# smooth Luzin theorem

For measurable functions $f(x)$, $g(x)$ on $[0,1]$ define the distance $\rho(f,g)$ as a Lebesgue measure of the set $\{x:f(x)\ne g(x)\}$. Then Luzin's famous theorem states that $C[0,1]$ is dense with respect to this metric in the set of all measurable functions.

The question is to describe the completion of $C^1[0,1]$ (that is, informally, for given function give a receipt how to recognize wether it belongs to completion or not.)

• Related: mathoverflow.net/q/34518/5701 and the answer which shows that whatever the completion is, it is not all of $C[0,1]$. – Vaughn Climenhaga May 6 '14 at 14:58
• Any such function must have an "$L^0$ derivative", in the sense that $\frac{1}{t}(f(\cdot + t) - f(\cdot))$ has a limit in measure as $t \to 0$. – Alexander Shamov May 6 '14 at 15:03

This holds precisely if $f$ is approximately differentiable a.e. This condition implies that $f|_A$ can be extended to a $C^1$ function for suitable sets $A$ of almost full measure by a Theorem from Federer's book, or see here.
The converse is obvious: If $f=g\in C^1$ on $A$, then $f$ is approximately differentiable at all points of density of $A$.