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.) 



smooth Luzin theorem
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](http://books.google.com/books?id=EzkG1lf2G8kC&pg=PA53&lpg=PA53&dq=approximately%20differentiable%20federer%27s%20theorem&source=bl&ots=UVv6s75W9B&sig=8nB2Y7Rio2XBJPuurVaC_EDyB84&hl=en&sa=X&ei=FH5pU_OFMoaBogTh_IKYBQ&ved=0CDsQ6AEwAw#v=onepage&q=approximately%20differentiable%20federer%27s%20theorem&f=false).


The converse is obvious: If $f=g\in C^1$ on $A$, then $f$ is approximately differentiable at all points of density of $A$.


