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)

  • 2
    $\begingroup$ A priori, there only must exist a Radon measure $\mu$ such that $$\int_{\Omega} f\,d\nu = \int_{\Omega} f\,d\mu$$ for all $f \in C_0^{\infty}(\Omega)$. But in nice spaces like open subsets of $\mathbb{R}^n$, every Borel measure that is finite on all compact sets is inner and outer regular, see e.g. Theorem 2.18 in Real and Complex Analysis. $\endgroup$ – Daniel Fischer May 16 '17 at 14:56
  • $\begingroup$ Question 1: Do we have that your Radon measure $\mu$ is unique? $\endgroup$ – user36548 May 16 '17 at 15:05
  • $\begingroup$ Yes, it is unique. But since we're in a nice space, it actually follows that $\mu = \nu$. I forgot about that at first. $\endgroup$ – Daniel Fischer May 16 '17 at 15:07
  • $\begingroup$ Question 2: in practice, if we already have a Borel measure $\mathcal{v}$; how can we prove it is Radon? It seems that checking inner, outer regularity are not easy. $\endgroup$ – user36548 May 16 '17 at 15:10
  • 2
    $\begingroup$ There are (locally compact) spaces less nice than open subsets of $\mathbb{R}^n$, and then a Borel measure that is finite on all compact subsets need not be regular. On the less nice spaces, we need the regularity assumption to get the useful results about Radon measures. // In practice, it is usually not very difficult to prove or disprove regularity when you have a measure concretely given. $\endgroup$ – Daniel Fischer May 16 '17 at 15:17

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