Consider a continuous function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ and a null set $A \subset \mathbb{R}^2$. We assume that $$\forall (x,y) \in A^2, \ \ |f(y)-f(x)| \le ||y-x||^2$$ In addition, we suppose that $f(A)$ is measurable. Is $f(A)$ also a null set (in $\mathbb{R}$) ?

It is well known that not every continuous function maps null sets to null sets : the devil staircase does not have Lusin N property. But what happens when we assume a Hölder condition of order 2 ?

If we also assume that the Hausdorff dimension of A is $\dim _H (A) < 2$, then classically, as $f$ is Hölder continuous of order 2 on $A$, we have $\dim_H \big( f(A) \big) \le \frac{1}{2} \dim_H(A)$. So $\dim_H \big( f(A) \big) < 1$. All measurable subsets of the line with positive measure have Hausdorff dimension 1, and we assumed that $f(A)$ is measurable, thus $f(A)$ is a null set.

However, (see for instance Measure 0 sets on the line with Hausdorff dimension 1, on mathoverflow, or Are there sets of zero measure and full...),

There exists null sets $A \subset \mathbb{R}^2$ with Hausdorff dimension $\dim_H(A) = 2$.

Is it possible to give an answer in the general case $\mu (A) = 0$ ?

Subsidiary question : is it really necessary to assume that $f(A)$ is measurable ? Is it not implied by the previous hypothesis ?

Edit : Note that $f$ is supposed Holder continuous of order $2$ only on $A$, so this condition does not a priori prevent $f$ from being constant on no open set.


Yes, $f(A)$ is null. By definition of Hausdorff measure and the coincidence of Lebesgue and Hausdorff measure on measurable sets of $\mathbb{R}^n$, since $A$ is null then for every $\delta>0$ we have $$H^2_\delta(A):=\inf\left\{\sum_i diam(E_i)^2:A \subseteq \bigcup E_i,\, diam(E_i)\leq\delta\right\}=0.$$ Now if $\{E_i\}$ covers $A$, then $\{f(E_i)\}$ covers $f(A)$, and moreover $diam(f(E_i))\leq diam(E_i)^2$ by the Hölder condition. This implies that $$H^1_{\delta^2}(f(A))\leq \inf\left\{\sum_i diam(f(E_i)):A\subseteq \bigcup E_i,\,diam(E_i)\leq \delta\right\}\leq H^2_\delta (A)=0.$$ Now taking the supremum as $\delta\to 0$ we obtain $H^1(f(A))=0$.

(More in general $H^s(f(A))\leq C(|f|_{C^{0,\alpha}},\alpha)H^{s\alpha}(A)$ for a $\alpha$-Hölder continuous function $f$, which is slightly stronger than saying that the dimension can increase at most by a factor $\frac{1}{\alpha}$.)

Measurability follows from the fact that $f(A)$ is Hausdorff null, and Hausdorff measure coincides with Lebesgue outer measure. You can have a look at this other question for a related result: Image of $A\subset \mathbb R^d$ under a Lipschitz function is $H^d$ measurable.

  • $\begingroup$ Thank you for your answer ! $\endgroup$ – charmd Jan 7 '17 at 16:21
  • $\begingroup$ @charMD You're welcome. $\endgroup$ – Del Jan 10 '17 at 21:07

If $c>1$ and $f: \mathbb{R}^k \to \mathbb{R}$ satisfy condition $|f(u) -f(v) |\leq |u-v|^c $ then we have $$\left|\frac{f(x+h) -f(x) -0h }{|h|}\right| \leq |h|^{c-1}$$ letting $h\to 0$ we obtain that $f$ is differentiable on whole $\mathbb{R}^k$ and its derivative is equal to $0.$ Thus $f$ is constant on whole $\mathbb{R}^k$. So image of any set by $f $ is one point set and hence measure zero.

  • $\begingroup$ No, $A$ might be everywhere disconnected, this does not hold. The Holder condition is assumed only on $A$. $\endgroup$ – charmd Jan 5 '17 at 20:55
  • $\begingroup$ Holder continuous functions of order $>1$ are locally constant. With a really simple example, if I take $A = \{ (0,0), (1,0),(0,-1)\}$, $f:(x,y)\mapsto x$ satisfies the hypothesis, without being constant $\endgroup$ – charmd Jan 5 '17 at 21:02
  • $\begingroup$ Of course you're right. I've made mistake. $\endgroup$ – MotylaNogaTomkaMazura Jan 6 '17 at 10:18

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