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I understand the construction and proof that no measure on the Vitali set can meet the conditions that a measure should: correspond to length for intervals; be countably additive; be monotone; be translation invariant.

The Carathéodory construction starts with a pre-measure $\mu_0$ on a semi-ring on a set $X$ and extends it to an outer measure $\mu^*$ on the power set $\mathscr P (X)$ then provides a restriction that measurable sets must satisfy, that if $M$ is measurable then $\mu^*(A) = \mu^* (A \cup M) + \mu^*(A \cup M^c)$ for all $A \in \mathscr P (X)$.

So, my question is if one starts with the Lebesgue pre-measure defined as the length of half open intervals, and $V$ is the vitali set, is there a demonstrable set $ A \subset \mathbb R$ such that $\mu^*(A) \ne \mu^* (A \cup V) + \mu^*(A \cup V^c)$ ?

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HINT:

Note that for every $B\subset \mathbb{R}^n$ its outer measure also equals $$\mu^{\star}(B) = \inf \{ \mu(M) \ | \ M \textrm{ measurable}, M \supset B\} $$

Let's consider for every subset $A$ of $\mathbb{R}^n$ its inner measure defined as

$$\mu_{\star}(B) = \sup \{ \mu(M) \ | \ M \textrm{measurable} , M \subset B\}$$

It's fairly easy to show that for every $A$ measurable and any $B\subset A$ we have $$\mu(A) = \mu^{\star} (A\backslash B) + \mu_{\star} (B)$$

Consider now a Vitali set $V$, i.e. such a set so that $$\mathbb{R}^n = \sqcup_{q \in \mathbb{Q}^n} (V+q)$$ Then we have

  1. $\mu_{\star}(V)= 0$.

  2. $\mu^{\star}(V) > 0$.

For the first statement: Take a measurable $M \subset V$, and let's show that $\mu(M) = 0$. We may assume $M$ bounded. We have $$\sum_{q \in \mathbb{Q}^n \cap [0,1]^n} \mu( M+q) = \mu( \sqcup_{q \in \mathbb{Q}^n\cap [0,1]^n}\mu( M+q)) < \infty$$ so all terms ( being equal) are $0$, that is $\mu(M)=0$.

For the second: $$\sum_{q \in \mathbb{Q}^n}\mu^{\star}(V+q) \ge \mu^{\star}( \cup_{q \in \mathbb{Q}^n} M+q) = \mu^{\star}(\mathbb{R}^n) = \infty$$ so $\mu^{\star}(V) > 0$.

Now it's easy to find $A$ as required. Let's assume that $V$ is contained in a subset $A$ of finite measure. Since

$$\mu(A) = \mu^{\star}(A\backslash V) + \mu_{\star}(V)$$ we get $\mu^{\star}(A\backslash V) = \mu(A)$. Since $\mu^{\star}(V) >0$ we get the strict inequality.

Observation: if $B$ is a subset of a measurable set $A$ of finite measure and $$\mu(A) = \mu^{\star}(A\backslash B) + \mu^{\star}(B)$$ then $B$ is measurable. Indeed, we get $\mu_{\star}(B) = \mu^{\star}(B)$ and so $B$ is measurable (easy to prove, use completeness of the measure).

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  • $\begingroup$ Thanks for the detailed response. From my starting point, it will take me a while to work through it. $\endgroup$ – Tom Collinge Aug 24 '17 at 7:55

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