This is exercise $1$ from chapter 5 in Bauer's book:
Let $\mu=\varepsilon_{\omega}$ be the dirac premeasure on a ring $\mathcal{R}$ in $\Omega$. Suppose there exist a sequence $(B_n)\subset \mathcal{R}$ such that $\bigcap_{n\in \mathbb{N}}B_n=\{\omega\}$ and $\bigcup_{n\in \mathbb{N}}B_n=\Omega$. Prove that the outer measure $\mu^*$ generated by $\mu$ assigns to every set $A\subset \Omega$ the value $1$ or $0$ according as $\omega\in A$ or $\omega\in A^{c}$.
Note: $\mu^*(A):=\inf\big\{\sum_{n=1}^{\infty}\mu(A_n): (A_n)\subset \mathcal{R}, A\subset\bigcup_{n\in \mathbb{N}}A_n \big\}$ for $A\subset \Omega$.
My work so far:
We know that $\mu^*=\mu$ on $\mathcal{R}$. I first show that $\mu^*(\{\omega\})=1$. Since for any sequence $(A_n)\subset \mathcal{R}$ such that $\{\omega\}\subset\bigcup_{n\in \mathbb{N}}A_n$ we must have $\omega \in A_n$ for some $n$, we see that $1\leq\mu^*(\{\omega\})$. On the other hand, by monotonicity of outer measures we get $\mu^*(\{\omega\})\leq \mu^*(B_n)=\mu(B_n)=1$.
Now if $\omega\in A\subset \Omega$, then again by monotonicity we get $1=\mu^*(\{\omega\})\leq \mu^*(A)$. But how can I show the reverse inequality?
Next we have $\Omega\setminus \{\omega\}=\bigcup_{n\in \mathbb{N}}B_n\setminus \bigcap_{n\in \mathbb{N}}B_n=\bigcup_{(n_1,n_2)\in \mathbb{N}\times\mathbb{N}}B_{n_1}\setminus B_{n_2}$ where $B_{n_1}\setminus B_{n_2}\in \mathcal{R}$ and $\mu(B_{n_1}\setminus B_{n_2})=0$. Hence by countable subadditivity of outer measures we get $\mu^*(\Omega\setminus\{\omega\})=0$. Then, if $\omega\notin A\subset \Omega$, monotonicity allows us to conclude $\mu^*(A)=0$.
Any help is greatly appreciated.