Proving that some measure which satisfies a certain condition is equivalently zero. Let $\Omega$ be an open set in $\mathbb{R}^n$ and $\mu$ be a positive Borel measure on $\Omega$ with $\mu(K) < \infty$ for every compact $K \subset \Omega$.
If $$\int_\Omega \phi \,d\mu = 0$$ for $\forall \phi \in C_c^\infty(\Omega)$, then $$\mu(E) = 0$$ for every Borel set $E \subset \Omega$? I'd appreicate any help!
 A: Let $E_n := E \cap K_n$, where $K_n$ is compact set such that $\bigcup K_n = \Omega$ and $K_n \subset K_{n+1}$.
Then, $\chi_{E_n} \in L^1(\mu)$ and there is a $f \in C_c(\Omega)$ such that $\Vert f -  \chi_{E_n}\Vert_1 < \epsilon$. Also, since $f \in C_c(\Omega)$, there is a $g_k \in C_c^\infty(\Omega)$ such that $g_k\to f_n$ uniformly.
Therefore, $\mu(E_n) = \int_\Omega \chi_{E_n} = 0$ and accordingly $\mu(E)$ = 0.
A: From the hypothesis it follows that $\mu$ is regular. Suppose the claim is false. Let $\epsilon>0$ and $\{K_n\}_n $ be an exhaustion of $\Omega$ by compact sets. Then, some compact set in this collection, say $K$ satisfies $\mu(E\cap K)\neq 0. $ So we may assume that $E\subseteq K$ to begin with.
Now, we may choose an open set $V$ such that $E\subseteq K\subseteq V\subseteq  \Omega$ and $\mu(V\setminus K)<\epsilon$ and such that $V$ is contained in one of the compact sets in $\{K_n\}_n $. (Indeed, let $\{U_{\alpha}\}_{\alpha\in I}$ be a cover of $K$ by open sets such that $U_{\alpha}\subseteq \Omega$. Passing to a finite subcover, we have that the union $\bigcup_{i=1}^nU_{\alpha_i}$ is a bounded open set, so $V_1:=V\cap \bigcup_{i=1}^nU_{\alpha_i}$ is a bounded open set that satisfies $\mu(V_1\setminus K)<\epsilon$).
To finish, note that Urysohn gives us a continuous $f:\Omega\to [0,1]$ such that $f(K)=1$ and $f(V^c)=0$ which we may smooth to a $C^{\infty}$ function $\phi.$  Then, $\phi$ is compactly supported in $\Omega$ and so
$\mu(E)=\mu(K)+\mu(E\setminus K)\le \int_{\Omega}\phi d\mu+\epsilon=\epsilon$ which is a contradiction.
