measures on the family of locally measurable subsets I am contemplating over the exercise about the ways to extend a measure to a collection of locally measurable subsets. To be precise:
Let $(X,\mathcal M,\mu)$ be a measure space. Assume that $\mu$ is semifinite (i.e. each measurable set of infinite measure contains measurable sets of arbitrarily large finite measure). Let $\mathcal C$ denote the collection of locally measurable subsets of $X$. (Recall that a subset $E\subseteq X$ is called locally measurable if for each $B\subseteq \mathcal M$ with $\mu(B)<\infty$, the intersection $E\cap B$ belongs to $\mathcal M$.) It can be shown that $\mathcal C$ is a $\sigma$-algebra, so one can extend measure $\mu$ from $\mathcal M$ to $\mathcal C$ in at least two different ways: for every $E\in\mathcal C$ define
$\bar{\mu}(E)=\mu(E)$ if $E\in\mathcal M$ and $\bar{\mu}(E)=\infty$ if $E\notin\mathcal M$;
$\underline{\mu}(E)=\sup\{\mu(B)\mid B\in\mathcal M,\, B\subseteq E\}$.
One can show that each of these functions is indeed a measure on $\mathcal C$ making $(X,\mathcal C)$ a saturated measure space.
The question is to give an example showing that $\bar\mu$ and $\underline\mu$ are different.
 A: The example of such measures is given in Exercise 1.3.16, part f, in Folland's "Real Analysis", as was pointed out by Quinn Culver. The construction there is as follows:

Let $X_1$, $X_2$ be disjoint uncountable sets, $X=X_1\cup X_2$, and $\mathcal M$ the $\sigma$-algebra of all countable and co-countable ["co-countable" means that the complement is countable] sets in $X$. Let $\mu_0$ be counting measure on the set $\mathcal P(X_1)$ of all subsets of $X_1$ and define a new measure $\mu$ on $\mathcal M$ by $\mu(E)=\mu_0(E\cap X_1)$ (where the counting measure $\mu_0(A)$ equals cardinality of $A$ if $A$ is finite and $\infty$ otherwise). Then the following is true:
(i) $\mu$ is a measure on $\mathcal M$;
(ii) all subsets of $X$ are locally measurable with respect to $\mu$;
(iii) the two measures $\bar\mu$ and $\underline\mu$ defined as in the original question above, are different.

The proof of (i) is straightforward. To prove (ii), let's take any subset $E\subseteq X$. To show that $E$ is locally measurable, one needs to take arbitrary $F\in\mathcal M$ such that $\mu(F)<\infty$ and show that $E\cap F\in\mathcal M$. Since $F\in \mathcal M$, $F$ is either countable or co-countable. But since $\mu(F)=\mu_0(F\cap X_1)<\infty$, and $X_1$ is uncountable, $F$ cannot be co-countable, so $F$ must be countable. Hence $E\cap F$ is countable and thus $E\cap F\in\mathcal M$.
For (iii), take $E=X_2$. Then $\underline\mu(E)=\sup\{\mu(B)\mid B\in M, B\subseteq E\}=0$, since if $B\subseteq X_2$, $B\cap X_1=\varnothing$, and $\mu(B)=0$. But $\bar\mu(E)=\infty$, as $E\notin\mathcal M$.
