In Rudin's Real and Complex Analysis, in Theorem $1.36$, Rudin states the following:
Let $(X, \mathcal{M}, \mu)$ be a measure space, let $\mathcal{M}^*$ be the collection of all $E \subset X$ for which there exist sets $A$ and $B \in \mathcal{M}$ such that $A \subset E \subset B$ and $\mu(B-A) = 0$, and define $\mu(E) = \mu(A)$ in this situation. Then $\mathcal{M}^*$ is a $\sigma$-algebra, and $\mu$ is a measure on $\mathcal{M}^*$.
Proofs that $\mathcal{M}^*$ is a $\sigma$-algebra and that $\mu$ is a well-defined measure over $\mathcal{M}^*$ then follow, which I understand fairly well. The problem is the final line of the proof:
The countable additivity of $\mu$ on $\mathcal{M}^*$ is obvious
Now, clearly we would expect this to hold, and I attempted the following (admittedly messy) proof:
We desire to prove that $\mu\!\left(\bigcup E_i\right) = \sum\mu(E_i)$ for any collection $\{E_i\}$ of sets, where $E_i \in \mathcal{M}^*$ for each $i$. If each $E_i$ is a member of $\mathcal{M}$ then this clearly holds. Let $\{E_i\}$ be any collection of sets contained in $\mathcal{M}^*$, so that for each $E_i$ there exist sets $A_i, B_i \in \mathcal{M}$ such that $A_i \subset E_i \subset B_i$. Thus:
$$\begin{align}
\sum\mu(A_i) &= \mu\!\left(\bigcup A_i\right) \le \mu\!\left(\bigcup E_i\right) \le \mu\!\left(\bigcup B_i\right) =\sum\mu(B_i)
\end{align}$$
We thus have that
$$\sum\mu(A_i) \le \mu\!\left(\bigcup E_i\right) \le \sum\mu(B_i)$$
Subtracting $\sum\mu(A_i)$ from each term,
$$0 \le \mu\!\left(\bigcup E_i\right) - \sum\mu(A_i) \le \sum\mu(B_i) - \sum\mu(A_i) \color{red}{\stackrel{?}{=}} \sum\mu(B_i - A_i) = 0$$
Yielding
$$\mu\!\left(\bigcup E_i\right) - \sum\mu(A_i) = 0 \\\implies \mu\!\left(\bigcup E_i\right) = \sum\mu(A_i) = \sum\mu(E_i)$$
Given that each $A_i$ and $B_i$ are finite in measure I feel alright with this proof, but worry if any of the sets have infinite measure; specifically, I fear that my rearrangements of the sums might not be valid. Is there a simple proof I am missing, or if not, a way to patch up any holes in my proof?
One important point you are missing is that $\mu $ is not given to be a finite measure. So it is important to avoid the indeterminate form $\infty -\infty$. Here is the correct proof: let $\{E_i\}$ be a disjoint sequence in $\mathcal M^{*}$ and choose $A_i,B_i$ as you have done. Note that $A_1,A_2,...$ are disjoint. We have $\mu (\cup_i E_i) \leq \mu (\cup_i B_i) \leq \mu (\cup_i A_i)+ \mu( \cup_i(B_i \setminus A_i)) \leq \mu (\cup_i A_i)+\sum_i \mu(B_i \setminus A_i)=\sum_i \mu(A_i)+0$ since $\mu$ is a measure on $\mathcal M$. Since $\mu (A_i) \leq \mu (E_i)$ we get $\mu (\cup_i E_i) \leq \sum_i \mu (E_i)$. The revese inequality is easier and I will let you write that (avoiding $\infty - \infty$).