Let $v \in L^¹(a,b)$ and define the function $$u(x):=\int_{x_0}^xv(\xi)\ d\xi, \ \ x \in [a,b], x_0 \in [a,b]$$
How can I show that $v$ is the generalized derivative of $u$ on $[a,b]$ without making use of the fundamental theorem of calculus on $u$? I appreciate any hint.
This is not in general true. By integration by parts, you can show that $v$ is a weak derivative of $u$. If $u$ has a classical derivative, then $v$ must be that derivative. However, there's no guarantee that $u$ here has a classical derivative.
Integration by parts: $$ \int_a^b f(x)g(x)dx = \left(\int^x f(x) g(\xi)d\xi\right)\biggr\rvert_a^b - \int_a^b\int^x f'(x) g(\xi)d\xi dx $$
Let $f\in C^\infty([a,b])$ such that $f(a)=f(b)=0$ and let $g(x) = v(x)$, and you will recover the definition of weak derivative.