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.