For $f: \mathbb R \to \mathbb R$ a measurable function, we say $g: \mathbb R \to \mathbb R$ is a *modification* of $f$ if $f = g$ a.e.


Suppose $f$ Is a measurable function that is differentiable a.e.


We say that a modification $g$ of $f$ is *maximally differentiable* if whenever $h$ is another modification of $f$, we have that for all $x \in \mathbb R$, if $h$ is differentiable at $x$, then so is $g$.


**Question:** Does every measurable, differentiable a.e. function $f$ admit a modification that is maximally differentiable?



Does every differentiable a.e. function admit a maximally differentiable representative?
$\DeclareMathOperator\*\appliminf{app-liminf}\DeclareMathOperator\*\applimsup{app-limsup}\DeclareMathOperator\*\applim{app-lim}\DeclareMathOperator\*\essliminf{ess liminf}\DeclareMathOperator\*\esslimsup{ess limsup}$The answer is indeed yes. Further, the assumption that $f$ be differentiable a.e. is unnecessary. The main idea for this solution was suggested by Sam Forster on another website.


Define the function $g$ by


$$g(x) := \frac{1}{2} \bigl (\appliminf\_{y \to x} f(y) + \applimsup\_{y \to x} f(y) \bigr),$$


whenever the above two limits are finite, and $g(x) = f(x)$ otherwise, where here $\appliminf$ and $\applimsup$ denote the [approximate limit supremum and limit infimum](https://encyclopediaofmath.org/wiki/Approximate_limit).


We note that $g$ has the following properties:


1. $g(x) = \applim\_{y \to x} f(y)$ wherever the limit exists. In particular the limit exists wherever $f$ is differentiable and equals $f(x)$.
2. $g$ depends only on the equivalence class of $f$ modulo null sets.


Now I claim that this $g$ is maximally differentiable. To see this, let $h$ be another representative of $f$ and assume $h$ is differentiable at $x$. Then for every $\varepsilon > 0$, there exists $\delta > 0$, and $L \in \mathbb R$ such that


$$\left\lvert\frac{h(y) - h(x)}{y - x} - L\right\rvert \leq \varepsilon$$


whenever $\lvert y-x\rvert < \delta$.


I claim that the same holds for $g$, thus $g$ is differentiable as well at $x$. To see this, note that by (2) above, we may write


$$g(x) := \frac{1}{2} \bigl (\appliminf\_{y \to x} h(y) + \applimsup\_{y \to x} h(y)\bigr ).$$


Next, we note that for all $y$ with $\lvert y - x\rvert < \delta$,


$$\frac{\essliminf\_{z \to y} h(z) - h(x)}{y - x} \leq \frac{g(y) - g(x)}{y - x} \leq \frac{\esslimsup\_{z \to y} h(z) - h(x)}{y - x}$$


where we have applied (1) to write $g(x) = h(x)$. This implies


\begin{gather\*}
\frac{(L - \varepsilon)(y-x)}{y - x} \leq \frac{g(y) - g(x)}{y - x} \leq \frac{(L + \varepsilon)(y -x)}{y - x} \\
L -\varepsilon \leq \frac{g(y) - g(x)}{y - x} \leq L + \varepsilon.
\end{gather\*}


Since $\varepsilon$ was arbitrary, we conclude that $\lim\_{y \to x} \frac{g(y) - g(x)}{y - x} = L$ and so $g$ is differentiable at $x$ as claimed.


