# 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.

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I guess you can take the measure which assigns to a set $A$ the value $\mu(A)/2$, where $\mu$ is the usual Lebesgue measure on $\mathbb R$. Am I missing something? –  Martin Sleziak Nov 8 '11 at 19:37
@MartinSleziak Maybe it is that simple, I'll run with this and see how it goes, thanks. –  Anastasia Nov 9 '11 at 2:50
I am no mercenary... but I do remember there was a bounty here! I do not remember from whom... :-( –  André Caldas Nov 24 '11 at 2:41

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.

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Thanks! So I gather that $[1,2]$ is not $\rho^*$-measurable? Do you mind expanding on how to use the definition of $\rho^*$-measurable to conclude that $[1,2]$ is not measurable? I don't follow your conclusions as quickly. –  Anastasia Nov 10 '11 at 19:23
@Anastasia: Notice that for the outer measure, $\rho^*([1,2]) = \rho^*([-2,-1]\cup[1,2]) = 1$ (why?), while for the inner measure, $\rho_*([1,2]) = \rho_*(\emptyset) = 0$ (why?). –  André Caldas Nov 11 '11 at 4:10
@Anastasia: Or, take $I = [-2,-1]\cup [1,2]$. Then, $I \cap [1,2] = [1,2]$ and $I \setminus [1,2] = [-2,-1]$. As in my preview comment, $\rho^*([1,2]) + \rho^*([-2,-1]) = \rho^*(I) + \rho^*(I) = 2 \rho^*(I) \neq \rho^*(I)$. Notice that $I$ is the smaller set in your semi-ring that contains $[1,2]$. It is also the smaller that contains $[-2,-1]$. –  André Caldas Nov 11 '11 at 4:15
Thank you Andre. –  Anastasia Nov 13 '11 at 4:31
By the way, if you have a measure $\mu$ on $\Omega$ and $\Omega_1$ and $\Omega_2$ are disjoint sets in bijection with $\Omega$, you could have done what you did in $\Omega_1 \cup \Omega_2$! What would be the measurable sets then? Actually, if $S$ is any non-empty set, you could have done what you did in $S \times \Omega$. The previews case is just the case for $S = \{1,2\}$. If $S$ is not countable, what would be the "null sets" in $S \times \Omega$? –  André Caldas Nov 14 '11 at 13:01