I am currently preparing for a measure theory exam and struggling with the following problem:
Consider the algebra
\begin{align} \mathfrak A = \{ A \subseteq \mathbb R: |A| < \infty ~ \text{or} ~ |A^C| < \infty \} \end{align}
and set function
\begin{align} \mu(A) = \begin{cases} |A|, & \text{if} ~ |A| < \infty, \\ -|A^C|, & \text{if} ~ |A^C| < \infty. \end{cases} \end{align}
I want to show that the set function $\mu$ is sigma additive on the algebra $\mathfrak A$, but not extendable to a signed measure on the $\mathfrak A$-generated sigma algebra.
I have tried blunt calculation for the first part - perhaps there is a necessary clever trick - and don't know what to do for the second part.
Any Hints?
Okay, so I think I have found a solution.
For the first part, we take a sequence $(A_n)_{n \in \mathbb N}$ of disjoint elements of $\mathfrak A$ such that $A := \bigcup_{n \in \mathbb N} A_n \in \mathfrak A$.
Case ($|A| < \infty$):
For all $n \in \mathbb N$ it holds that $|A_n| < \infty$. Thus, we get
\begin{align} \mu(A) = |A| = \sum_{n \in \mathbb N} |A_n| = \sum_{n \in \mathbb N} \mu(A_n). \end{align}
Case ($|A| = \infty$):
Because $A \in \mathfrak A$, we get $|A^\complement| < \infty$.
Now, it cannot happen, that for all $n \in \mathbb N$ we have $|A_n| < \infty$, because then $|A| = \aleph_0$ and due to $|\mathbb R| > \aleph_0$ we get $|A^\complement| = \infty$. Hence, there must exist an $m \in \mathbb N$ such that $|A_m| = \infty$ and because $A_m \in \mathfrak A$, it holds that $|A_m^\complement| < \infty$, i.e. $\mu(A_m) = -|A_m^\complement|$. Also, because $(A_n)_{n \in \mathbb N}$ are disjoint, for all $n \in \mathbb N \setminus \{ m \}$ there holds $A_n \subseteq A_m^\complement$, and thus $|A_n| < \infty$, i.e. $\mu(A_n) = |A_n|$.
Because $A_m \subseteq A$ and $A \setminus A_m \subseteq A$ we have
\begin{align} A^\complement = ((A \setminus A_m) \cup A_m)^\complement = A_m^\complement \setminus (A \setminus A_m) \subseteq A_m^\complement. \end{align}
Using $A \setminus A_m = \bigcup_{n \in \mathbb N \setminus \{ m \}} A_n \subseteq A_m^\complement$, this implies that
\begin{align} A_m^\complement = A_m^\complement \setminus (A \setminus A_m) ~\dot \cup~ (A \setminus A_m) = A^\complement ~\dot \cup~ (A \setminus A_m). \end{align}
Thus, putting it all together, we get
\begin{multline} \sum_{n \in \mathbb N} \mu(A_n) = \sum_{n \in \mathbb N \setminus \{ m \}} \mu(A_n) + \mu(A_m) = \sum_{n \in \mathbb N \setminus \{ m \}} |A_n| - |A_m^\complement| \\ = |A \setminus A_m| - |A_m^\complement| = -|A^\complement| = \mu(A). \end{multline}
For the second part, we need that $\mathbb N, \mathbb N^\complement, \mathbb R \in \mathfrak A_\sigma(\mathfrak A)$. Then we get
\begin{align} \mu(\mathbb N) & = \sum_{n \in \mathbb N} \mu(\{ n \}) = \sum_{n \in \mathbb N} |\{ n \}| = \infty, \quad \text{but} \quad \mu(\mathbb N^\complement) = \mu(\mathbb R) - \mu(\mathbb N) = -|\emptyset| - \infty = -\infty. \end{align}