# Regular Measures

Let $$\Sigma$$ be the Borel $$\sigma$$-algebra of $$\mathbb{R}$$. We define a measure $$\mu$$ on $$\Sigma$$. For every Borel set $$A$$ we define $$\mu(A)$$ to be the number of elements from the set $$\{\frac{1}{n}: n\in\mathbb{N}\}$$ which are in $$A$$. The question asks to prove this measure is not regular. From what I know $$\mu$$ is called regular if for any measurable set $$A$$ the following holds:

$$\mu(A)$$=sup{$$\mu(K)$$: $$K\subseteq A$$, $$K$$ is compact and measurable}

The problem is that I got the impression that the statement I need to prove is simply wrong. Let $$A\in\Sigma$$. If $$\mu(A)<\infty$$ then we can take $$K=A\cap\{\frac{1}{n}:n\in\mathbb{N}\}$$. This is a compact subset of $$A$$ which has exactly the same measure as $$A$$. If $$\mu(A)=\infty$$ then we can write $$A\cap\{\frac{1}{n}:n\in\mathbb{N}\}=\{x_m\}_{m=1}^\infty$$ using the fact that this is a countable set. And then for every $$m\in\mathbb{N}$$ we define $$K_m=\{x_1,x_2,...,x_m\}$$. We get that $$K_m$$ is a measurable compact subset of $$A$$ which has measure $$m$$. Because we define that for all $$m\in\mathbb{N}$$ we get that for any $$M>0$$ there is a measurable compact subset $$K\subseteq A$$ with $$\mu(K)>M$$. Hence the supremum of the measures of the compact subsets of $$A$$ is $$\infty$$, just like $$\mu(A)$$. So $$\mu$$ is actually regular.

Am I missing something here, or is the statement really wrong?

## 1 Answer

Your proof is correct.

But if we use the definition from Wikipedia, which has the additional condition $$\mu(A)=\inf\{\mu(O): O\supseteq A, O\;\text{is open and measurable}\}.$$ It turns out that the measure $$\mu$$ does not satisfy this condition (leave a comment if you need help to understand why this is the case).

(Note: The definition of a regular measure probably varies in the literature. Even in wikipedia an alternative version is mentioned. You have to check your source material which definition of regular measure is used.)

• Thanks for the answer. I see why according to your definition $\mu$ is not regular-because if we take $A=\{0\}$ then it has meausre $0$ but any open set containing $A$ has infinite measure. Well, in class we defined regularity as I wrote it in the question. Though the professor is not the one who writes the exercises so maybe the person who does just didn't know how he defined a regular measure. Thanks for your help. – Mark Nov 23 '18 at 17:25