Let $A$ be an uncountable Borel subset of $\mathbb{R}^n$, and consider the Lebesgue measure on $\mathbb{R}^n$. Assume the axiom of choice (if you need it). Does there exist a Lebesgue measurable set $B \subset A$, which has zero Lebesgue measure and is not Borel?

I have no idea of the answer. I know Lebsegue measure theory, and more generally, abstract measure theory, but not the the theory of Polish spaces and analytic sets (the so called descriptive set theory, which could be the key to answer the question). Any help is welcome.

  • $\begingroup$ Now, I am not sure (wait for more expert people), but analytic sets that are not Borel should make the job. $\endgroup$
    – Kolmin
    Commented Apr 14, 2016 at 18:25
  • $\begingroup$ Shouldn't be that deep. And Lebesgue sets are a proper sub-family of the analytic sets. $\endgroup$ Commented Apr 16, 2016 at 13:04
  • 1
    $\begingroup$ There are many similar questions here, for example this. $\endgroup$
    – iiivooo
    Commented Apr 18, 2016 at 11:20
  • $\begingroup$ @Bungo I know that the answer is yes in the special case of the Cantor set, but the issue is to ascertain the existence of such a $B$ for a generic uncountable Borel set $A$. $\endgroup$ Commented Apr 20, 2016 at 22:26

2 Answers 2


Yes, by a simple counting argument:

  • If $B$ is Borel and uncountable, it has cardinality $2^{\aleph_0}$. EDIT: see below.

  • The number of subsets of $B$ is therefore $2^{2^{\aleph_0}}$.

  • "Most" of those will not be Borel, since there are only continuum-many Borel sets.

  • However, if $B$ is null, all of them will be measurable.

So the result follows since every uncountable Borel set contains an uncountable null Borel set.

A proof of that last claim, that every uncountable Borel set has an uncountable null Borel subset: Every uncountable Borel set has a perfect subset (https://en.wikipedia.org/wiki/Perfect_set_property), so we might as well assume our $A$ is perfect: that is, closed and without isolated points. (Actually, all we need is that $A$ be closed.)

Since $A$ is perfect, it's easy to show that there is a sequence $\{S_i\}$ of finite sets of disjoint closed intervals such that

  • $S_1$ contains a single interval, $S_2$ contains two, . . . , $S_n$ consists of $2^{n-1}$-many intervals.

  • Each $I\in S_n$ contains exactly two $J_0, J_1\in S_{n+1}$.

  • Every interval in $S_n$ has length at most $2^{-2^n}$.

  • For each $I\in S_n$, the intersection $I\cap A$ is uncountable.

Now consider the "limit set" $X=\bigcap_{i\in\mathbb{N}} (\bigcup S_i)$. It's a good exercise to show that $X$ is a closed uncountable subset of $A$ with measure zero.

EDIT: Based on the comments below, I think I should probably explain the first bullet point above: that every uncountable Borel set has cardinality continuum. (One way this is phrased is "the Continuum Hypothesis holds for Borel sets.") Of course, on the face of it this seems to need CH (the statement $\vert\mathbb{R}\vert=\aleph_1$); so the whole point is that there's something special about Borel sets which prevents "weird" behavior.

Although don't put too much faith in the Borel sets! Weird behavior is still possible, especially in the absence of the axiom of choice: see this paper by Arnold Miller, which constructs a Dedekind-finite infinite Borel set, and also this question.

First off, it's worth proving the much simpler claim that every uncountable closed set has size continuum. While this is fairly easy, it's not trivial; and even pushing it up to the next level of the Borel hierarchy ($G_\delta$ - countable intersections of open sets) is somewhat hard. So proving it for all Borel sets is a daunting task.

The solution is via games, specifically the perfect set game. For a set $A\subseteq\mathbb{R}$, the game $PSG(A)$ is defined as follows:

  • It's a game of perfect information between two players, Large (I) and Small (II).

  • On turn $n$, player I plays a pair of disjoint intervals $I_{n, 0}$ and $I_{n, 1}$; player II then picks one of them (which becomes $J_n$).

  • We demand that $I_{n+1, i}\subseteq J_n$, that is, the intervals are nested appropriately.

  • After $\omega$-many turns, we've defined a sequence of intervals $J_0\supseteq J_1\supseteq...$. Let their intersection be $J$. Player I wins iff $J\cap A\not=\emptyset$.

We can prove that

  • Player I has a winning strategy iff $A$ contains a perfect subset, and

  • Player II has a winning strategy iff $A$ is countable.

So the goal is to show:

  • If $A$ is Borel, then $PSG(A)$ is determined: one player or the other has a winning strategy.

This is a consequence, for instance, of Borel determinacy (see https://en.wikipedia.org/wiki/Borel_determinacy_theorem), but that proof is very hard. In particular, this wasn't how the result was originally proved. Instead, the original proof is to "unfold" the game $PSG(A)$ for $A$ Borel into an open game on a certain set $X$; then by Gale-Stewart, that game is determined, so we are done. Note that this is actually how the proof of full Borel determinacy goes, too - but that unfolding process is much harder. This argument actually proves that analytic sets have the perfect set property; I'm not aware of any proof that the Borel sets have the perfect set property which doesn't immediately apply to analytic sets too (besides the argument via full Borel determinacy, which I mean come on that's just silly).

I'm not going to include this whole argument, although it's really cool, since it's pretty long - see the "Games people play" section in Kechris' book for a great exposition. But hopefully the above gives some context for the result.

  • $\begingroup$ Doesn't this assume the continuum hypothesis? $\endgroup$
    – Jason
    Commented Apr 20, 2016 at 22:51
  • $\begingroup$ @Jason No - why would it? (CH is the statement $2^{\aleph_0}=\aleph_1$, it doesn't have any bearing here.) $\endgroup$ Commented Apr 20, 2016 at 22:52
  • $\begingroup$ @Jason Note that it is outright provable in ZFC that the Borel hierarchy has height $\omega_1$ (key step - union of countably many countable sets is countable), and that each level in the Borel hierarchy consists of continuum-many sets; so in ZFC we prove that there are only continuum-many Borel sets. $\endgroup$ Commented Apr 20, 2016 at 22:58
  • 1
    $\begingroup$ Hmm, I was thinking more of your first statement. Why must an uncountable Borel set have cardinality of the continuum? $\endgroup$
    – Jason
    Commented Apr 20, 2016 at 23:06
  • $\begingroup$ @Jason Borel sets have the perfect set property, provably in ZFC. (This follows - overkill! - from Borel determinacy, via the Perfect Set Game.) $\endgroup$ Commented Apr 20, 2016 at 23:49

I fill the gaps in the proof sketched by Noah Schweber that every uncountable Borel set of $\mathbb{R}^n$ has an uncountable null Borel subset (since the proof is too long I can't post it as a comment). As Noah noted, since very analytic set has the perfect set property, we can assume that $A$ is a perfect set. More generally, I will prove that if $A$ is a closed and uncountable subset of $\mathbb{R}^n$, then we can find $X \subseteq A$, with $X$ perfect set of zero Lebesgue measure. I call a cell a set which is defined, for some $a_i \leq b_i$ for $i=1, \dots, n$, as consisting of all $x=(x_1, \dots, x_n) \in \mathbb{R}^n$, with $a_i \leq x_i \leq b_i$ for all $i=1, \dots, n$.

First, we claim there exists a sequence $(S_i)$, where each $S_i$ is a finite family of pairwise disjoint cells such that:

  • each $S_k$ contains $2^{k-1}$ cells;

  • each $I \in S_k$ contains exactly two elements $J_0, J_1$ of $S_{k+1}$;

  • every $I \in S_k$ has sides whose length is at most $2^{-2^k}$;

  • for each $I \in S_k$, the set $I \cap A$ is uncountable.

To prove this, for every $v=(v_1, \dots, v_n)$, with $v_1, \dots, v_n\in \mathbb{Z}$, and every nonnegative integer m, let $I_{m,v}$ the dyadic cell consisting of all $x=(x_1, \dots, x_n) \in \mathbb{R}^n$ such that for all $i=1, \dots, n$ we have \begin{equation} 2^{-m}v_i \leq x_i \leq 2^{-m}(v_i + 1). \end{equation} Assume that for some $m$ and $v$, the set $I_{m,v} \cap A$ is uncountable. Then we claim that there are infinitely many integers $p > m$ such that, for each such $p$ there are $s, t \in \mathbb{Z}^n$, such that $I_{p,s}$ and $I_{p,t}$ are disjoint, both contained in $I_{m,v}$, and such that both $I_{p,s} \cap A$ and $I_{p,t} \cap A$ are uncountable. If not, there would exist some integer $N$ such that, for all $p > N$, there exists a dyadic cell $C_p$ of side of length $3 \times2^{-p}$ which contains all the cells of side $2^{-p}$ which have an uncountable intersection with $A$. We would get a sequence of compact sets $(C_p)_{p > N}$, with $C_{p+1} \subset C_p$, so that $\bigcap_{p>N} C_p$ contains exactly one point. But then $A$ would be countable, a contradiction.

From the property just proved, we can easily construct our sequence $(S_i)$.

Now, define $X= \bigcap_{i \in \mathbb{N}} (\bigcup S_i)$. Clearly $X$ is closed, since each $\bigcup S_i$ is. Moreover, if $x \in X$, then for each $k$, $x$ belongs to some $I \in S_k$. Since each point $y \in S_k$ has a distance from $x$ which is not greater than $n \times 2^{-2^k}$, $x$ turns out to be a limit point of a sequence of points of $A$, and since $A$ is closed, we conclude that $x \in A$. Finally, let $J_0, J_1$ be the elements of $S_{k+1}$ contained in $I$, and $x \in J_0$. Then there is some $y \in J_1 \cap X$ (since a decreasing sequence of compact sets has nonempty intersection). The distance between $x$ and $y$ is not greater than $n \times 2^{-2^{k}}$. We conclude that $x$ is a limit point of $X$, so that $X$ is perfect. To see that $X$ has zero Lebesgue measure just note that each $S_k$ has measure at most $2^{k-1} \times 2^{-n 2^k}$. QED


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .