If $X$ is a compact metric space and $\mu$ is a Borel probability measure on $X$, then the space $C(X)$ of continuous real-valued functions on $X$ is a closed nowhere dense subset of $L^\infty(X,\mu)$, and hence bounded measurable functions are generically discontinuous. Nevertheless, Luzin's theorem says that every measurable function is in fact continuous on a set of arbitrarily large measure. This allows us to gain continuity from measurability at the cost of ignoring a small portion of $X$.

Question: Are there any analogues of Luzin's theorem that allow us to go from continuity to Hölder continuity?

A direct analogue would be a statement that given a continuous function $f\in C(X)$ and an arbitrary $\epsilon>0$, there exists a set $X_\epsilon \subset X$ such that $\mu(X_\epsilon) > 1-\epsilon$ and the restriction of $f$ to $X_\epsilon$ is Hölder continuous. (For my purposes, it would be all right if the Hölder exponent and coefficient become arbitrarily bad as $\epsilon\to 0$.)

Another possible analogue, and one that I am actually more interested in, would be a statement that given a continuous function $f\in C(X)$ and an arbitrary $\epsilon>0$, there exists a set $X_\epsilon \subset X$ such that the restriction of $f$ to $X_\epsilon$ is Hölder continuous (again with arbitrarily bad exponent and coefficient) and instead of an estimate on the measure of $X_\epsilon$, we have $$ \dim_H(X_\epsilon) > \dim_H(X) - \epsilon, $$ where $\dim_H$ is Hausdorff dimension.

Full motivation: Ideally I would like to consider the setting where $T\colon X\to X$ is a continuous map, and obtain a similar statement about the restriction of a continuous potential $f\in C(X)$ to a set of large topological pressure, $$ P_{X_\epsilon}(f) > P_X(f) - \epsilon, $$ such that $f$ restricted to $X_\epsilon$ has the Walters property, which deals with variation on Bowen balls rather than on metric balls. But the purely static version stated above for Hausdorff dimension seems like a good place to start. Does anybody know of any results in this direction? Or counterexamples showing that such a theorem can't be true in full generality?

Edit: I've accepted Anonymous's answer, which shows quite nicely that the direct analogue (using measures) fails. However, I remain very interested in the indirect analogue (using dimensions), which seems to still have a chance of holding, so any information in that direction would be welcomed.

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    $\begingroup$ Great question! Did you do mathscinet search? There's a paper "The Luzin approximation of functions from the classes $W^p_\alpha$ on metric spaces with measure." and two other papers cited in the review. I'm not sure if it's very relevant. $\endgroup$ Aug 4, 2010 at 19:33
  • $\begingroup$ The following answer to a different MO question gives some positive results for certain classes of functions: mathoverflow.net/a/296845/5701 $\endgroup$ Apr 8, 2018 at 3:23

2 Answers 2


Unfortunately, no, it is not possible to go from continuity to Hölder continuity in Luzin's theorem. At least, not in the sense of your first statement. We can give counterexamples to the following.

...given a continuous function $f\in C(X)$ and an arbitrary $\epsilon > 0$, there exists a set $X_\epsilon\subset X$ such that $\mu(X_\epsilon) > 1−\epsilon$ and the restriction of $f$ to $X_\epsilon$ is Hölder continuous.

We can't even make the restriction of f to $X_\epsilon$ satisfy any given modulus of continuity. Letting $\omega\colon[0,1]\to\mathbb{R}^+$ be continuous and strictly increasing with $\omega(0)=0$, there exist continuous functions $f\colon[0,1]\to\mathbb{R}$ such that $\vert f(x)-f(y)\vert/\omega(\vert x-y\vert)$ is unbounded over $x\not=y$ belonging to any set $S\subseteq[0,1]$ of Lebesgue measure greater than 1/2. Taking, e.g., $\omega(x)=e^{-\vert\log x\vert^{1/2}}=x^{\vert\log x\vert^{-1/2}}$ will show that f is not Hölder continuous on any set of Lebesgue measure greater than 1/2.

The idea is to construct a continuous function $f\colon[0,1]\to\mathbb{R}$ and a sequence of positive real numbers $\epsilon_n\to0$ such that $\vert f(x+\epsilon_n)-f(x)\vert/\omega(\epsilon_n)\ge n$ on a subset of $[0,1-\epsilon_n]$ of measure at least $1-2\epsilon_n$.

We can construct this by applying the Baire category theorem to the complete metric space $C$ of continuous functions $[0,1]\to\mathbb{R}$ under the supremum norm. For any $f\in C$ and $K,\epsilon>0$, let $S(f,K,\epsilon)$ denote the set of $x\in[0,1-\epsilon]$ such that $\vert f(x+\epsilon)-f(x)\vert > K\omega(\epsilon)$. Then, $$ A(K,\epsilon)=\left\{f\in C\colon\mu\left(S(f,K,\delta)\right)>1-2\delta{\rm\ some\ }0 < \delta < \epsilon\right\} $$ is an open subset of $C$. It is also dense. To see this, first choose a continuously differentiable $f\in C$ and set $g(x)=f(x)+1_{\{\lfloor x/\delta\rfloor{\rm\ is\ even}\}}K(\omega(\delta)+\sqrt{\delta})$. Choosing $\delta$ small enough, the inequality $\vert g(x+\delta)-g(x)\vert > K\omega(\delta)$ will hold on $[0,1-\delta]$, and there will then be a continuous function $\tilde g$ equal to g outside a set of measure $\delta$ and satisfying $\Vert\tilde g-f\Vert\le K(\omega(\delta)+\sqrt{\delta})$. So, choosing $\delta$ small enough, we have $\tilde g\in A(K,\epsilon)$ and $\tilde g$ as close to $f$ as we like, showing that $A(K,\epsilon)$ is indeed dense in $C$.

The Baire category theorem says that $$ A=\bigcap_{n=1}^\infty A(n,1/n) $$ is nonempty, and any $f\in A$ satisfies the requirements mentioned above.

Now, suppose that $S\subseteq[0,1]$ has measure greater than 1/2. Choosing a random variable X uniformly in $[0,1]$, $$ \begin{align} &\mathbb{P}(X,X+\epsilon_n\in S)\ge\mathbb{P}(X\in S)+\mathbb{P}(X+\epsilon_n\in S)-1\ge 2\mu(S)-1-\epsilon_n,\\\\ &\mathbb{P}(X\in S(f,n,\epsilon_n))\ge 1 - 2\epsilon_n. \end{align} $$ This gives $X,X+\epsilon_n\in S$ and, simultaneously, $X\in S(f,n,\epsilon_n)$ with probability at least $2\mu(S)-1-3\epsilon_n$. For large enough n, this is positive. Therefore, there exist $x,x+\epsilon_n\in S$ with $\vert f(x+\epsilon_n)-f(x)\vert/\omega(\epsilon_n) > n$ for arbitrarily large n.

  • $\begingroup$ Very nice argument! I'm still quite curious about the other version of the question (whether you can gain Holder continuity on a set of large Hausdorff dimension), but this puts to rest any hope of a measure-theoretic analogue. $\endgroup$ Nov 9, 2010 at 19:48
  • $\begingroup$ Vaughn - subsets of the unit interval with Hausdorff dimension greater than $1-\epsilon$ can still be very small (for a suitable meaning of "small"), so it does seem reasonable that your question is true in that case. I'm not sure about a proof though. $\endgroup$
    – Anonymous
    Nov 9, 2010 at 20:18

I'm posting some references given to me by Jon Chaika as a CW answer, which addresses the question of whether we can get a set $X_\alpha$ such that the restriction to $X_\alpha$ is Hölder continuous. It seems this question was posed by Márton Elekes in a 2004/5 paper, where a partial answer was given: for every $\alpha\in (0,1]$, a typical continuous function $[0,1]\to\mathbb{R}$ is not $\alpha$-Hölder on any set of Hausdorff dimension greater than $1-\alpha$. The converse direction was provided in the 2009 PhD thesis of András Máthé, and appeared in a 2013 paper: for every $\alpha$ and every Borel measurable $f\colon [0,1]\to \mathbb{R}$, there is a set $X_\alpha\subset [0,1]$ with Hausdorff dimension $1-\alpha$ such that $f|_{X_\alpha}$ is $\alpha$-Hölder.


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