An uncountable measurable subset of $\Bbb R$ containing no nonempty perfect set

Question

$\newcommand\R{\Bbb R}$Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set?

Of course, such a set $S$, if it exists, must be of measure $0$.

"The axiom of choice implies the existence of sets of reals that do not have the perfect set property", so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?

On the other hand, it "is well-known that assuming the countable axiom of choice that every uncountable Borel set contains a [nonempty] perfect set." So, a set $S$ in question, if it exists, cannot be Borel.

Answer 1

A set $B\subseteq\mathbb R$ is a Bernstein set if $B$ contains no nonempty perfect set while having nonempty intersection with every nonempty perfect set; in other words, if neither $B$ nor $\mathbb R\setminus B$ contains a nonempty perfect set. The existence of Bernstein sets is an easy consequence of the axiom of choice (see below); the axiom of choice is essential here since Bernstein sets are not Lebesgue measurable.

Let $B$ be a Bernstein set and let $C$ be the Cantor set. Then $C\cap B$ and $C\setminus B$ are sets of Lebesgue measure zero, neither of which contains a nonempty perfect set, and it's clear that at least one of them is uncountable. In fact $C\cap B$ and $C\setminus B$ both have cardinality $\mathfrak c=2^{\aleph_0}$ because $C$ can be partitioned into $\mathfrak c$ nonempty perfect sets (since $C$ is homeomorphic to $C\times C$).

Construction of Bernstein sets. Since there are $\mathfrak c$ nonempty perfect subsets of $\mathbb R$, and since each of them has cardinality $\mathfrak c$, we can enumerate them in a transfinite sequence $(P_\alpha:\alpha\lt\mathfrak c)$ and we can recursively choose $a_\alpha,b_\alpha\in P_\alpha\setminus\bigcup_{\xi\lt\alpha}\{a_\xi,b_\xi\}$ with $a_\alpha\ne b_\alpha$; then $\{a_\alpha:\alpha\lt\mathfrak c\}$ and $\{b_\alpha:\alpha\lt\mathfrak c\}$ will be Bernstein sets. In a similar way we can construct $\mathfrak c$ disjoint Bernstein sets.

Answer 2

Yes, AC gives us a continuums-sized measure 0 set without the perfect set property.

For the construction of just any continuums-sized subset of $\mathbb{R}$, what matters is that $\mathbb{R}$ has cardinality $\mathfrak{c}$, that there are $\mathfrak{c}$-many perfect subsets of $\mathbb{R}$ and that each of those itself has cardinality $\mathfrak{c}$. This allows us to minimally well-order all our requirements (the perfect subsets of $\mathbb{R}$), and then to deal with them one by one without every running out of options as to what to do about it.

Since we know continuums-sized measure 0 subsets of $\mathbb{R}$ (eg the Cantor middle third set), we can run the construction of a continuums-sized set without perfect subsets inside our favourite continuums-sized measure 0 set, and obtain a continuums-sized measure 0 set without the perfect set property.

Answer 3

Assuming the axiom of choice, there is a (may I say very natural) uncountable set of real numbers that is measure-zero with regard to any $\sigma$-finite, complete, regular measure that measures all the Borel sets (so this includes the Lebesgue measure). In other words it is absolutely measurable. Furthermore this set has no perfect subset.

This was known by Luzin and Sierpinski (maybe Lebesgue too) as early as 1918. Below we identify $\mathbb{R}$ with the Cantor space $2^\omega$.

Theorem. (ZFC) Say two reals $x,y$ from $\mathsf{WO}$, the set of reals coding well-orderings, are equivalent, just in case code well-orderings of the same ordertype. Let $A\subseteq 2^\omega$ be such that it contains one and only one element from each equivalence class. Then $A$ is measurable (of measure zero) and doesn't contain any perfect subset.

This can be proved using an amusing forcing argument in Fenstad, Jens Erik; Normann, Dag, On absolutely measurable sets, Fundam. Math. 81, 91-98 (1974). ZBL0275.02057. The article attributes the argument to Dag Normann's Thesis. It's very amusing because it makes no appeal to Shoenfield absoluteness or poset combinatorics, but rather uses the peculiar property that forcing doesn't add ordinals. To be fair, all the forcing/measure-theoretic machinery can be found in the construction of Solovay's model.

The modern forcing proof. Let $A$ be the set choosing a code for each countable ordinal. Clearly $|A|=\omega_1$. First observe that $A$ has no perfect subset: this is because any perfect subset $P$ will be (with possibly an extra singleton) a Borel subset of $\mathsf{WO}$ and so by boundedness must be countable, which is impossible.

For measurability, let $M$ be a countable transitive model of (enough of) $ZFC$. So now $A = W_0\cup W_1$, where $W_0$ codes the ordinals in $M$ and $W_1$ codes those not in $M$. Now, $W_0$ is a countable set of reals, and hence has measure zero. Next we show $W_1$ can be covered by a countable union of measure zero sets, which implies that $A$ has measure zero.

Consider random forcing over $M$. We claim that any real $r\in W_1$ is not random over $M$. This is because if it were, then the generic extension $M[r]$ of $M$ would have the same ordinals as $M$, and hence the ordinal coded by $r$ is in $M$, contradicting that $r\in W_1$.

Now since each $r\in W_1$ fails to be random over $M$, by Solovay's characterization of random-genericity, $r$ belongs to a Borel set of measure zero coded in $M$. But there can be only countably many such sets, so $W_1$ is covered by a countable union of measure zero sets. $\square$

The Luzin-Sierpinski proof. First notice that $\mathsf{WO}$ can be partitioned into Borel set $\{P_\alpha\mid\alpha<\omega_1\}$, where each $P_\alpha$ is the set of reals coding well-ordering of type $\alpha$. Second, since $\mathsf{WO}$ is $\Pi^1_1$, it is measurable, and by usual properties of Lebesgue measure, $\mathsf{WO} = \bigcup_{n\in\omega} N\cup M_n$, where $N$ has measure zero and each $M_n$ is closed.

By $\Sigma^1_1$-boundedness, each $M_n$ is bounded in $\mathsf{WO}$. Write $\alpha_n$ as the least upper bound of (the ordinals coded in) $M_n$. Note that this implies that for all $\beta>\alpha_n$, we have $M_n\cap P_\beta=\emptyset$. In other words, $M_n=\bigcup_{\alpha<\alpha_n} M_n\cap P_\alpha$.

But now observe that, since $P_\alpha\cap A$ only has a single element, $M_n\cap A$ is at most countable and hence measure zero. Therefore, $A = A \cap \mathsf{WO} = \bigcup_{n\in\omega} (A\cap N) \cup (A\cap M_n)$. This writes $A$ as a countable union of measure zero sets, and hence $A$ has measure zero. $\square$

The classical proof can be found in:

Lusin, N.; Sierpiński, W., Sur quelques propriétés des ensembles mesurables ((A))., Krak. Anz. 1918, 35-48 (1918). ZBL46.0296.04.