Zero integration for any function in $C_c(X)$ would imply that the measure is zero?

Question

Let $X$ be any locally compact Hausdorff space, and $\mu$ be a positive finite regular (both inner and outer) measure on $X$ such that $\operatorname{supp}(\mu)$ is compact. We have that $$\int_X f \ d\mu = 0$$ for each $f\in C_c(X)$.

Then can we conclude that $\mu =0$?

I was trying to find a counter-example, but am stuck at some technical issue.

Answer 1

Yes, you may.

Note that $I_\mu : C_0(X) \to \mathbb{R}$ given by $I_\mu (f) = \int_X f d\mu$ is a continuous linear functional. Since $C_c(X)$ is dense in $C_0(X)$ and $I_\mu$ is zero on a dense set, it must extend uniquely to the zero functional on space $C_0(X)$. But the zero functional is represented by integration against the zero measure, so $\mu = 0$.

Answer 2

The existing answer uses implicitly the Markov-Riesz theorem, but one can also argue directly (in particular, this statement could be used in the proof of the Markov-Riesz theorem).

Let $K\subset X$ be compact. By Urysohn's lemma there exists a function $f\in C_c(X)$ such that $0\leq f\leq 1$ and $f|_K=1$. Thus $$ 0\leq\mu(K)=\int_X1_K\,d\mu\leq \int_X f\,d\mu=0 $$ and hence $\mu(K)=0$.

For arbitrary measurable $A\subset K$ the inner regularity of $\mu$ implies $$ \mu(A)=\sup\{\mu(K)\mid K\subset A\text{ compact}\}=0. $$ Note that neither finiteness nor outer regularity of $\mu$ are needed.