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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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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. –  Andrey Gogolev Aug 4 '10 at 19:33
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up vote 11 down vote accepted

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.

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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. –  Vaughn Climenhaga Nov 9 '10 at 19:48
    
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. –  Anonymous Nov 9 '10 at 20:18
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