I have been asked to show that the Heaviside function $H := \chi_{[0,+ \infty)}$ does not admit weak derivative in $L^1_{loc}(\mathbb{R})$. Here's my reasoning:
By definition the weak derivative of $H$ is a function $w$ s.t. $\forall \varphi \in C^\infty_c(\mathbb{R})$
$$ \int_{\mathbb{R}}\varphi w dx = - \int_{\mathbb{R}}\varphi' H dx = -\lim_{R\to \infty} \int_0^R \varphi'dx = \varphi(0) = \int_{\mathbb{R}} \varphi \delta_0$$
where $\delta_0$ is the Dirac's delta centered in $0$.
Thus the only thing to show is that
$$ \exists w \in L^1_{loc}(\mathbb{R}) :\forall \varphi \in C^\infty_c(\mathbb{R}) \int_{\mathbb{R}}\varphi w dx = \int_{\mathbb{R}}\varphi \delta_0 \Longrightarrow wdx = \delta_0$$
i.e. $\delta_0$ admits Radon-Nykodim derivative w.r.t. the Lebesgue measure (which we know is not the case) so to conclude by contraposition.
Let $\mathcal{L}_w^1 := w dx$ be the measure associated to $w$. The idea is that if, for every finite measure borelian $E$ we have
$$ \mathcal{L}_w^1(E) = \delta_0(E)$$
then we are done, as this would imply $\mathcal{L}_w^1 = \delta_0$.
Let $(\rho_h)_h$ be a sequence of mollifiers and define $u_h := \rho_h * \chi_E$. By the properties of mollifiers we have
$$ u_h \xrightarrow{L^1} \chi_E \\ u_h \in C^\infty_c(\mathbb{R})$$
thus
$$ \mathcal{L}_w^1(E) = \int_\mathbb{R} \chi_E ~ \mathcal{L}_w^1 = \lim_{h \to \infty} \int_\mathbb{R} u_h ~ \mathcal{L}_w^1 = \lim_{h \to \infty} \int_\mathbb{R} u_h ~ \delta_0$$
If I could bring the limit inside the integral than I would have done, but since the convergency is only in $L^1$ I shouldn't be able to apply dominated convergency. How can I conclude the proof?
You really only need to consider your first equation in order to obtain a contradiction. Suppose that there exists $w \in L^{1}_{\mbox{loc}}(\mathbb{R})$ such that $$ \int w\varphi dx = \varphi(0),\;\;\;\varphi \in \mathcal{C}_{c}^{\infty}(\mathbb{R}). $$ That equation is inconsistent. To see why, find $\varphi\in\mathcal{C}_{c}^{\infty}(\mathbb{R})$ for which $\varphi(0)=1$. Then $\varphi_{\epsilon}(x)=\varphi(x/\epsilon)$ is $1$ at $0$ for all $\epsilon > 0$, remains uniformly bounded as $\epsilon\downarrow 0$ and converges pointwise everywhere to $0$ except at $0$. So you can apply the Lebesgue dominated convergence theorem to $(w\varphi_{\epsilon})$ on some finite interval in order to conclude that $$ 0=\lim_{\epsilon\downarrow 0}\int w\varphi_{\epsilon}dx = \lim_{\epsilon\downarrow 0}\varphi_{\epsilon}(0)=1. $$