Show that a set function is sigma additive on an algebra, but not extendable to a signed measure on the generated sigma algebra. 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?
 A: 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}
