Concrete application of Caratheodory extension I am working on the following problem:
Question:

Let $\Omega$ be an uncountable set. Denote by $\mathfrak{R}$ the ring
of finite subsets of $\Omega$. Consider the measure $\mu: \mathfrak{R}
 \rightarrow [0, + \infty]: E \mapsto 0$. What is in this case the
outer measure $\mu^{*}$ (induced by $\mu$) and the associated
$\sigma$-algebra $\mathfrak{M}_{\mu^{*}}$? What is the
$\sigma$-algebra $\mathfrak{M}$ generated by $\mathfrak{R}$? Is the
extension of the measure $\mu$ from $\mathfrak{R}$ to $\mathfrak{M}$
obtained via the Caratheodory procedure the only possible extension?

Attempt: First recall some definitions. By definition, $\mathfrak{M}_{\mu^{*}} = \left\{ E \subset \Omega \mid E \ \text{is} \ \mu^{*}\text{-measurable} \right\}$. And a set $E \subset \Omega$ is $\mu^{*} \text{-measurable}$  if for all $A \subset \Omega$, we have $$ \mu^{*} (A) = \mu^{*} (A \cap E) + \mu^{*} (A \cap E^c). $$ Also, recall how $\mu^{*}$ is defined: For each $E \subset \Omega$ we have $$ \mu^{*} (E) := \text{inf} \left\{ \sum_n \mu(E_n) \mid (E_n)_n \in \mathfrak{R} \ \text{is a sequence with} \ E \subset \bigcup_n E_n \right\}. $$
In this specific question, I think we have $$  \mu^{*} (E) =  \begin{cases} 0 & \text{if} \ E \ \text{is finite or countable infinite} \\ + \infty & \text{if} \ E \ \text{is uncountable} \end{cases} $$
However, I am stuck with finding $\mathfrak{M}_{\mu^{*}}$ and $\mathfrak{M}$. I know from a theorem that $\mathfrak{M}_{\mu^{*}}$ will contain $\mathfrak{R}$. I also know that in general, $\mathfrak{M}_{\mu^{*}}$ will be a bigger set than $\mathfrak{M}$. By definition, $\mathfrak{M}$ is the smallest $\sigma$-algebra containing $\mathfrak{R}$.
Any help is appreciated.
 A: Let us go step by step
$1$. The $\sigma$-algebra $\mathfrak{M}$ generated by $\mathfrak{R}$.
Since every countable subset of $\Omega$ is the countable union of finite sets (singletons), we have that every countable subset of $\Omega$ is in $\mathfrak{M}$.
Then it is easy to see that
$$\mathfrak{M} =\{ E \subseteq \Omega : E \textrm{ or } E^c \textrm{ is countable }\}$$
$2$. The outer measure $\mu^{*}$.
It is exactly as you wrote: For each $E \subset \Omega$ we have $$ \mu^{*} (E) := \text{inf} \left\{ \sum_n \mu(E_n) \mid (E_n)_n \in \mathfrak{R} \ \text{is a sequence with} \ E \subset \bigcup_n E_n \right\}. $$
In this specific question, we have $$  \mu^{*} (E) =  \begin{cases} 0 & \text{if} \ E \ \text{is finite or countable infinite} \\ + \infty & \text{if} \ E \ \text{is uncountable} \end{cases} $$
because an uncountable subset can not be covered by a countable collection of finite subsets.
$3$. The $\sigma$-algebra $\mathfrak{M}_{\mu^{*}}$.
We have that, for all $E \subset \Omega$, $E \in \mathfrak{M}_{\mu^{*}}$  if and only if for all $A \subset \Omega$, we have $$ \mu^{*} (A) = \mu^{*} (A \cap E) + \mu^{*} (A \cap E^c). $$
For any $E \subset \Omega$, we have:
If $A$ is uncountable, then $A \cap E$ or $ A \cap E^c$ is also uncountable and we get
$$ \mu^{*} (A) =+\infty = \mu^{*} (A \cap E) + \mu^{*} (A \cap E^c). $$
If $A$ is countable, then $A \cap E$ and $ A \cap E^c$ are both countable and we get
$$ \mu^{*} (A) =0 = 0+0 = \mu^{*} (A \cap E) + \mu^{*} (A \cap E^c). $$
So $\mathfrak{M}_{\mu^{*}}= 2^\Omega =\{ E : E \subseteq \Omega\}$.
$4$. Is the extension of the measure $\mu$ from $\mathfrak{R}$ to $\mathfrak{M}$ obtained via the Caratheodory procedure the only possible extension?
Answer: No.
Let $\mu_1$ be the extension of the measure $\mu$ from $\mathfrak{R}$ to $\mathfrak{M}$ obtained via the Caratheodory procedure. Then
$$  \mu_1(E) =  \begin{cases} 0 & \text{if} \ E \ \text{is countable} \\ + \infty & \text{if} \ E^c \ \text{is countable} \end{cases} $$
Another extension would be $\mu_2$ defined by $\mu_2(E)=0$, for all $E \in \mathfrak{M}$.
Note tha, since $\Omega$ is uncountable, $\mu$ is not $\sigma$-finite. So it does not need to have a unique extension to the $\sigma$-algebra $\mathfrak{M}$ generated by $\mathfrak{R}$.
