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?
 A: @Ramiro: Thanks for your answers!
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.
