Construction of outer measure where the closed unit interval is not measurable and outer measure there is not equal to set function

Question

Occasioned by some exercises I saw about creating outer measures in which [0,1] is not measurable because of the definition of the set function that generates the outer measure, I was wondering if given an σ-cover constisting of the closed intervals [a,b] in $\mathbb{R}$ we could create an outer measure m* with a proper function ρ such that
m*([0,1]) $\neq$ ρ([0,1]) AND [0,1] is not m*-measurable

I couldn’t find something online and the main thing that’s been bothering me is that because [0,1] is a closed interval it can be written as a cover of itself and therefore something like that would be true m*([0,1]) $\leq$ ρ([0,1]). How can I define ρ to get rid of the equality?

Thanks in advance

Answer 1

Example 1: Consider $(\Bbb R, \Sigma , \mu)$ where $\Sigma$ is the $\sigma$-algebra of countable / co-countable subsets of $ \Bbb R$ and $\mu(E) = 0$ if $E$ is countable and $\mu(E) = 1$ is $E$ is co-countable in $\Bbb R$.

It follows immediately that $[0,1] \notin \Sigma$ (because $[0,1]$ is not countable nor co-countable). Moreover: $$ \mu^*(\Bbb R) =1 < 1 +1 = \mu^*(\Bbb R \cap [0,1]) + \mu^*(\Bbb R \cap [0,1]^C)$$ So $[0,1]$ is not measurable (in Caratheodory sense). That is $[0,1]$ is not $\mu^*$- measurable.

Remarks:

1. Please, note that $\mu$ is not the Lebesgue measure and $\mu^*$ is not the Lebesgue outer measure.
2. Since $\Bbb R$ is co-countable, $\mu^*(\Bbb R)= \mu(\Bbb R) =1$.
3. Since $[0,1] \subseteq \Bbb R$, we have that $\mu^*([0,1])\leqslant \mu^*(\Bbb R)=1 $. On the other hand, since any countable cover of $[0,1]$ by sets in $\Sigma$ must contain at least one co-countable set, we have $\mu^*([0,1]) \geqslant 1$. So $\mu^*([0,1]) = 1$.
4. By a similar argument, we have that $\mu^*(\Bbb R \cap [0,1]^C)=1$.

Example 2: Consider $\Bbb R$ and let $C$ be the semi-algebra of all intervals (including single point intervals and the empty interval). Let $\rho: C \rightarrow [0, +\infty]$ be a function defined defined: $\rho(I) = 0.1$, if $I \subsetneq [0,1]$ and $\rho(I)=1$ if $I=[0,1]$ or $I \not \subset [0,1]$.

Note that $\rho$ is not a pre-measure, it is not $\sigma$-additive and it is not even finitely additive.

Let $\mu^*$ be the outer measure induced by $\rho$. It easy to see that: $\mu^*([0,1]) = 0.2 < 1 =\rho([0,1])$. I is also easy to see that $\mu^*(\Bbb R) = 1$ and $\mu^*(\Bbb R \cap [0,1]^C) = 1$. It follows that $$\mu^*(\Bbb R) = 1 < 1.2 = 0.2 + 1 = \mu^*(\Bbb R \cap [0,1]) + \mu^*(\Bbb R \cap [0,1]^C)$$ So $[0,1]$ is not $\mu^*$-measurable.

Note that since $\rho$ is not $\sigma$-additive, $\mu^*$ is not an extension of $\rho$, which we can clearly see from the fact that $\mu^*([0,1]) = 0.2 \neq 1 =\rho([0,1])$.