# function a.e. differentiable and it's weak derivative

Note - I am just starting to learn about theory of distributions, so this may be a trivial question, if so I'd be grateful for a reference, nevertheless the question is the following: suppose I have a function $f$ such that $f \in L^1$ and $f$ is differentiable almost everywhere (in the strong sense) and I'm trying to find whether or not it's weak derivative can be represented by a function $g \in L^1_{\operatorname{loc}}$.

Is it obvious that if such $g$ exists then we must have $g = f'$ a.e.? of course I know it doesn't have to be true that if $f'$ exists a.e. then it's the weak derivative of $f$ since the weak derivative can be for example $\delta_0$ but if all I'm interested in is whether or not the weak derivative of $f$ can be represented as an integration (with respect to the Lebesgue measure) against some function then is it enough to just check whether or not $f'$ works?

Useful fact: a function with locally integrable weak derivative (i.e., a function of Sobolev class $W^{1,1}_{\rm loc}$) is approximately differentiable almost everywhere, and the approximate derivative agrees with the weak derivative. This is stated on page 8 here with a reference to the book by L. Ambrosio, N. Fusco, D.Pallara, Functions of bounded variation and free discontinuity problems. Another reference is the book by L.C. Evans and R.F. Gariepy Measure theory and fine properties of functions.
If a function has strong (pointwise) derivative at $x$, then this derivative is also the approximate derivative. Hence, if a function is $W^{1,1}_{\rm loc}$ and differentiable a.e., then its weak derivative is represented by the pointwise derivative.
• so we require $f$ to be integrable in the compact sets...this is the best that can be done? to deduce that there is a representative of the generalized derivative we assume that there is a representative in each compact set? – Javier Apr 22 '18 at 14:47