Let $D \subset \mathbb R^N$ be an open set and $u \in H_0^1(D)$ such that $- \Delta u \geq 0$ (as a distribution). Then $- \Delta u$ defines a Radon measure. If $w \in C_c^\infty(D)$, one can write $$ \int_D w \ d(- \Delta u) = \langle w, - \Delta u \rangle_{C_c^\infty \times \mathcal D'} = \int_D \nabla w \nabla u. \qquad (*) $$
So far, so good. Now, my book (Shape Variation and Optmization, by A. Henrot and M. Pierre) says the following:
If $w \in H_0^1(D)$ is continuous and has compact support, one may approximate it by convolution in $H^1$ and uniformly by functions in $C_c^\infty(D)$ and pass to the limit in $(*)$ to obtain $$ \int_D w \ d(-\Delta u) = \int_D \nabla w \nabla u \leq \|w\|_{H^1} \|u\|_{H^1}. $$
What I understand:
Using the uniform convergence, one shows that $$ \exists \ \int_D w \ d(- \Delta u) = \lim_n \int_D \varphi_n \ d(-\Delta u), $$ by the Dominated Convergence Theorem.
By the $H^1$ convergence, there is a $C^\infty$ sequence $w^\varepsilon$ such that $$ \int_D w^\varepsilon \ d(- \Delta u) = \int_D \nabla w^\varepsilon \nabla u \to \int_D \nabla w \nabla u \leq \|w\|_{H^1} \|u\|_{H^1}. $$
My doubt is, how to put this information together to prove the claim? Or am I making some mistake?
Moreover, why does it hold that $$ \int_K d(- \Delta u) = \int_K - \Delta u $$ on a compact set $K$?
Thanks in advance.