For which functions $f$ is $f^{(0)}(x) = f(x)$?

Question

Classically, for $0<\alpha<1$ the fractional derivative of order $\alpha$ is given by: $$ D^{\alpha}f(x) = \frac{1}{\Gamma(1-\alpha)}\frac{d}{dx}\int_0^x \frac{f(t)}{(x-t)^{\alpha}}\,dt$$ When $\alpha=0$ we can extend this to get $$ D^0f(x) = \frac{d}{dx}\int_0^x f(t)\,dt $$For continuous $f$, this reduces to $f(x)$ by the Fundamental Theorem of Calculus. However, this expression is undefined for functions like $\mathbf{1}_{\mathbb{Q}}$ or non-integrable functions. This feels weird: even if a function $f$ isn't differentiable, taking zero derivatives feels like we aren't 'doing anything' to $f$ and so I would expect to get $D^0$ to act as an identity, as in $ D^0 f(x)=f(x)$ for any $f$. However, the formula would seem to suggest that only integrable $f$ satisfy $f^{(0)}=f$. Can these views be reconciled?