Is this closed interval measurable?  A common example of a semiring of sets is the family of half open interals $(a,b]\subseteq\mathbb{R}$. Also, the premeasure $\rho((a,b])=b-a$ is well known to extend to a measure on a $\sigma$-algebra. 
With a little tinkering, I believe the "mirror images" across the origin of these intervals also form a semiring. That is, the sets of form $[-b,-a)\cup(a,b]$ for $0<a<b$ are also a semiring. Say $I_{a,b}=[-b,-a)\cup(a,b]$. If I put nearly the same premeasure $\rho(I_{a,b})=b-a$ on this semiring, then I think that $\rho$ can be extended to a measure on a $\sigma$-algebra by taking the measure which sends a set $A$ to $\mu(A)/2$ for the usual Lebesgue measure $\mu$ on $\mathbb{R}$. (I hope this is correct?)
Is there a way to tell if a closed interval $[a,b]$ is $\rho^*$ measurable? I'm interested in seeing maybe an example first to figure this out. Take an interval $[1,2]$ for example. I know that $[1,2]$ is $\rho^*$ measurable if for any $I\subseteq\mathbb{R}$, then $$\rho^*(I)=\rho^*(I\cap[1,2])+\rho^*(I\setminus[1,2]).$$ 
My feeling is that $[1,2]$ is $\rho^*$ measurable just by testing it with a few subsets $I$ of the real line. Is there a way to prove or disprove whether this is true?
Thank you.
 A: It seems to me that you are basically talking about the Lebesgue measure on
$$
  \mathbb{R}^+
  =
  \{x \in \mathbb{R} | x > 0\}.
$$
Notice that the $\sigma$-algebra generated by these "symmetric sets" is fully composed of "symmetric sets"!! I would say that the $\rho^*$-measurable sets are simply the sets of the form
$$
  -A \cup A \cup N,
$$
where $A$ is a Borel set of $\mathbb{R}^*$ and $N$ is a Lebesgue null set.
This, because the Lebesgue null sets are exactly the same as $\rho^*$ null sets.
So, the $\rho^*$-measurable closed intervals would be of the form $[-a,a]$.
The measure $\rho$ is the same as $\frac{\mu}{2}$ where $\rho$ is defined.
But the Lebesgue measurable sets is a much wider family.

As requested, let's show that $[1,2]$ is not $\rho^*$-measurable as per
the provided definition.
To make things simpler, let's prove that $(1,2]$ is not $\rho^*$-measurable.
Take $I = [-2,-1) \cup(1,2]$.
Notice that $I$ is the smaller set in the generating semiring such that
$(1,2] \subset I$.
Therefore,
$$
  \rho^*((1,2]) = \rho^*(I) = 1.
$$
Likewise,
$$
  \rho^*([-2,-1)) = \rho^*(I) = 1.
$$
That is,
$$
  \rho^*(I) \neq \rho^*(I \cap (1,2]) + \rho^*(I \setminus (1,2]).
$$
If you really want to prove for $[1,2]$, just notice that $\{1\}$
has null outer measure and is therefore $\rho^*$-measurable.
If $[1,2]$ was $\rho^*$-measurable, then $[1,2] \setminus \{1\} = (1,2]$
would also be $\rho^*$-measurable.
But we already know that it is not!

Edit: Added proof that $[1,2]$ is not $\rho^*$-measurable.
