# Inclusions for certain types of measures

Let's use the following definitions:

Definition. A measure $\mu: \mathcal P(X) \to [0, \infty]$ is what some authors call a outer measure, i.e.

(1) $\mu(\emptyset) = 0$.

(2) If $A, A_k \subset X$ for $k \in \mathbb N^\times$ and $A \subset \bigcup_{k=1}^\infty A_k$, then $\mu(A) \leq \sum_{k=1}^\infty \mu(A_k)$.

Now, let's distinguish certain types of measures:

Definition.

(1) A measure $\mu: \mathcal P(X) \to [0,\infty]$ is called regular, if for any $A \subset X$, there exists a $\mu$-measurable set $B \subset X$, such that $\mu(A) = \mu(B)$.

(2) A measure $\mu: \mathcal P(\mathbb R^n) \to [0, \infty]$ is called Borel, if any Borel set $B \subset X$ is $\mu$-measurable.

(3) A measure $\mu: \mathcal P(\mathbb R^n) \to [0, \infty]$ is called Borel-regular, if $\mu$ is Borel and for any set $A \subset \mathbb R^n$, there is a Borel set $B \subset \mathbb R^n$, such that $\mu(A) = \mu(B)$.

(4) A measure $\mu: \mathcal P(\mathbb R^n) \to [0, \infty]$ is called Radon, if $\mu$ is Borel-regular and for any compact set $K \subset \mathbb R^n$, we have $\mu(K) < \infty$.

By this definitions, we clearly have $$\{ \text{Radon measures} \} \subset \{ \text{Borel-regular measures} \} \subset \{ \text{Borel measures} \} \subset \{ \text{Measures on } \mathbb R^n \}$$ and $$\{ \text{Regular measures on } X \} \subset \{ \text{Measures on } X \} \; .$$

I would like to create examples, to show that these inclusions can be strict.

So by taking $\mu$ the counting measure on $\mathbb R^n$, we have for any $A, B \subset \mathbb R^n$ with $\text{dist}(A, B) > 0$ $$\mu(A \sqcup B) = \mu(A) + \mu(B) \; ,$$ so by the Caratheodory criterium is $\mu$ Borel. Since finite sets are Borel, it is clear, that $\mu$ is Borel-regular. But $\mu$ is not Radon, since $\mu(\overline {\mathbb B}(0,1)) = \infty$.

Unfortunately, I could not find any examples for the other inclusions, i.e.

• A measure on a set $X$, which is not regular.
• A Borel measure, which is not Borel-regular.
• A measure on $\mathbb R^n$, which is not Borel.

Are there any easy examples for these cases? If so, how do they look like?

• An outer measure on a set $X$, which is not regular. Let $X=\{a,b,c\}$. Let $\mu: \mathcal P(X) \to [0,\infty]$ be such that $\mu(\emptyset)=0$; $\mu(A)=1$, if $A$ has $1$ element and $\mu(A)=2$, if $A$ has $2$ or $3$ elements. It is easy to check that $\mu$ is an outer measure. The only $\mu$-measureable sets are $\emptyset$ and $X$. So take $A=\{a\}$. We have $\mu(A)=1$, but there there is no $\mu$-measureable set $B$ such that $\mu(B)=\mu(A)$. – Ramiro Dec 23 '15 at 21:25
• An outer measure on $\mathbb R^n$, which is not a Borel outer measure. Let $\mu: \mathcal P(\mathbb R^n) \to [0,\infty]$ be such that $\mu(\emptyset)=0$ and, for all $A\in \mathcal P(\mathbb R^n)$, if $A \neq \emptyset$, $\mu(A)= 1$. It is easy to check that $\mu$ is an outer measure. The only $\mu$-measureable sets are $\emptyset$ and $\mathbb R^n$. – Ramiro Dec 23 '15 at 21:26

I have found another example of an outer measure, which is not regular:

Let $\{ a_n \}_{n=0}^\infty$ be a strictly increasing sequence in $\mathbb R$, which satisfies $$0 = a_0 < \frac 1 2 < a_1 < a_2 < \cdots < 1 \quad \text{and} \quad \lim_{n \to \infty} a_n = 1 \; .$$ We define $\mu: \mathcal P(\mathbb N) \to [0, \infty]$ as $$\mu(A) = \begin{cases} a_{\# A} \, , & A \text{ is finite} \\ 1 \, , & A \text{ is infinite}\end{cases} \; .$$

We easily see, that $\mu$ is an outer measure.

Now, if $A \subset \mathbb N$ is finite, $A \neq \emptyset$, then $$1 = \mu(\mathbb N) \neq \underbrace{ \mu(\mathbb N \cap A) }_{> \frac 1 2} + \underbrace{ \mu(\mathbb N \backslash A)}_{=1} > \frac 3 2 \; ,$$ so $A$ is not $\mu$-measurable.

If $A \subset \mathbb N$ is infinite, $A \neq \mathbb N$, then $$1 = \mu(\mathbb N) \neq \underbrace{ \mu(\mathbb N \cap A) }_{=1} + \underbrace{ \mu(\mathbb N \backslash A) }_{> \frac 1 2 } > \frac 3 2 \; ,$$ so $A$ is not $\mu$-measurable.

We have shown, that the only $\mu$-measurable subset of $\mathbb N$ are $\emptyset$ and $\mathbb N$.

Let $A \subset \mathbb N$ be finite, $A \neq \emptyset$. We have $\mu(A) \in \left( \frac 1 2 , 1 \right)$, and the only $\mu$-measurable subset $B \supset A$ is $B = \mathbb N$, but $$\mu(\mathbb N) = 1 \neq \mu(A) \; .$$ This shows, that $\mu$ is not regular.

I still have no example for a Borel measure, which is not Borel-regular, but I think I will post this question in a new thread.