I managed to prove $(a),(b),(c),(d),(e)$ from the following exercise in Folland's book:
I'm currently attempting $(f)$. In particular, I am stuck at showing that $\mu$ is countably additive. So let $(E_n)_n$ be a disjoint sequence in $\mathcal{B}_{\Omega^*}$. I want to show that $$\mu(\bigcup_n E_n) = \sum_n \mu(E_n)$$
My progress:
I can show that if for some $n$, $E_n \cup \{\omega_1\}$ contains an uncountable closed subset then $\mu(\bigcup_n E_n) = 1 = \mu(E_n) = \sum_n \mu(E_n)$ using $(d)$.
Hence, I may assume that for all $n$ we have that $E_n \cup \{\omega_1\}$ contains no uncountable closed subset. If I can show that $\bigcup E_n \cup \{\omega_1\}$ contains no uncountable closed subset, I will be done but I'm stuck at showing this. How can I show this?
Note that it follows from parts d and e that if $E\in\mathscr{B}_{\Omega^*}$, exactly one of $E$ and $\Omega^*\setminus E$ contains an uncountable closed set, $\mu(E)=0$ iff $E$ is disjoint from some uncountable closed set.
For $n\in\omega$ let $F_n=E_n\cup\{\omega_1\}$, and suppose that $\mu(F_n)=0$ for each $n\in\omega$. Then for each $n\in\omega$ there is an uncountable closed $C_n\subseteq\Omega^*$ such that $F_n\cap C_n=\varnothing$. Let $C=\bigcap_{n\in\omega}C_n$; $C$ is an uncountable closed set by part d, and clearly $C\cap\bigcup_{n\in\omega}F_n=\varnothing$, so $\mu\left(\bigcup_{n\in\omega}F_n\right)=0$.