Saturated measure defined as a supremum of a semifinite measure and countable unions Here is what I am working on:

Suppose that $\mu$ is semifinite. For E in $\overline{M}$, define $\underline{\mu}(E)=\sup\{\mu(A):A$ in $M$ and $A \subseteq E$$\}$. Then $\underline{\mu}$ is a saturated measure on $\overline{M}$ that extends $\mu$.

I know that I need to establish that $\underline{\mu}$ is a measure, and then show that it is a saturated measure. I'm stuck on showing that the measure of a countable union of sets is equal to the sum of their measures. In particular, I know that if I have $\bigcup_{i=1}^∞ S_i$ where each $S_i \epsilon \overline{M}$, and some $S_k$ has $\underline{\mu}(S_k)=∞$, then clearly both the sum and the measure of the union will be $∞$, and if every $S_i$ is in M, then the sum of the measures is clearly the measure of the union since this is the case for the already-established measure $\mu$. I don't know what to do, however, if a few of the $S_i$'s happen to be in $\overline{M}$ but not in $M$. I wish I could just create a sequence of $T_i$ where each $T_i$ is the subset of $S_i$ maximal under the condition that $T_i$ is in M, but the problem is that I don't think the measure of such a $T_i$ under $\mu$ is necessarily the same as $\underline{\mu}(S_i)$ when $S_i$ is not in M since this latter measure is defined as a supremum, and suprema are not necessarily themselves elements of the set they are taken from.
(To be clear, $\overline{M}$ is the set of all locally measurable subsets of $X$, where a set $E ⊆ X$ is called locally measureable if $E ∩ A$ is in $M$ for all $A$ in $M$ such that $μ(A) < ∞$, and a measure is saturated if $M = \overline{M}$ with respect to that measure)
 A: As Weltschmerz was trying to point out to me, the proof involves using $\epsilon/2^i$.
If $\underline{\mu}(S_i) < ∞$ for every i, suppose $\epsilon>0$. Since $\underline{\mu}$ is defined as a supremum, for each i there exists a set $T_i \subseteq S_i$ so that $T_i \in M$ and $\underline{\mu}(T_i) = \mu(T_i) = \underline{\mu}(S_i) - \epsilon/2^i$. So 
$\sum_{i=1}^∞ \underline{\mu}(S_i)$ $=\sum_{i=1}^∞ (\underline{\mu}(T_i) + \epsilon/2^i)$ $=\sum_{i=1}^∞ \mu(T_i) + \sum_{i=1}^∞ \epsilon/2^i$ $=\mu(\bigcup_{i=1}^∞ T_i)+\sum_{i=0}^∞ \epsilon/2^{i+1}$ $=\mu(\bigcup_{i=1}^∞) T_i + (\epsilon/2)/(1-1/2)$ $=\mu(\bigcup_{i=1}^∞ T_i) + \epsilon$. 
Since $\bigcup_{i=1}^∞ T_i$ is a subset of $\bigcup_{i=1}^∞ S_i$ for any $\epsilon > 0$ and an element of $M$, and $\underline{\mu}(\bigcup_{i=1}^∞ S_i)$ $=\sup\{\mu(A):A$ in $M$ and $A \subseteq \bigcup_{i=1}^∞ S_i\}$, $\mu(\bigcup_{i=1}^∞ T_i) + \epsilon$ approaches $\underline{\mu}(\bigcup_{i=1}^∞ S_i)$ as $\epsilon$ approaches $0$, so $\sum_{i=1}^∞ \underline{\mu}(S_i)$ $=\underline{\mu}(\bigcup_{i=1}^∞ S_i)$, as desired.
(Did I miss anything?)
A: To me you only demonstrated that $\underline{\mu}(\bigcup S_i) \geq \sum\underline{\mu}(S_i)$. Now you need to show that $\underline{\mu}(\bigcup S_i) \leq \sum\underline{\mu}(S_i)$ and for that you need the semifinite property of $\mu$ for the case where $\exists A \subset \bigcup S_i, A \in {\cal M}$ and $\mu(A) = \infty$ exactly like Weltschmerz was pointing out to you.
