Equivalence of antiderivative in L1 sense and in the usual sense We say that$\ f$ is differentiable w.r.t to $L_1$ if there exists a$\ g$ such that:
$$
\lim_{h\to 0}\left\Vert\frac{f(x+h)-f(x)}{h} - g(x)\right\Vert_1 = 0 
$$
where $\Vert \cdot \Vert_1$ is the $L_1$ norm. Since $f$ is in $L_1$, the corresponding$\ g$ must be in$\ L_1$ too, and so by Lebesgue, it has an antiderivative $G$ which is differentiable a.e, with $G'(x)=g(x)$.
My question is: does $f=G$ a.e? 
Here is my line of thought: if $G$ is in $L_1$, it can be shown that 
$$
\hat{g}{(t)} = 2\pi it\hat{G}{(t)} =  2\pi it\hat{f}{(t)},
$$
which then implies that $f=G$ a.e. and so, in order to show that $f=G$ a.e, it is enough to show that$\ G$ is in$\ L_1$, and that's where i got stuck.
 A: Most antiderivative of $g$ are not in $L^1(\mathbb{R})$, in your case only one antiderivative $G_0$ will be in $L^1(\mathbb{R})$, the one actually equal to $f$. All the other antiderivatives $G$ are equal to $G_0 + c$, with $c \neq 0$, which is not in $L^1(\mathbb{R})$.
To proof that there exist one antiderivative $G_0$ in $L^1(\mathbb{R})$.
You start by noticing that your $L^1(\mathbb{R})$ differentiability imply differentiability in the distributional sens so, for $\phi \in \mathcal{C}_{comp}^\infty(\mathbb{R})$, we have
$$
\langle f',\phi \rangle = \langle g,\phi \rangle. \qquad (1)
$$
Fix $G$ an antiderivative of $g$, you have $G' = g$ in the distributional sens.
The equation $(1)$ become
$$
\langle f',\phi \rangle = \langle G',\phi \rangle \implies \langle (f-G)',\phi \rangle = 0
$$
and than imply $f-G = c$ a constant.
Choosing $G_0 = G + c$, we have $G_0 = f \in L^1(\mathbb{R})$.
You can further show that there exists a constant $c_0$ such that
$$
G_0(x) = c_0 + \int_0^x g(y)\, dy.
$$
