\al$, where
$\text{diam  }U_\la$ denotes the
 diameter of $U_\la$\,, and, if we define the norm of a covering
$\mathcal{U}$ by $||\mathcal{U}|| = 
\sup_{U\in\mathcal{U}} (\text{diam }U)$, then the Hausdorff measure of
dimension $\al$ of $X$
 is $m_\al(X) = \underset{\substack{\mathcal{U} \text{ covering of } X\\
||\mathcal{U}||\to0}}{\liminf} 
\,m_\al(\mathcal{U})$.

It is not difficult to see that there is a unique $d \in [0,+\infty]$ such
that if $\al > d$ then $m_\al(X)=0$
 and if $\al < d$ then $m_\al(X) = +\infty$. This number $d$ is called the
{\it Hausdorff dimension\/} of $X$.
 It is easy to see that if $X \subset \re^n$ then its Hausdorff dimension $d
=: HD(X)$ belongs to $[0,n]$.

 Sard himself proved that if $C_p(F)=\{x\in\re^n \mid\text{ rank}(DF(x))\le p\}$ then for any $\varepsilon>0$ there is $k\in\na$ such that if
$F$ is $C^k$ then $F(C_p(F))$ has zero Hausdorff measure of dimension $p+\varepsilon$ ([S2]). This result was made more precise by Federer ([F]), who proved that if $k\in\na$ then the Hausdorff measure of dimension $p+\frac{n-p}k$ of $F(C_p(F))$ is zero. We should also mention the works of Church ([Ch1], [Ch2]), which gave more results about the structure of the set of critical values of differentiable maps. Later, Yomdin ([Y]) proved that the Hausdorff dimension of $F(C_p(F))$ is at most $p+\frac{n-p}{k+\alpha}$, provided that $F\in C^{k+\alpha}$, where $k\in\na$ and $0\le\alpha< 1$. More recently, Bates ([B2]) proved that if $F\in C^{k+\alpha}$ with $k\in\na$, $0<\alpha\le 1$ and $p+\frac{n-p}{k+\alpha}=m$ then $F(C_p(F))$ has zero Lebesgue measure in $\re^m$ (this in particular improves the hypothesis of the classical Morse-Sard theorem from $F\in C^{n-m+1}$ to $F\in C^{n-m+\text{Lips.}}$, i.e., $F\in C^{n-m}$ and $D^{n-m}F$ Lipschitz.

The aim of this work is to generalize the mentioned results by proving a general version of the Morse-Sard Theorem
involving Hausdorff measures. Let $k \ge 1$ be an integer and $\a \in [0,1]$.
We say that a function $F\colon U\subset\re^n \to \re^m$ is of class
$C^{k+(\a)}$ at a subset $A$ of $U$ if $F$ is $C^k$ in $U$ and for
each $x \in A$ there are $\ve_x > 0$, $K_x > 0$ such that $|y-x| < \ve_x
\Rightarrow |D^k\,F(y)-D^k\,F(x)| \le K_x|y-x|^\a$ (this is less restrictive
than supposing $F \in C^{k+\a}$). Our main result is the following

\vskip .1in

\noindent
{\sc Theorem:} Let $F\colon U\subset\re^n \overset{C^k}{\longrightarrow} \re^m$
and let $p < m$ be an integer. If $C_p(F) := \{x \in U \mid
\text{rank}\,(DF(x)) \le p\}$ and if $F$ is of class $C^{k+(\a)}$ at $C_p(F)$
then the Hausdorff $\left(p + \frac{n-p}{k+\a}\right)$-measure of $F(C_p(F))$
is zero.

\vskip .1in

In particular, if $k+\a = \frac{n-p}{m-p}$\,, we recover the result of [B2],
with a weaker hypothesis. We remark that if $p + \frac{n-p}{k+\a} < m$, the
Hausdorff $\left(p + \frac{n-p}{k+\a}\right)$-measure is not the Lebesgue
measure or a product measure in $\re^m$, and so we can not use Fubini's Theorem. This difficulty is
solved in the present paper by replacing the use of Fubini's theorem by a careful decomposition of the critical set, combined with a parametrized strong version of the
main lemma of Morse's paper([M]) (Theorem 2.1).

We shall also give examples that show that our result is quite sharp, by giving
counterexamples to slight changes of the hypothesis or of the conclusion.


\section{Functions whose zeros include a given set}

We shall prove here a version of Theorem 3.6 of [M] and Lemma 3.4.2 of [F],
which will be fundamental for the later results.


\begin{thm}\label{thm2.1}
 Let $k \ge 1$, $\a \in [0,1]$, $n > p$ and $A
\subset U \subset \re^n$, where $U$ is an open set. Then there are sets
$A_1,A_2\dots, \subset A$ such that $A = \bigcup_{i=1}^\infty \,A_i$, where for each
$i=1,2,\ldots$ there is a function $\psi_i\colon B_i\times V_i
\overset{C^1}{\longrightarrow} U$ where $B_i$ is a ball in some $\re^{r_i}$,
$r_i \ge 0$ and $V_i$ is a ball in $\re^p$ such that $\psi_i(x,y) =
(\widetilde\psi_i(x,y),y)$, and $|\psi_i(x_1,y_1)-\psi_i(x_2,y_2)| \ge 
|(x_1,y_1)-(x_2,y_2)|$, $\forall\, (x_1,y_1),(x_2,y_2)\in B_i\times V_i$
 and $A_i \subset \psi_i(B_i\times V_i)$, with the following property: We can
write $A_i = A_i^\prime \cup A_i^\twoprime$ so that
$\psi_i^{-1}(A_i^\twoprime)$ has measure zero in $B_i\times V_i$\,, and if
$f\colon U \to \re$ vanishes in $A$ and $f$ is $C^{k+(\a)}$ at $A$ we have:

\begin{itemize}

\item[{$\bullet$}] $\underset{(x,y_0)\to (x_0,y_0)}{\limsup}\,
\dfrac{f(\psi_i(x,y_0))}{|x-x_0|^{k+\a}} < +\infty$, $\forall\,(x_0,y_0) \in
B_i\times V_i$ such that $\psi_i(x_0,y_0) \in A_i$,

\item[{$\bullet$}] $\underset{(x,y_0)\to(x_0,y_0)}{\lim}\,
\dfrac{f(\psi_i(x,y_0))}{|x-x_0|^{k+\a}} = 0$, $\forall\, 
(x_0,y_0) \in B_i\times V_i$ such that $\psi_i(x_0,y_0)) \in A_i^\prime$.
\end{itemize}
\end{thm}

\begin{pf}
Let us consider first the case $k=1$ and $df(x)\cdot v=0$
$\forall\, x \in A$, $v \in \re^{n-p}\times\{0\}$. In this case we
take $A = (A'\cap A) \cup A''$ where $A'$ is the set of density points of $\ov
A$ in the direction of $\re^{n-p}\times\{0\}$ ($(x,y)\in A' \Rightarrow
\underset{\ve\to0}{\lim}\dfrac{m((B_\ve(x)\times\{y\})\cap\ov
A)}{m(B_\ve(x))} = 1$, where $m$ is the $(n-p)$-dimensional measure). The
measure of $A'' = A-A'$ is zero, since it is zero in each plane $\re^{n-
p}\times\{y\}$.

For $(x_0,y_0) \in A$ take $B((x_0,y_0),\ve(x_0,y_0))$ a ball contained in $U$ and $\psi = Id|_{B((x_0,y_0),\ve(x_0,y_0))}$. We have 
$\underset{(x,y_0)\to(x_0,y_0)}{\limsup}\,\dfrac{f(x,y_0)}{|x-x_0|^{1+\a}} <
+\infty$, since $f(x,y_0) = f(x,y_0) - f(x_0,y_0) = df(tx_0+(1-t)x)(x-x_0)$, $t
\in (0,1) \Rightarrow |f(x,y_0)| \le K_{x_0}|x-x_0|^{1+\a}$\,. For $(x_0,y_0)
\in A'$,
$$
\lim_{\delta\to0} \frac{1}{\vol(S^{n-p-1})} \int_{S^{n-p-1}}
\bigg(\frac{1}{\delta}\int_0^\delta \chi_{\ov A}(x_0+ t v,y_0)dt\bigg)dv = 1, 
$$
so $\forall\,\ve > 0$\,\,$\exists \delta_0 > 0$ s.t. $|x-x_0| < \delta_0
\Rightarrow \exists\, v \in S^{n-p-1}$ with $\lv v - \dfrac{x-x_0}{|x-x_0|}\rv
< \ve$ and \linebreak 
$\lv\dfrac{1}{|x-x_0|}\int_0^{|x-x_0|} \chi_{\ov A}(x_0+tv,y_0)dt-
1\rv < \ve$, so, if $\tilde x = x_0 + |x-x_0|v$, \linebreak 
$|f(x,y_0)-f(x_0,y_0)| \le
|f(x,y_0)-f(\tilde x,y_0)| + |f(\tilde x,y_0) - f(x_0,y_0)|$, but $f(x,y_0) -
f(\tilde x,y_0) = \linebreak 
 df(\theta t + (1-\theta)\tilde x,y_0)\cdot(x-\tilde x),
\theta \in (0,1) \Rightarrow |f(x,y_0)-f(\tilde x,y_0)| \le K_{x_0}|x-x_0|^\a
\cdot \ve|x-x_0| = \linebreak 
 \ve K_{x_0}|x-x_0|^{1+\a}$ and $f(\tilde x,y_0) - f(x_0,y_0)
= \int_0^{|\tilde x-x_0|} df(x_0+tv,y_0)\cdot v\,dt \le \linebreak $ 
$K_{x_0}|x-x_0|^\a \cdot m\{t \in [0,|\tilde x-x_0|] \Big|  \dfrac{\po f}{\po
x}(x_0+ tv,y_0)\ne0\} \le$ $ K_{x_0}|x-x_0|^\a \cdot \ve|x-x_0| = \ve
K_{x_0}|x-x_0|^{1+\a}$.

So $|f(x,y_0)| = |f(x,y_0)-f(x_0,y_0)| \le 2\ve K_{x_0}|x-x_0|^{1+\a}
\Rightarrow
\underset{(x,y_0)\to(x_0,y_0)}{\lim}\,\dfrac{f(x,y_0)}{|x-x_0|^{1+\a}} = 0$.

We can take a countable subcovering of $A$ by the $B((x_0,y_0),
\ve(x_0,y_0))$ to finish the proof in this case.

Consider now the case $k \ge 1$, $n$ arbitrary. We have $A = A^*\cup A^{**}$
where $A^* = \{x \in A \mid \linebreak 
\exists\,g\colon U
\overset{C^k}{\longrightarrow} \re$, $g|_A\equiv 0, \exists\,v \in \re^{n-
p}\times\{0\}$, $dg(x)\cdot v\ne0\}$. $A^{**} = A\backslash A^*$. If $(x_0,y_0)
\in A^*$ there is $g$ as above, so there is $\ve > 0$ such that $g^{-1}(0) \cap
B_\ve(x_0,y_0)$ is contained in the image of $\psi\colon B\times V
\overset{C^k}{\longrightarrow} U$ where $B$ is a ball in $\re^{n-p-1}$, as
in the statement, and $A \subset g^{-1}(0)$. Taking a countable subcovering of
$A^*$ by these balls we reduce the proof in this case to a case with smaller
$n$. If $k=1$, the result was yet proved for $A^{**}$. If $k > 1$ , and
assuming by induction the result for $k-1$, we have 
\begin{align*}
\, &A^{**} = \bigcup_{i=1}^\infty\,A_i^{**}, A_i^{**} =
(A_i^{**})' \cup (A_i^{**})'', A_i^{**} \subset \psi_i(B_i\times V_i),
\psi_i \in C^1,\\
&\psi_i(x_0,y_0) \in A_i^{**} \Rightarrow \underset{x\to x_0}{\limsup}
\frac{||df(\psi_i(x,y_0))|_{\re^{n-p}\times \{0\}}||}{|x-x_0|^{k-1+\a}} <
+\infty \Rightarrow \underset{x\to
x_0}{\limsup}\frac{|f(\psi_i(x,y_0))|}{|x-x_0|^{k+\a}} < +\infty
\end{align*}
and
$$
\psi_i(x_0,y_0) \in (A_i^{**}) \Rightarrow \underset{x\to x_0}{\lim}
\frac{||df(\psi_i(x,y_0))|_{\re^{n-p}\times\{0\}}||}{|x-x_0|^{k-1+\a}} = 0
\Rightarrow \underset{x\to x_0}{\lim}\frac{f(\psi_i(x,y_0))}{|x-x_0|^{k+\a}}
 =0,
$$
both by the mean value theorem, and the proof is finished by induction.
\qed
\end{pf}

\begin{Cor}\label{cor2.2}
 Let $k \ge 1$, $\a \in [0,1]$, $n > p$ and $A
\subset U \subset \re^n$, where $U$ is an open set. Then there are sets
$A_1,A_2\dots, \subset A$ such that $A =
\bigcup_{i=1}^\infty\,A_i$, where for each
$i=1,2,\ldots$ there is a function $\psi_i\colon B_i\times V_i
\overset{C^1}{\longrightarrow} U$ where $B_i$ is a ball in some $\re^{r_i}$,
$r_i \ge 0$ and $V_i$ is a ball in $\re^p$ such that $\psi_i(x,y) =
(\widetilde\psi_i(x,y),y)$, and $|\psi_i(x_1,y_1)-\psi_i(x_2,y_2)| \ge 
|(x_1,y_1)-(x_2,y_2)|$, $\forall\, (x_1,y_1),(x_2,y_2)\in B_i\times V_i$
 and $A_i \subset \psi_i(B_i\times V_i)$, with the following property: We can
write $A_i = A_i^\prime \cup A_i^\twoprime$ so that
$\psi_i^{-1}(A_i^\twoprime)$ has measure zero in $B_i\times V_i$\,, and if
$f\colon U \to \re$  is $C^{k+(\a)}$ at $A$ and $D_xf\equiv 0$ in $A$ we have:

\begin{itemize}

\item[{$\bullet$}] $\underset{(x,y_0)\to(x_0,y_0)}{\limsup}\,
\dfrac{|f(\psi_i(x,y_0))- f(\psi_i(x_0,y_0)) |}{|x-x_0|^{k+\a}} < +\infty$, $\forall\,(x_0,y_0) \in
B_i\times V_i$ such that  $\psi_i(x_0,y_0) \in A_i$,

\item[{$\bullet$}] $\underset{(x,y_0)\to(x_0,y_0)}{\lim}\,
\dfrac{|f(\psi_i(x,y_0))- f(\psi_i(x_0,y_0)) |}{|x-x_0|^{k+\a}} = 0$, $\forall\, 
(x_0,y_0) \in B_i\times V_i$ such that  $\psi_i(x_0,y_0)) \in A_i^\prime$.

\end{itemize}
\end{Cor}

\begin{pf}
 If $k \ge 2$ this is an immediate consequence of Theorem
2.1 applied to $D_xf$ and of the mean value theorem. If $k=1$ this can be proved
exactly as the case $k=1$ of the Theorem 2.1. \qed

\end{pf}

\begin{Cor}\label{cor2.3}
 In the statements of Theorem 2.1 and Corollary
2.2, for any $x \in B_i$ s.t. $\psi_i(x) \in A_i$ there are $\ve_x > 0$, $K_x >
0$ such that $|y-x| < \ve_x \Rightarrow |f(\psi_i(y)) - f(\psi_i(x))| \le
K_x|y-x|^{k+\a}$, \linebreak 
and for any $\ve > 0$ there is a $\delta > 0$ so that
$\frac{\la(\psi_i^{-1}(A_i) \cap B_r(x))}{\la(B_r(x))} > 1-\delta \Rightarrow$ \linebreak
$|f(\psi_i(y)) - f(\psi_i(x))| \le \ve K_xr^{k+\a}$, if $r \le \ve_x$ and $|y-x|
\le r$ ($\delta$ depends only on $\ve$ and $n$, but not on $f$ or on $x$).

\end{Cor}

\begin{pf}
 This is only a more precise formulation of the results
proved in the demonstration of the theorem. 
\qed
\end{pf}

\begin{rem}\label{rem2.4}
 For $k=0$ we have the same results, except the
statement $\underset{y\to x}{\lim}\,\frac{f(\psi_i(y))}{|y-x|^{k+\a}} = 0$,
for each $x \in B_i$ such that $\psi_i(x) \in A_i^\prime$.
\end{rem}

\section{The main results}

\begin{lem}\label{lem3.1}
Let $A \subset \re^m$ with $\la(A) < \infty$ and let
$\U$ be a family of balls $B_r(x)$, $x \in A$ such that for each $x \in A$
there is an $\ve_x > 0$ such that $r \le \ve_x \Rightarrow B_r(x) \in \U$. Then
for each $\ve > 0$ there are $x_n \in A$, $r_n > 0$ with $B_{r_n}(x_n) \in \U$
and $A \subset
\bigcup_{n=1}^{\infty}\,B_{r_n}(x_n)$
such that $\sum_{n=1}^\infty \la(B_{r_n}(x_n)) < \la(A) + \ve$.

\end{lem}

\begin{pf}
 This lemma is essentially the Vitali covering theorem
from measure theory. Take $U \supset A$ an open set with $\la(U) < \la(A) +
\frac{\ve}{2}$. If we choosed $B_{\tilde r_1}(x_1),\ldots,B_{\tilde r_n}(x_n)$,
define $s_n = \sup\{r > 0 \mid \exists\, x\in A\,\,\text{ s.t. }\,\, r <
\frac{\ve_x}{5}$, $B_r(x) \subset U$ and $B_r(x) \cap (B_{\tilde r_1}(x_1) \cup
\cdots\cup B_{\tilde r_n}(x_n)) = \emptyset\}$. Choose $B_{\tilde
r_{n+1}}(x_{n+1})$ such that $\tilde r_{n+1} > \frac{s_n}{2}$, $\tilde r_{n+1}
< \frac{\ve_{x_{n+1}}}{5}$, $B_{\tilde r_{n+1}}(x_{n+1}) \subset U$ and
$B_{\tilde r_{n+1}}(x_{n+1}) \cap (B_{\tilde r_1}(x_1) \cup\cdots\cup
B_{\tilde r_n}(x_n)) = \emptyset$. Since the $B_{\tilde r_i}(x_i)$ are disjoint and
contained in $U$ we have $\sum_{i=1}^\infty \la(B_{\tilde r_i}(x_i)) <
\la(A) + \frac{\ve}{2}$\,, and so there is a $n_o \in \na$ such that
$\sum_{i=n_o}^\infty \la(B_{5\tilde r_i}(x_i)) < \frac{\ve}{2}$\,. We
take $B_{r_i}(x_i) = B_{\tilde r_i}(x_i)$, $i < n_o$ and $B_{r_i}(x_i) =
B_{5\tilde r_i}(x_i)$, $i \ge n_o$. 

Clearly we have $\sum_{i=1}^\infty
\la(B_{r_i}(x_i)) < \la(A) + \ve$. To prove that $A \subset
\bigcup_{n=1}^{\infty}\,\ov{B_{r_n}(x_n)}$, take $x
\in A$ and 
$r = \min\{\tilde r_{n_o}, \ve_x/5, d(x,U^c \cup \bigcup_{i<n_o}\, B_{r_i}(x_i))\}$. If $r > 0$, take $n \ge n_o$ such that
$s_n < r \le s_{n-1}$ (we have $r \le \tilde r_{n_o} \le s_{n_o-1})$, and note
that $s_n < r \Rightarrow B_r(x)\cap(B_{\tilde r_1}(x_1) \cup\cdots\cup
B_{\tilde r_n}(x_n)) \ne \emptyset \Rightarrow \exists \,\, i\le n$ such that
$B_r(x) \cap B_{\tilde r_i}(x_i) \ne \emptyset$. We have $n \ge n_o$ since $r
\le d(x,B_{\tilde r_i}(x_i))$, and $\tilde r_i > \frac{s_{n-1}}{2} \ge \frac
r2$, since $i\le n$. Therefore, we have $x \in B_{5\tilde r_i}(x_i)$. If $r=0$
then $x \in \ov{B_{r_i}(x_i)}$ for some $i < n_o$. This proves that $A \subset
\bigcup_{n=1}^{\infty}\,\ov{B_{r_n}(x_n)}$. 
Taking $\tilde r_n= \left( \frac{\la(A)+\ve}{\sum_{i=1}^\infty
\la(B_{r_i}(x_i))} \right)^{1/2m}\cdot r_n$,
 we have $A \subset
\bigcup_{n=1}^{\infty}\,B_{\tilde r_n}(x_n)$, with
$\sum_{n=1}^\infty \la(B_{\tilde r_n}(x_n)) = (\la(A)+\ve)^{1/2}(\sum_{i=1}^\infty \la(B_{r_i}(x_i)))^{1/2} < \la(A) + \ve$. \qed

\end{pf}

\begin{rem}\label{rem3.1}
 In the Lemma 3.1 we can replace a family of balls
$B_r(x)$ by a family of cubes $C_r(x) =
\prod_{i=1}^{m}\,[x_i-r, x_i+r]$, where $x =
(x_1,\ldots,x_m)$, using the same proof.

\end{rem}

\begin{lem}\label{lem3.2}
 Let $F\colon U\subset \re^n \to \re^m$ be a function, $A
\subset U$ and $d > 0$ such that for any $x \in A$ there are $\ve_x > 0$, $K_x
> 0$ such that $m_d(F(B_\ve(x)\cap A)) \le K_x.\la(B_\ve(x))$, $\forall\, \ve <
\ve_x$, where $m_d$ is the Hausdorff measure of dimension $d$, and there is 
$A^\prime \subset A$ such that $\la(A\backslash A^\prime) = 0$ and
$\underset{\ve\to0}{\lim}\,\frac{m_d(F(B_\ve(x)\cap A))}{\la(B_\ve(x))} =
0$, $\forall\, x \in A^\prime$. Then $m_d(F(A)) = 0$.

\end{lem}

\begin{rem}\label{rem3.2}
 The same result is true if we replace $B_\ve(x)$ by
$C_\ve(x)$.

\end{rem}

\begin{rem}\label{rem3.3}
 We can replace the condition ``$m_d(F(B_\ve(x) \cap
A)) \le K_x\,\la(B_\ve(x))$, $\forall\, \ve < \ve_x$'' by ``$F(B_\ve(x) \cap
A)$ can be covered by balls $B_{\delta_i}(y_i)$, $i \in \na$, with
$\sum_{i=1}^\infty \delta_i^d \le K_x\,\la(B_\ve(x))$, $\forall\,
\ve < \ve_x$'', and the condition
``$\underset{\ve\to0}{\lim}\,\frac{m_d(F(B_\ve(x)\cap A)}{\la(B_\ve(x))} =
0$, $\forall x \in A^\prime$\,'' by ``$F(B_\ve(x) \cap A)$ can be covered by
balls $B_{\delta_i^{(\ve)}}(y_i)$, $i \in \na$ with
$\underset{\ve\to0}{\lim}\,
\frac{\sum_{i=1}^\infty(\delta_i^{(\ve)})^d}{\la(B_\ve(x))} = 0$,
$\forall\, x \in A^\prime$\,''. The proof remains essentially the same, and
Remark 3.2 is still valid.

\end{rem}

\begin{rem}\label{rem3.4}
 If we replace the conditions of this lemma by
``$F(B_\ve(x) \cap A)$ can be covered by balls $B_{\delta_i}(y_i)$, $i \in
\na$, with $\sum_{i=1}^\infty \delta_i^d \le k\,\la(B_\ve(x))$,
$\forall\, \ve < \ve_x$ (note that here $k$ does not depend on $x$), and
$\la(A) < \infty$'', then we can conclude, using the same proof, that
$m_d(F(A)) \le k\,\la(A)$.

\end{rem}

\begin{pf}
 We may suppose that $A$ has finite Lebesgue measure,
since $A$ is a countable union of sets with finite measure, and a countable
union of sets with Hausdorff $d$-measure zero has Hausdorff $d$-measure zero.
Moreover, since $A =\bigcup_{k=1}^{\infty}\,A_k$,
where $A_k = \{x \in A \mid K_x \le k\}$, we may suppose $K_x \le K$,
$\forall\, x \in A$. Let $C$ be the Lebesgue measure of $A$.

Let $\ve > 0$. For each $x \in A^\prime$ take $\delta_x > 0$ such that
$B_{\delta_x}(x) \subset U$ and $r \le \delta_x \Rightarrow
\frac{m_d(F(B_r(x)\cap A))}{\la(B_r(x))} \le \frac{\ve}{2(C+1)}$. By the Lemma
3.1 we can cover $A^\prime$ by
$\bigcup_{n=1}^{\infty}\,B_{r_n}(x_n)$ with 
$\sum_{n=1}^\infty \la(B_{r_n}(x_n)) < C+1$ and $r_n \le \delta_{x_n}
\Rightarrow \sum_{n=1}^\infty m_d(F(B_{r_n}(x_n) \cap A)) \le
\frac{\ve}{2(C+1)}\cdot (C+1) = \frac{\ve}{2} \Rightarrow m_d(F(A^\prime)) \le
\frac{\ve}{2}$\,. By Lemma 3.1 we can cover $A\backslash A^\prime$ by 
$\bigcup_{n=1}^{\infty}\,B_{\tilde r_n}(\widetilde
x_n)$ such that $B_{\tilde r_n}(\widetilde x_n) \subset U$ and $\tilde r_n <
\ve_{\widetilde x_n}$, $\forall\, n\in\na$, with $\sum_{n=1}^\infty
\la(B_{\tilde r_n}(\widetilde x_n)) < \frac{\ve}{2K} \Rightarrow
\sum_{n=1}^\infty m_d(F(B_{\tilde r_n}(\widetilde x_n))) \le
\frac{\ve}{2K}\cdot K = \frac{\ve}{2} \Rightarrow m_d(F(A\backslash A^\prime))
\le \frac{\ve}{2} \Rightarrow m_d(F(A)) \le \frac{\ve}{2} + \frac{\ve}{2} =
\ve$. Since $\ve > 0$ is arbitrary we have $m_d(F(A)) = 0.$ \qed

\end{pf}

We first use Lemma 3.2 to prove the following strong version of Constantin's result ([Co]), that does not suppose continuity of the derivatives. Here we do not suppose differentiability in every point, but only in the set of critical points under consideration.

\begin{thm}\label{thm3.3}
 Let $F\colon X \subset \re^n \to \re^n$ be a
function, and let $A = \{x \in X \mid DF(x)$ exists and is not surjective$\}$.
Then $\la(F(A)) = 0$.

\end{thm}

\begin{pf}
 It is a simple consequence of Lemma 3.2, since if $x \in
A$ then $\underset{r\to0}{\lim}\,\frac{\la(F(B_r(x)))}{\la(B_r(x))} = 0$.
Indeed, $x \in A \Rightarrow F(x+h) = F(x) + DF(x).h + r(h)$, where 
$\underset{h\to0}{\lim}\,\frac{r(h)}{|h|} = 0$. Let $K = ||DF(x)||$, and let
$\ve \in (0,1)$. Let $\delta > 0$ such that $|h| \le \delta \Rightarrow
\frac{|r(h)|}{|h|} < \frac{\ve}{2(K+1)^{n-1}}$. Then, if $|h| \le \delta$,
$F(x+h)-F(x)$ belongs to an $\frac{\ve.|h|}{2(K+1)^{n-1}}$ neighbourhood of a
ball of radius $K|h|$ in a subspace of $\re^n$ of dimension $n-1$ (a fixed
subspace of $\re^n$ of dimension $n-1$ which contains the image of $DF$), and
thus belongs to the orthogonal product of a ball of radius $(K+1)|h|$ in this
subspace by an interval of radius $\frac{\ve|h|}{2(K+1)^{n-1}}$. Therefore,
$\la(F(B_r(x)) \le \frac{\ve.r.r^{n-1}(K+1)^{n-1}}{(K+1)^{n-1}}v_{n-1} = \ve
r^n v_{n-1}$, where $v_{n-1}$ is the volume of the unitary ball in $\re^{n-1}$,
and, since $\ve > 0$ is arbitrary,
$\underset{r\to0}{\lim}\,\frac{\la(F(B_r(x)))}{\la(B_r(x))} = 0$. 
\qed
\end{pf}

\begin{thm}\label{thm3.4}
 Let $F\colon U \subset \re^n
\overset{C^k}{\longrightarrow}\, \re^m$ be a function of class
$C^{k+(\a)}(\a \in (0,1])$ at $C_p(F) := \{x \in U \mid \text{ rank } (DF(x))
\le p\}$. Then the Hausdorff measure of dimension $d = p + \frac{n-p}{k+\a}$ of
$F(C_p(F))$ is zero, $\forall\, p < \text{min}\{m,n\}$.

\end{thm}

\begin{pf}
 Since $C_p(F) =
\bigcup_{r=0}^{p}\,\{x \in U \mid \text{ rank }(DF(x))
= r\}$, and $r + \frac{n-r}{k+\a} \le p + \frac{n-p}{k+\a}$ for $0 \le r \le
p$, we may restrict our attention to $\widetilde C_p(F) = \{x \in U \mid \text{
rank }(DF(x)) = p\}$. If $x_o \in C_p(F)$, we have, after a change of
coordinates of class $C^k$, $F(z,y) = (z,G(z,y))$, with $(z,y) \in
\re^p\times\re^{n-p}$ and $G(z,y) \in \re^{m-p}$, in a neighbourhood $V$ of
$x_o = (z_o,y_o)$. We shall restrict our attention to this neighbourhood. We
have $x = (z,y) \in C_p(F) \Leftrightarrow D_y\,G(z,y) = 0$. We can apply the
results of the section 2 (Theorem 2.1, Corollary 2.3 and Remark 2.1) to the
function $D_y\,G$, and obtain the decomposition $A
=\bigcup_{i=1}^{\infty}\,A_i$, $A_i \subset
\psi_i(V_i\times B_i)$, where $A=\{(z,y)\in V \mid D_yG(z,y)=0\}$. Let us fix such an $A_i$. 

Since $\psi_i^{-1}(A_i) =
\bigcup_{m\in\na}\,\{x \in \psi_i^{-1}(A_i) \mid \ve_x \ge \frac
1m, K_x \le m\}$, we may suppose $\ve_x \ge \frac 1M$, $K_x \le M$, $\forall\,
x \in \psi_i^{-1}(A_i)$, for some fixed $M$ and also that $V$ has finite Lebesgue measure $\la(V)$. 

With these assumptions, we shall
prove that there is a constant $K_o$ such that for any $X \subset V$, $\nu >
0$, we can cover $F(A_i \cap X)$ by balls $B_{\delta_i}(p_i)$ so that
$\sum_{i=1}^\infty \delta_i^d \le K_o(\la(X)+\nu)$. For this, given a point $x \in A_i \cap X$ and an $\ve <
\frac{1}{2\sqrt n\,M}$\,, we can divide the cube $C_\ve(x) = C_\ve(z)\times
C_\ve(y)$ into $([\ve^{1-(k+\a)}] + 1)^p$ boxes $C_\delta(\widetilde z_i)
\times C_\ve(y)$, $\delta < \ve^{k+\a}$. If there is some point $(z_i,y_i)$ in 
$(C_\delta(\widetilde z_i)\times C_\ve(y)) \cap (A_i\cap X)$, then for any
point $(z_i^\prime,y_i^\prime)$ in $(C_\delta(\widetilde z_1) \times C_\ve(y))
\cap (A_i \cap X)$, we have $|F(z_i^\prime,y^\prime) - F(z_i,y_i)| \le
|F(z_i^\prime,y^\prime) - F(z_i,y^\prime)| +|F(z_i,y^\prime) - F(z_i,y_i)| \le
K^\prime\delta + |F(z_i,y_i^\prime) -
F(z_i,y_i)|$ (where $K^\prime$ is $\sqrt p$ times a Lipschitz constant of
$F\big|_V$ which we may suppose to exist) $\le K^\prime\delta +
|G(z_i,y_i^\prime)-G(z_i,y_i)|$.

Observe now that $(z_i,y_i) = (z_i, \widetilde\psi_i(p_1))$ and $(z_i^\prime,y_i^\prime) =
(z_i, \widetilde\psi_i(p_2))$, for some $p_1$, $p_2$ in $\{z_i\}\times B_i$ with $|p_1-p_2| \le |y_i-y_i'|\le 2\ve \sqrt n$. Let $\ga\colon [0,1] \to V_i\times B_i$ be a
straight path
joining $p_1$ and $p_2$. Then $G(z_i,y_i^\prime) - G(z_i,y_i) = \int_0^1
\frac{\po G}{\po y}(\widetilde\ga(t))\cdot\widetilde\ga^\prime(t)\,dt$, where
$\widetilde\ga := \psi_i\circ\ga$. We have $\frac{\po G}{\po y}(\widetilde\ga(0))=0$, so
\begin{align*}
\lV \frac{\po G}{\po y}(\widetilde\ga(t))\rV &= \lV \frac{\po G}{\po y}(\widetilde\ga(t))- \frac{\po G}{\po y}(\widetilde\ga(0)) \rV \\
&\le M|p_1-p_2|^{k+\a-1} \le M(2\ve\sqrt n)^{k+\a-1} \Rightarrow \lV \frac{\po G}{\po y}(\widetilde\ga(t))\rV |\ga'(t)| \\
&\le K''\ve^{k+\a},
\end{align*}
for some constant $K^\twoprime$. Indeed, $|\widetilde\ga^\prime(t)|$ is limited by a constant
multiple of $|p_1-p_2|\le 2\sqrt n \ve$. So
$$
|G(z_1,y_i')-G(z_i,y_i)|\le \int_0^1 \lv \frac{\po G}{\po y}(\widetilde \ga(t)) \circ \widetilde\ga(t)\rv \,\, dt \le K''\ve^{k+\a}
$$
and
$$
|F(z_i',y')-F(z_i,y_i)|\le K'\delta+ K''\ve^{k+\a}\le K_0'\cdot \ve^{k+\a},
$$
where $K_0'=K'+K''$.

Therefore, $F(C_\delta(\widetilde z_i) \times C_\ve(y))$ is
contained in some ball $B_{\delta_i}(q_i)$, with $\delta_i\le K_0'\cdot\ve^{k+\a}$, so 
\begin{align*}
\, & \sum_i \delta_i^d \le 
([\ve^{1-(k+\a)}]+1)^p(K_o^\prime\,\ve^{k+\a})^d \\
& = (K_o^\prime)^d([\ve^{1-
(k+\a)}]+1)^p(\ve^{k+\a})^{p + \frac{n-p}{k+\a}} \\
&\le \widetilde K_o\,\ve^n
\end{align*}
for some
constant $\widetilde K_o$. So, $F(C_\ve(x))$ can be covered by balls $B_{\delta_i}(q_i)$
with $\sum_i \delta_i^d \le K_o.\la(C_\ve(x))$, and by the Lemma 3.2,
remarks 3.2 and 3.4, we can conclude that $m_d(F(A_i\cap X)) \le K_o\,\la(X)$ where
\linebreak 
$K_0=2^n\widetilde K_0$,
and so we can cover $F(A_i\cap X)$ by balls $B_{\delta_i}(p_i)$ so that \linebreak
$\sum_{i=1}^\infty \delta_i^d \le K_o(\la(X)+\nu)$, as we stated.

We shall prove now that there is an $A_i^\prime \subset A_i \subset V$ with
$\la(A_i\backslash A_i^\prime) = 0$ such that \linebreak $F(C_\ve(x) \cap A_i)$ can be
covered by balls $B_{\delta_i^{(\ve)}}(W_i)$, $i \in \na$ with
$\underset{\ve\to0}{\lim}\,\frac{\sum_i(\delta_i^{(\ve)})^d}
{\la(C_\ve(x))} = 0$, $\forall x \in A_i^\prime$. This will imply our theorem,
by the Lemma 3.2, remarks 3.2 and 3.3, since we have proved above that
$m_d(F(C_\ve(x) \cap A_i) \le K_o\,\la(C_\ve(x))$, $\forall\, \ve <
\frac{1}{2\sqrt n\,M}$\,. For this, since $A_i \subset \psi_i(V_i\times B_i)$,
$B_i \subset \re^{r_i}$, $r_i \le n-p$, we may suppose $r_i = n-p$ and $\psi_i =$
identity, because $r_i < n-p \Rightarrow \la(A_i) = 0$ and we can take
$A_i^\prime = A_i$. Let us take $A_i^\prime$ equal to the set of the density
points of $A_i$. Given a point $x \in A_i^\prime$, and an $\eta^\prime > 0$, we
want to find an $\ve_o > 0$ such that $\ve < \ve_o \Rightarrow F(C_\ve(x) \cap
A_i)$ can be covered by balls $B_{\delta_i^{(\ve)}}(W_i)$, $i \in \na$ such
that $\sum_i (\delta_i^{(\ve)})^d \le \eta^\prime\,\la(C_\ve(x))$. Let
$\eta,\tilde \eta > 0$, $\tilde\ve < \frac{1}{2\sqrt n\,M}$ such that
$\frac{\la(C_\ve(x)\cap A_i)}{\la(C_\ve(x))} > 1-\tilde \eta^2$, $\forall\, \ve
\le \tilde\ve$. Divide the cube $C_\ve(x) = C_\ve(\widetilde z) \times
C_\ve(\widetilde y)$, $\ve < \tilde\ve$ into $N = ([\ve^{1-(K+\a)}
\eta^{-1}]+1)^p$ boxes $C_\delta(\widetilde z_i)\times C_\ve(\widetilde y)$,
$\delta < \eta\,\ve^{k+\a}$, $1 \le i \le N$. Then for at least
$(1-\tilde\eta)N$ values of $i$, there is a $z_i \in C_\delta(\widetilde z_i)$
such that  \linebreak $\la(\{y\in C_\ve(\widetilde y) \mid (z_i,y) \in
A_i\}/\la(C_\ve(\widetilde y)) > 1-\tilde \eta$ (here $\la$ is the Lebesgue
measure in $\re^{n-p}$), because $\la(C_\ve(x) \cap A_i) > (1-\tilde
\eta^2)\la(C_\ve(x))$. For such an $i$, take an $y_i$ such that $(z_i,y_i) \in
A_i$. Then, applying Theorem 2.1, Corollaries 2.2 and 2.3, given $\eta$ we can
choose $\tilde \eta$ so that $|F(z_i,y) - F(z_i,y_i)| < \eta\cdot\ve^{k+\a}
\Rightarrow |F(z,y)-F(z_i,y_i)| \le 2K^\prime \sqrt
n.\eta\ve^{K+\a}+\eta\ve^{K+\a} \le K^\twoprime \eta\ve^{k+\a}$, for some constant
$K^\twoprime$ and for any $z \in C_\ve(\widetilde z_i)$, where $K^\prime$ is a
Lipschitz constant for $F\Rightarrow F(C_\delta(\widetilde z_i)\times
C_\ve(\widetilde y))$ is contained in a ball \linebreak $B_{\delta_i}(Q_i)$,  with
$\sum_i \delta_i^d$ over these values of $i$ less than
$$
([\ve^{1-(K+\a)}\eta^{-1}]+1)^p(K^\twoprime \eta\ve^{k+\a})^d = 
(K^\twoprime)^d([\ve^{1-(k+\a)}.\eta^{-1}]+1)^p
(\eta\ve^{k+\a})^{p+\frac{n-p}{k+\a}} \le \widetilde K_o \eta^{\frac{n-
p}{k+\a}}.\ve^n
$$ 
for some constant $\widetilde K_o$. The union of the remaining
(at most $\tilde\eta N$) boxes has volume at most $\tilde \eta\ve^n \Rightarrow$
the union of the image of the intersection of $A_i$ with the union of these
boxes by $F$ is contained in a union of balls $B_{\delta_i}(Q_i)$ with
$\sum_i \delta_i^d \le 2K_o\tilde \eta\ve^n$, by the statement proved
before, and so $F(C_\ve(x))$ can be covered by balls $B_{\delta_i}(\widetilde
Q_i)$ with $\Sigma \delta_i^d \le (\widetilde K_o \eta^{\frac{n-p}{k+\a}} +
2K_o\tilde \eta)\ve^n$. Choosing $\eta$, $\tilde \eta$ so small that
$\widetilde K_o \eta^{\frac{n-p}{k+\a}} + 2K_o\tilde \eta \le \eta^\prime$, we
obtain the desired result with $\ve_o = \tilde \ve$. \qed
\end{pf}


\begin{rem}\label{rem3.5} In the cases of functions of class $C^k$ ($C^{k+\a}$
with $\a=0$) we have the same result. If $k \ge 2$, it follows from the case of
class $C^{k-1+(1)}$ of the theorem. If $k=1$, $p + \frac{n-p}{k+\a} = n$,
 \linebreak 
and the proof of the Theorem 3.3 shows that if $F\colon \re^n \to \re^m$ is a
function and \linebreak $C(F) = \{x \in U \mid DF(x) \,\,\text{ exists and rank }\,\,DF(x)
< n\}$ then $m_n(F(C(F))) = 0$, where $m_n$ is the Hausdorff measure of
dimension $n$.
\end{rem}

\section{Examples}

In this section we give some examples which show that the previous results are
quite sharp. In all these examples we shall use a certain kind of functions of
the real line that we shall describe below.

\begin{Def}\label{def4.1}
 Let $(\la_n)_{n\in\na}$ be a sequence of real
numbers with $0 < \la_i < \frac 12$\,, $\forall\, i\in\na$. The central Cantor
set $K_{\underline\la}$ is the Cantor set constructed as follows: We remove
from the interval [0,1] the central open interval $U_{1,1}$ of proportion $1-
2\la_1$, then we remove from the two remaining intervals the central open
intervals $U_{2,1}$ and $U_{2,2}$ of proportion $1-2\la_2$, and so on. After
the $r$-th step of the construction there will remain $2^r$ intervals of lenght
$\la_1\la_2\ldots\la_r$. The intersection of all these sets is the central
Cantor set $K_{\underline\la}$. The open intervals removed in the $r$-th step
of the construction have lenght $\la_1\la_2\cdots \la_{r-1}(1-2\la_r)$.

Let $\psi\colon \re\overset{C^\infty}{\longrightarrow}\,\re$ be a fixed
function such that $\psi(\re) \subset [0,1]$, $\psi(x)=0$, $\forall\,x \le 0$,
$\psi(x)=1$, $\forall x \ge 1$. Given two central Cantor sets $K_{\underline\la}$ and
$K_{\underline\mu}$, we construct the function $f_{\underline \la,
\underline\mu}\colon \re \to \re$ as follows: $f_{\underline \la,
\underline\mu}(x)=0$, $\forall\,x\le 0$, $f_{\underline \la,
\underline\mu}(x)=1$, $\forall x \ge 1$, and if $U_{i,j} = (a,b)$ and $v_{i,j}
= (c,d)$ are corresponding removed intervals in the constructions of
$K_{\underline\la}$ and $K_{\underline\mu}$, respectively, we define 
$f_{\underline\la,\underline\mu}(x) = c+(d-c) \psi\left(\frac{x-a}{b-
a}\right)$, $\forall\, x \in (a,b)$. We extend $f_{\underline\la,\underline\mu}$
to $K_{\underline\la}$ by continuity, obtaining
$f_{\underline\la,\underline\mu}(K_{\underline\la}) = K_{\underline\mu}$. It is
easy to check that if $g_r := \la_1\la_2\ldots\la_{r-1}(1-2\la_r)$ and $\tilde
g_r := \mu_1\mu_2\ldots\mu_{r-1}(1-2\mu_r)$ satisfy
$\underset{r\to\infty}{\lim}\,\frac{\tilde g_r}{g_r^k} = 0$ then
$f_{\underline\la,\underline\mu}$ is $C^k$ (if $k \ge 1$ is an integer).
Moreover, if $q > 1$, and $\underset{r}\to{\sup}\,\frac{\tilde g_r}{g_r^q} <
\infty$ then $f_{\underline\la,\underline\mu}$ is $C^{q-1,1}$ if $q$ is integer
and is $C^q$ (i.e., it is $C^{[q]+\{q\}}$, where $\{q\}=q-[q]\in(0,1)$)  otherwise. See [BMPV] for more details.

\end{Def}

\begin{exem}\label{exem4.1}
 Let $\la_n = \frac 12 - \frac{1}{2n}$, $\mu_n\equiv
a$. Then $\underset{n\to\infty}{\lim}\,\frac{\tilde g_n}{g_n^q} = 0$,
$\forall\, q < \frac{-\log a}{\log 2}$, and so
$f_{\underline\la,\underline\mu}$ is $C^q$, $\forall\, q < \frac{-\log a}{\log
2}$. On the other hand, $m_d(K_{\underline\mu}) = 1$ where $d = \frac{-\log
2}{\log a}$ (see [PT]). Moreover, since $a \in \left(0,\frac 12\right)$,
$\underset{n\to\infty}{\lim}\,\frac{\tilde g_n}{g_n} = 0$, and so
$f_{\underline\la,\underline\mu}^\prime(x)=0$, $\forall\, x \in
K_{\underline\la}$. If $F\colon \re^{n+p} \to \re^{n+p}$ is given by
$F(x_1,x_2,\ldots,x_n, x_{n+1},\ldots,x_{n+p}) =
(f_{\underline\la,\underline\mu}(x_1),
f_{\underline\la,\underline\mu}(x_2)\ldots,f_{\underline\la,\underline\mu}
(x_n),x_{n+1}\ldots,x_{n+p})$, then $F(C_p(F)) = F(K_{\underline\la}\times
K_{\underline\la} \times\cdots\times K_{\underline\la}\times\re^p) =
K_{\underline\mu}\times K_{\underline\mu} \times\cdots\times
K_{\underline\mu}\times \re^p$ that is a set with positive $(nd+p)$-measure.
This shows that given $q > 1$, $p > 0$ and $n > p$ there is a map $F\colon
\re^n \to \re^n$ such that $m_d(F(C_p(F))) > 0$, where $d = p + \frac{n-
p}{q}$\,, and $F$ is of class $C^{q^\prime}$ for each $q^\prime < q$.

\end{exem}

\begin{rem}\label{rem4.1}
 If $a = \frac{1}{2^n}$, $F(x_1,x_2,\ldots,x_n) =
f_{\underline\la,\underline\mu}(x_1) + 2f_{\underline\la,\underline\mu}(x_2)
+\cdots+ 2^{n-1}f_{\underline\la,\underline\mu}(x_n)$ gives an example of a
function $F\colon \re^n \to \re$ which is of class $C^q$, $\forall\, q <
n$\,\,$(q\in\re)$ such that $F(\re^n)$ contains an open set, since
$K_{\underline\mu} + 2K_{\underline\mu} +\cdots+ 2^{n-1} K_{\underline\mu} =
[0,2^n-1]$, which can be proved easily using representation in basis $2^n$.

\end{rem}

\begin{exem}\label{exem4.2}
 Let $\la_n = \frac 12 - \frac{1}{3n^2}$, $\mu_n = a -
 \frac{a}{3n}$, $a \in (0,1/2]$. Then
$\underset{n\to\infty}{\lim}\,\frac{\tilde g_n}{g_n^q} = 0$, where $q =
\frac{-\log a}{\log 2}$\,, and so $f_{\underline\la,\underline\mu}$ is $C^q$.
On the other hand we have $HD(K_{\underline\mu}) \ge \frac{-\log 2}{\log a}$\,.
Indeed, if $b < a$ and $\theta_n\equiv b$, $f_{\underline\mu,\underline\theta}$
is clearly $C^1$, and $f_{\underline\mu,\underline\theta}(K_{\underline\mu}) =
K_{\underline\theta} \Rightarrow HD(K_{\underline\mu}) \ge
HD(K_{\underline\theta}) = \frac{-\log 2}{\log b}$\,, $\forall\, b < a$. If
$F\colon \re^{n+p} \to \re^{n+p}$ is given by $F(x_1,x_2,\ldots,x_n,
x_{n+1},\ldots,x_{n+p}) =
(f_{\underline\la,\underline\mu}(x_1),\ldots,f_{\underline\la,\underline\mu}
(x_n),x_{n+1},\ldots,x_{n+p})$ then $F(C_p(F)) = K_{\underline\mu}
\times\cdots\times K_{\underline\mu} \times\re^p$, that is a set with Hausdorff
dimension $nd+p$, where $d = \frac{-\log 2}{\log a}$\,. This shows that given
$q \ge 1$, $p > 0$ and $n > p$ there is a map $F\colon \re^n \to \re^n$ such
that $HD(F(C_p(F))) = p + \frac{n-p}{q}$\,, and $F$ is of class $C^q$.
\end{exem}

\vskip .5in

\centerline{\Large\bf  References}

\begin{itemize}

\item[{[B1]}] S.M. Bates, {\it On the image size of singular maps II\/}, Duke Math.
Journal {\bf 68} (1992), 463--476.

\item[{[B2]}] S.M. Bates, {\it Toward a precise smoothness hypothesis in Sard's
theorem\/}, Proc. Amer. Math. Soc. {\bf 117} (1993), 279--283.

\item[{[BMPV]}] R. Bam\'on, C.G. Moreira, S. Plaza and J. Vera, {\it Differentiable structures of central Cantor sets\/}, Ergodic Th. Dyn. Systems {\bf 17} (1997), No. 5, 1027--1042.

\item[{[Ch1]}] P.T. Church, {\it Differentiable open maps on manifolds\/}, Trans. Amer. Math. Soc. {\bf 109} (1963), 87--100.

\item[{[Ch2]}] P.T. Church, {\it On points of Jacobian rank $k$ II\/}, Proc.
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