Assume that every point in $G$ has a neighborhood on which $f$ vanishes. Prove that $f$ vanishes on $G$

Question

Let $f$ be a generalized function on $\Omega$. Let $G$ be an open subset of $\Omega$. Assume that every point in $G$ has a neighborhood on which $f$ vanishes. Prove that $f$ vanishes on $G$

Let $\phi \in D(G)$. We show that $(f,\phi)=0$. Let $\text{Supp $\phi$} \subset K \subset G$, where $K$ is a compact set. For each $k \in K$ , there exists a $U_{k}$ on which $f$ vanishes and $\bar{U_k} \subset G$. Since $K$ is compact , $K \subset \cup_{i=1}^{n}U_{k_i}$. Then $G=U_{k_1}\cup U_{k_2}\cup..\cup U_{k_n} \cup (G-K)$

By Partition of Unity, there is a set of test functions $\phi_i$ such that

1. $ 0 \le \phi_i \le 1$

2. Supp $\phi_i \subset U_{k_i}$

3. $\sum_{i} \phi_i(x)=1$

Then $(f,\phi)=\sum_{i=1}^{n}(f,\phi \phi_i)+(f,\phi \phi_{n+1})=0$

The first indexed sum is $0$ since $f$ vanishes on each such $U_{k_i}$ and the last term is $0$ since supp$\phi \subset K$.

I have just one doubt. I need the closure of $G-K$ to be contained in $G$. I am not so sure about it. How do I make an adjustment to it??

Thanks for the help!!