Maximal Hausdorff dimension of the set on which derivatives do not agree Let $f, g: [0, 1] \to \mathbb R$ be functions that are differentiable a.e. with $f’ = g’$ almost everywhere. What is the maximal Hausdorff dimension $d$ (and corresponding Hausdorff $d$-measure) of the set on which $f$ and $g$ are both differentiable but $f’ \neq g’$?
 A: If by not agree you include that the derivative may not exist, you can get any dimension and measure that does not contradict the almost everywhere. Consider the worst case $d=1$:
Take the construction of the Cantor-set. If in the $n$-th step you remove the middle $\frac{1}{n+1}$-th part of each interval, you get a set $C$ of Hausdorff-dimension $1$ but with $\mathcal{H}^1(C) =0$. So if you construct the Cantor-function using the same intervals, you get a function $f$ with $f'=0$ a.e, so for $g(x):=0$ you get $f'=g'$ a.e. except on $C$.
A: If the derivatives exist for $x \in [0,1],$ then $f$ and $g$ must be continuous on some open set $A \subset [0,1].$ If $f \neq g$ on $A$ but $f' = g'$ a.e on $A$, then by integrating $f_1 = g_1 + C$, for some constant $C$ locally on open set $A$. However, because all the functions are continuous (locally on A) $f = g + C$ everywhere on $A$. Moreover, it's enough to consider the problem locally, if the derivatives exist. So the answer is $0,$ if the derivatives exist.
EDIT:(I now try to use FTC locally)
