*Note: Here $\mu$ denotes Lebesgue measure on $\mathbb R$.*


We say a function $f: \mathbb R \to \mathbb R$ is *uniformly Lebesgue differentiable* if there exists some measurable subset $E$ of $\mathbb R$ with $\mu(E^c) = 0$ such that the following holds:



> 
> For all $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x \in E$ and all $0 < r \leq \delta$, we have
> $$\frac{1}{2r} \int\_{B\_r(x)} |f(y) - f(x)| \, dy < \varepsilon.$$
> 
> 
> 


**Question:** Is it true that $f$ admits a uniformly continuous representative if and only if it is uniformly Lebesgue differentiable?


*Remark: The “only if” direction is clear, the “if” direction is the issue.*



Uniformly Lebesgue differentiable functions
Yes. Suppose $f$ is uniformly Lebesgue differentiable (ULD).


Let's first note that $f$ at least has a *continuous* representative, since the functions $f\_r(x) = \frac{1}{2r} \int\_{B\_r(x)} f(y)\,dy$ are each continuous (by dominated convergence if you like), and by the triangle inequality and the ULD condition, they are uniformly Cauchy and converge a.e. to $f$. Therefore we can replace $f$ by its continuous representative, disregard the null set $E^c$, and assume that the ULD condition holds for all $x$.


Now the idea is that the ULD condition implies that $f(y)$ must be close to $f(x)$ for "most" $y \in B\_r(x)$. If $x\_1, x\_2$ are sufficiently close, then $B\_r(x\_1), B\_r(x\_2)$ have a lot of overlap, and so "most" of the $y$ in their intersection have $f(y)$ close to both $f(x\_1)$ and $f(x\_2)$, so they are close to each other.


To make this precise, let $c$ be a small constant ~~to be chosen later~~ less than $\frac{1}{8}$. Now fix $\epsilon > 0$, and fix an $r$ so small that $\frac{1}{2r} \int\_{B\_r(x)} |f(y)-f(x)| \,dy < c \epsilon$ for all $x$. For arbitrary $x$, let
$$
A(x) = \left\{ y \in B\_r(x) : |f(y) - f(x)| \ge \frac{\epsilon}{2}\right\}
$$
then by Markov's inequality
$$
m(A(x)) \le \frac{2}{\epsilon} \int\_{B\_r(x)} |f(y) - f(x)|\,dy < \frac{2}{\epsilon} \cdot 2r \cdot c \epsilon = 4 c r.$$


Now suppose $|x\_1 -x\_2| < r$. Then we have $m(B\_r(x\_1) \cap B\_r(x\_2)) > (2-|x\_1-x\_2|)r > r$. On the other hand, $m(A(x\_1) \cup A(x\_2)) \le 8cr$. So ~~if we choose~~ since we chose $c < \frac{1}{8}$, then we have
$$m\left[(B\_r(x\_1) \cap B\_r(x\_2)) \setminus (A(x\_1) \cup A(x\_2))\right] \ge r - 8cr > 0$$
and hence this set is nonempty, so contains some $y$. We have $y \in B\_r(x\_1) \setminus A(x\_1)$ which means that $|f(y) - f(x\_1)| < \frac{\epsilon}{2}$, and similarly $|f(y) - f(x\_2)| < \frac{\epsilon}{2}$, so that $|f(x\_1) - f(x\_2)| < \epsilon$.


