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Is it possible to construct a non-measurable set in $[0,1]$ of a given outer measure $x \in [0,1]$? This will probably require the axiom of choice. Does anyone have a suggestion?

Edit: I forgot to mention that the set has to be in $[0,1]$ which seemed "clear" in my head, but which obviously isn't.

Edit 2: Now it is clear that I just need a non-measurable set with outer measure 1, from that I can make any other one. I have once seen somewhere that they do this with transfinite recursion (if I recall correctly), but I don't remember where and how because I didn't know that transfinite recursion was back then.

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3 Answers 3

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One can proceed directly with the Vitali construction, without need for any scaling.

Namely, just carry out the Vitali construction, but ensure that the resulting Vitali set is contained in the interval $[0,a]$. That is, declare that two reals are equivalent if their difference is rational, and observe that every real is equivalent to a real in the interval $[0,a]$. Let $V\subset [0,a]$ select exactly one element from each equivalence class. Observe that the rational translations $V+q$, working modulo $1$ so as to regard $V+q\subset [0,1]$, are disjoint and union up to the whole interval $[0,1]$. It follows easily that $V$ is not measurable and has inner measure $0$, since otherwise the translates modulo $1$ would have infinite measure inside $[0,1]$, which is impossible. Thus, the complement $[0,a]-V$ is non-measurable and has outer measure $a$, as desired.

I would like to note that this argument shows that the outer measure of the classical Vitali set is not determined by the usual features of that set. For example, if all you know about a Vitali set $V$ is that it is contained in $[0,1]$ and contains exactly one element of each equivalence class, then it follows that $V$ is non-measurable and has inner measure $0$, but for all you know, $V$ is actually contained in a very tiny interval $[0,\epsilon]$, and could have tiny outer measure.

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This was my answer to the question as originally phrased: Once you have one nonmeasurable set with finite outer measure, you can get all positive values by rescaling.

With the additional requirement that the set is contained in $[0,1]$, this argument will work as long as you have a nonmeasurable subset of outer measure $1$. Sierpiński and Lusin showed that $[0,1]$ can be decomposed into uncountably many nonmeasurable sets of outer measure $1$ in an article published in 1917, but I do not know the details. Robert Israel sketches a construction of a Vitali subset of [0,1] with outer measure 1 in a 1997 sci.math post.

(Oh, and strictly speaking the answer is no, you can't have $x=0$; I suppose you meant $x\in(0,1]$.)

As for the axiom of choice, I'm no expert but have been told that it is consistent with ZF that all subsets of the real line are Lebesgue measurable.

Peter LeFanu Lumsdaine points out in a comment that the result is due to Robert Solovay. The 1970 article is on JSTOR, and begins as follows:

alt text

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    $\begingroup$ Yes, it is indeed consistent with ZF that all sets are Lebesgue measurable — this is a beautiful construction due to Robert Solovay. So the construction of any non-measurable set does indeed require Choice. $\endgroup$ Dec 16, 2010 at 22:42
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    $\begingroup$ @Jonas T: I guess I should have guess the condition you just added to your question, that you want the set to be contained in $[0,1]$. OK, so far then my answer only reduces this to finding subsets with outer measure 1. Sierpinksi and Lusin gave examples of this in 1917, but I do not know the details: emis.de/cgi-bin/JFM-item?46.0294.01 $\endgroup$ Dec 16, 2010 at 22:46
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    $\begingroup$ @Jonas Meyer: Yes, I figured that out while you seemed to be typing it that the rescaling just works. I was wondering about intersecting the set with $[0,x]$ like Ross Millikan suggests. I don't see how that works. $\endgroup$
    – JT_NL
    Dec 16, 2010 at 23:03
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    $\begingroup$ @Jonas Meyer: I wonder if it is not possible to build a Vitali-like set with outer measure 1 using transfinite recursion. $\endgroup$
    – JT_NL
    Dec 16, 2010 at 23:21
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    $\begingroup$ @Jonas T: Ross's suggestion would work too, provided you have the outer measure $1$ example to begin with. If $E\cap[0,x]$ has outer measure less than $x$, then $m^*(E)=m^*(E\cap[0,x])+m^*(E\cap(x,1])\lt x+(1-x)=1$. $\endgroup$ Dec 16, 2010 at 23:24
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Let $V$ be a $\mathbb{Q}$ vector subspace of $\mathbb{R}$ of codimension at most countable. Then $\mu^*(V\cap I) = \mu(I)$ for every $I$ interval.

Proof:

We may assumed wlog $1\in V$ (just divide by some non-zero $t\in V$). Conclude $a V + b \subset V$ for all $a$, $b$ in $\mathbb{Q}$

Since a countable union of translates of $V$ covers $\mathbb{R}$ we have $\mu^*(V)>0$. So there exists an interval $I$ with rational ends such that $\mu^*(V\cap I) >(1-\epsilon) \mu(I)$. Using the rational invariance of $V$ we conclude $$\mu^*(V\cap I) >(1-\epsilon) \mu(I)$$ for every interval $I$ with rational ends. Therefore $$\mu^*(V\cap I) =\mu(I)$$ for every interval with rational ends. Since every finite interval is contained in one with rational ends, we have the above equality for every finite interval $I$.

Now, take $V$ a complement of $\mathbb{Q}$ in $\mathbb{R}$, that is $V\oplus \mathbb{Q} = \mathbb{R}$. Then $V$ is a VItali set. Consider some $x>0$. We have $\mu^*(V\cap [0,x])=x$. Complete $V\cap[0,x]$ to a Vitali subset $W$ of $[0,x]$ with $\mu^*(W)=x$.

$\bf{Added:}$ Now if $x\in (0,1]$, then $W= W_x \subset [0,x]\subset[0,1]$ is a Vitali subset of $[0,1]$ of exterior measure $x$.

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  • $\begingroup$ May I ask why is your $V$ a Vitali set? It seems to contain many duplicated representatives for every equivalence class. Isn't your $V$ actually just the irrationals? $\endgroup$
    – user760
    Nov 22, 2022 at 8:03
  • $\begingroup$ @user760: I think the definition of a Vitali set $V$ is any set $V$ such that every real number $x$ has a unique writing $x = v + q$. Now, $V$ does not have to be closed under addition, or closed under multiplication by rational scalars. But if it is ( so it's a vector space over $\mathbb{Q}$) we show above that $\mu^{\star}(V\cap I)= \mu(I)$ for every interval $I$. Now, fix any interval $J$. $V\cap J$ is not a Vitali set but can be completed to a VItali set $\subset J$. So we get a Vitali set $\subset J$ , of exterior measure $= \mu(J)$. $\endgroup$
    – orangeskid
    Nov 23, 2022 at 4:36
  • $\begingroup$ @user760: I think my write-up is not that clear. I answered a similar question on mathoverflow, maybe it's clearer there. Feel free to inquire again. Cheers! $\endgroup$
    – orangeskid
    Nov 23, 2022 at 4:38
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    $\begingroup$ Oh I just realized by "complement" you weren't talking about set complement, but algebraic complement..... $\endgroup$
    – user760
    Nov 23, 2022 at 5:05
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    $\begingroup$ @user760: Yes indeed! Now, in all this we use the axiom of choice. I've heard that there are model of set theory where all of the subsets of $\mathbb{R}$ are measurable. Interesting... $\endgroup$
    – orangeskid
    Nov 23, 2022 at 6:03

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