Consider the elliptic operator $$Lu=-\sum_{i,j} (a^{ij}u_{x_i})_{x_j},$$ where $a^{ij}\in L^\infty(\Omega)$ and $a^{ij}(x)\xi_i\xi_j\geq\lambda |\xi|^2$ for all $\xi\in\mathbb{R}^n$, $x\in\Omega$.
Definition. A function $u\in H^1(\Omega)$ is said to be a weak subsolution if $$\int_\Omega\sum _{i,j}a^{ij}u_{x_i}\phi_{x_j} \leq 0, \quad\forall\ \phi\in C_c^\infty(\Omega),\;\phi\geq 0.$$
Page 9 of this book says that, if $u$ is a weak subsolution, then the inequality above is also valid for any $\phi\in H_0^1(\Omega),\ \phi\geq 0$.
This is, probably, an elementary fact about density, but how can we prove it?
If $\phi\in H_0^1(\Omega)$ and $\phi\geq 0$, then there is a sequence $(\phi^n)$ in $C_0^\infty(\Omega)$ such that $\phi^n\geq 0$ and $\phi^n\to\phi$ in $H_0^1(\Omega)$, right? So,
$$\int_\Omega\sum _{i,j}a^{ij}u_{x_i}\phi^n_{x_j} \leq 0, \quad\forall\ n\in\mathbb{N}.$$
How can we justify the passage to the limit?
As a function of $\phi$, your last displayed equation's LHS is a bounded linear functional, so the weak convergence in the $H^1$ norm is enough to pass to limit!!