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Let $F$ be the set of bijective Borel-measurable functions $f \colon [0,1] \to [0,1]$ that preserve the Lebesgue measure.

Is it the case that for every non-Lebesgue-measurable set $A \subset [0,1]$, there exists a countable family $\{f_n\}_{n \in \mathbb{N}} \subset F$ such that $\ \bigcup_{n \in \mathbb{N}} f_n(A)\,$ contains a Lebesgue-measurable set of positive measure?

If so, then there is no natural way to extend the Lebesgue measure to include more null sets.

(Conversely, if not, then it seems reasonable to regard all the counterexemplary sets as "kind-of-null sets".)

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    $\begingroup$ Your title question is the opposite of the question asked in the post. I'd recommend an edit to change one of them or the other, in order to avoid confusion. $\endgroup$ – Joel David Hamkins Jun 20 '17 at 11:58
  • $\begingroup$ Regarding the conversely part, you will have trouble proving countable additivity of the resulting notion of "sort-of-null" sets. See also this question. $\endgroup$ – jonathanverner Jun 20 '17 at 13:23
  • $\begingroup$ I've edited the title as suggested. (I guess I had to edit the title rather than the question, since a negative answer has already been given.) $\endgroup$ – Julian Newman Jun 20 '17 at 15:08
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The answer is no, by a construction using the axiom of choice.

We shall build a counterexample set $A$ by a transfinite recursive process of length continuum. At each stage, we shall promise that certain elements are in $A$, in order to ensure that $A$ will be non-null, and that other elements are not in $A$, in such a way so as to prevent $\bigcup_n f_n(A)$ from containing a particular positive-measure Borel set. But at each stage, we will have made fewer than continuum many such promises altogether, and this fact will enable the construction to proceed to later stages.

To begin the construction, observe that there are continuum many Borel functions and therefore continuum many countable families $\{f_n\}$ of bijective Borel functions that you consider, and also there are continuum many Borel sets. So let us fix a well-ordered enumeration $\langle \{f_n^\alpha\}_n,B_\alpha\rangle$, for $\alpha<\mathfrak{c}$, of all pairs of such objects.

At stage $\alpha$, we consider first the possibility that $B_\alpha$ might be a measure-zero Borel set containing the set $A$ we aim to construct. In order to prevent this, if $B_\alpha$ is measure zero, then there must be some $c_\alpha\notin B_\alpha$ about which we have not yet made any promises, and we promise now that $c_\alpha\in A$. This will ensure that $A$ is not contained in this particular Borel measure-zero set, and therefore, since all such Borel measure-zero sets will eventually be considered, it will ensure that $A$ does not have measure zero.

Next, still at stage $\alpha$, we consider the possibility that $B_\alpha$ might be a positive-measure set contained in $\bigcup_n f^\alpha_n(A)$. Since we have made so far fewer than continuum many promises about $A$, it follows that we have promised fewer than continuum many elements altogether to be in this union. But if $B_\alpha$ has positive measure, then it must have size continuum, and so there is a real $b_\alpha\in B_\alpha$ about whose pre-images $a_\alpha=(f^\alpha_n)^{-1}(b_\alpha)$ we have not yet made any promises. Let's now promise that none of these particular $a_\alpha$ are in $A$, which is countably many additional promises at this stage. This will ensure that $\bigcup_nf^\alpha_n(A)$ does not contain this particular positive-measure Borel set.

By design, the construction ensures that $A$ is not measure zero, yet there is no positive-measure Borel set contained in $\bigcup_n f_n(A)$ for any Borel bijections $f_n$. And so $A$ has the features necessary to be the desired kind of counterexample.

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