In Halmos' Measure Theory, section 16, exercise 2 deals with the extension of a $\sigma$-finite measure $\mu$ defined on a $\sigma$-ring $S$ to any set $M$ in the hereditary $\sigma$-ring induced by $S$. In the context of Lebesgue measure, this means we can extend the measure to include any non-measurable set in its domain, with the caveat that the resulting extension is not translation invariant, as otherwise extending the measure to any Vitali set would result in the familiar contradiction.

Exercise 3 in the same section of Halmos then asks to perform a similar extension for any finite collection $\{M_1, M_2, \dots, M_n\}$, noting that it is not known whether it is possible to do an analogous extension for an infinite sequence $\{M_n\}$.

I have a couple of questions surrounding exercise 3:

  1. It seems to me (and I must be wrong, but don't understand why) that being able to perform repeated extensions would allow to get the contradiction in the Vitali sets. In particular, let $V$ be a Vitali set with positive outer measure, let $\langle q_k \rangle$ be an ordering of the rationals in $[-1, 1]$, and let $V_k$ be the $q_k$ translate of $V$. We know $[0,1] \subseteq \bigcup V_k \subseteq [-1,2]$. Now suppose we extend the Lebesgue measure to $V_1$: my understanding is that we can extend it in such a way that it assumes any value between the inner and outer measure of the set involved, so let $m(V_1) = \epsilon >0$. Since outer measure is translation invariant, we can extend $m$ to the first $n = \lceil 3/\epsilon\rceil +1$ translates of $V$ to each have measure equal to $\epsilon$, and then extend the measure to the union $\bigcup_{k=n+1}^\infty V_k$ in any way. Then $m(\bigcup V_k) = m(\bigcup_{k=1}^n V_k) + m(\bigcup_{k=n+1}^\infty V_k) > 3 = m([-1,2])$ What is the mistake here? I imagine it has something to do with the fact that the extension is not translation invariant, but it seems that you can force it to be for a finite collection of translates, assuming the result in exercise 3.

  2. Halmos' book was published in 1974. Has the question of extensions to a countable collection of non-measurable sets been resolved since then?

Addendum: I'm including the statements in exercises 2 and 3 for context and clarity, as well as a sketch of the proof for 2, though I believe David C. Ullrich has already answered the first question, for which it's most relevant.

  1. Let $\mu$ be a $\sigma$-finite measure on a $\sigma$-ring $S$ of subsets of a set $X$, and let $\mu^*$ and $\mu_*$ be the outer measure and the inner measure, respectively, induced by $\mu$ on $H(S)$ [the hereditary $\sigma$-ring induced by $S$]. Let $M$ be any set in $H(S)$, and let $\tilde(S)$ be the $\sigma$-ring generated by the class of all sets in $S$ together with $M$. The chain of assertions [in the exercise] is designed to lead up to a proof of the assertion that $\mu$ may be extended to a measure $\tilde{\mu}$ on $\tilde{S}$.

  2. If $\mu$ is a $\sigma$-finite measure on a $\sigma$-ring $S$ and if $\{M_1, \dots, M_n\}$ is a finite class of sets in the hereditary $\sigma$-ring $H(S)$, then $\{M_1, \dots, M_n\}$ may be adjoined to $S$ and a measure $\tilde{\mu}$ may be defined on the generated $\sigma$-ring $\tilde{S}$ so that it is an extension of $\mu$ on $S$. The analogous statement, for an infinite sequence $\{M_n\}$ of sets in $H(S)$, is not known.

Exercise 2 involves the following:

  • Finding a measurable cover $H$ and a measurable kernel $G$ of $M$.

  • Showing that every element in $\tilde{S}$ is of the form $(E \cap M) \cup (F \cap M^c)$, with $E, F \in S$.

  • Showing that the intersection of any set in $\tilde{S}$ with $(H-G)^c$ belongs to $S$.

The extension of the measure can then be defined as $\mu([(E \cap M) \cup (F \cap M^c)] \cap D^c) + \alpha\mu(E \cap D) + \beta\mu(f \cap D)$, where $D = H-G$ and $\alpha + \beta = 1$.

In the special case of the measure in question being Lebesgue measure on the Lebesgue measurable sets of $\mathbb{R}$ and the set $M$ being a Vitali set with outer measure 1, inner measure 0, the interval $(0,1)$ forms a measurable cover of $M$, $\emptyset$ is a measurable kernel, and $M$ can be assigned any measure between 0 and 1.

  • $\begingroup$ Halmos forgot to mention that under continuum hypothesis the answer is no. This was shown by Banach and Kuratowski in 1929. The consistency of the other direction is due to Solovay and also Carlson. $\endgroup$ – hot_queen Oct 1 '15 at 23:21

I don't know that theorem and you don't give a precise statement (the words "my understanding is" make things a little fuzzy, for example) but I bet I can explain anyway. Psychic powers.

Whatever the theorem states exactly, you seem to be deducing the following by applying it twice:

Obviously Impossible Theorem (OIT) [technical hypotheses] Given two non-measurable sets $M_1,M_2$, each of inner measure $0$ and outer measure $1$, there is an extension assigning to each set any value between $0$ and $1$.

That's obviously false; for example what if $M_1\subset M_2$?

(Q: But in my example $M_1$ and $M_2$ are disjoint! A: So? That doesn't change the fact that OIT is obviously impossible. See below.)

So what goes wrong with proving OIT by applying the actual theorem twice? Presumably the actual theorem is something like this:

Presumed Actual Theorem (PAT) [technical hypotheses] Given a non-measurable set $M$, of inner measure $0$ and outer measure $1$, there is an extension assigning to $M$ any value between $0$ and $1$.

The question that springs to mind is exactly what the outer measure and inner measure referred to in PAT are. They can't be defined in terms of open supersets and compact subsets, since there are no such things in the general context of the theorem.

It "must" be that in PAT the inner and outer measure are defined in terms of measurable subsets and supersets. What else could it be? You might confirm or deny that that's the definition in PAT.

Assuming so, this answers the question. Suppose you're proving OIT. You extend the measure to give $M_1$ whatever value you want between $0$ and $1$. But now when you extend again to include $M_2$ you're extending the already-extended measure. There are more measurable sets around than there were when you handled $M_1$, and in particular it may be that the outer and inner measure of $M_2$ are no longer $1$ and $0$.

(Even if your $M_1$ and $M_2$ are disjoint, it could be, and presumably is the case, that adding $M_1$ to the measurable sets changed the inner and outer measure of $M_2$...)

  • $\begingroup$ I've added some clarification, but I think it's unnecessary, I believe your answer is correct. It's easy to see where my example goes wrong. Let $V$ be a Vitali set with outer measure 1, inner measure 0, and extend the measure to it and its translates by -1 and 1 (call them $V_0, V_1, V_2$). Roughly speaking, each extension is adding measurable sets around (0,1) and its translates by -1 and 1, so each of these $V_i$ can be given measure 1. However, $(V_0 \cup V_1 \cup V_2)^c \cap [-1, 2]$ is measurable now, has measure zero, and contains the union of the remaining translates. $\endgroup$ – user28898 Aug 26 '15 at 20:01

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