What is wrong with my argument below? As follows I will show that any Borel measure that has a finite measure on any compact subset must be a Radon measure. This seems wrong! but I don't know why!
Definition: $\mu$ is called a Radon measure on an open set $\Omega\subset\mathbb{R}^{n}$ if $\mu$ is defined on Borel subsets, finite on compact subsets and inner, outer regular.
Fact: We also know that if $T$ is a distribution of order zero then $T$ is a Radon measure.
Now, let $\mathcal{v}$ be only a Borel measure and so that $\mathcal{v}(K)$ is finite, for any compact subset $K$. It is clear that: $$f\in C_{0}^{\infty}(\Omega)\rightarrow\int_{\Omega} f d\mathcal{v}$$ is a distribution of order zero (by definition). Therefore we must have that $\mathcal{v}$ is a Radon measure.
So there is sth. wrong here, because we get a Radon measure without inner,outer regularity assumption. (For simplicity, every measure is non-nagative)