Are weak derivatives and distributional derivatives different?

Given a real function $$f\in L^1_{\text{loc}}(\Omega)$$, we define both weak or distributional derivatives by $$\int f'\phi = - \int f \phi'$$ for all test functions $$\phi$$.

Now, take $$\Omega = (-1,1)$$, and $$f(x) = I_{x>0}$$, an indicator function. Then, according to Example 2 of Section 5.2 of Evans' PDE book, there is no weak derivatives. But, it is well known that $$f' = \delta_0$$ as a distribution. In fact, every distribution has its derivative according to Rudin's book on Functional Analysis, see Section 6.1.

At this far, can anybody clarify the following questions?

1. weak derivatives is stronger than distributional derivatives? If yes, how strong?
2. Is $$\delta_0$$ a $$L^1_{loc}(-1,1)$$, a locally integrable function? see also this question.
3. Most PDE books use weak derivatives, not distributional one?

(1) One usually wants to be weak derivatives to be functions, that is distributions represented by $\def\loc{\mathrm{loc}}$$L^1_\loc-functions, that is we say that f is weakly differentiable iff there is an f' \in L^1_\loc such that$$ \int_\Omega f\phi' = -\int_\Omega f'\phi\quad \phi\in C_c^\infty(\Omega) $$Now some functions, such as I_{x>0} don't have a weak derivative. As every distribution T \in \mathcal D'(\Omega) has derivatives of any order given by T^{(n)}(\phi) = (-1)^n T(\phi^{(n)}) in this case, the distributional derivatives cannot be represented by functions, see (2). (2) No. \delta cannot be represented by a function g \in L^1_\loc(\Omega). To see this, suppose g \in L^1_\loc(\Omega) were representing \delta, that is$$ \int_{-1}^1 g\phi = \delta(\phi) = \phi(0) $$for all \phi \in C^\infty_c(-1,1). Let \phi_n be a sequence in C^\infty_c such that 0\le \phi_n \le 1, \mathop{\rm supp} \phi_n \subseteq[-\frac 1n, \frac 1n], \phi_n(0) = 1. Then \phi_n \to 0 almost everywhere, hence$$ 1 = \phi_n(0) = \int_{-1}^1 \phi_n g \to 0$\$ contradiction.