I got a function $f:[a,b] \to \mathbb{R}$ where $f \in L^1(a,b)$. It is differentiable a.e. with derivative $f' \in L^1(a,b)$.
How can I show that $f$ has a weak derivative?
I know that $\int f\varphi'$ and $-\int f' \varphi$ both make sense but why they are equal?
In general, what you are asking is wrong. To see this, recall that (in dimension 1) every function with a weak derivative is (locally absolutely) continuous (has such a representative).
But if we take e.g. $f = \chi_{[0,2]}$, then $f$ is differentiable a.e., but has no continuous version, hence has no weak derivative.
You can even see this explicitly: the derivative if $f$ is $f' =0$ a.e., but
$$ \int f \cdot \varphi' dx = \int_0^2 \varphi' dx = \varphi(2) - \varphi(0) \neq 0= \int f' \cdot \varphi dx $$ for suitable $\varphi \in C_c^\infty$.
EDIT: the claim also remains false if $f$ is assumed to be continuous. A counter example is provided by the Cantor function (http://en.m.wikipedia.org/wiki/Cantor_function), which is an increasing, continuous function that is differentiable a.e. (with derivative $0$), but not absolutely continuous; hence, it has no weak derivative.