Are their simple/natural examples of real-valued Borel-measurable random variables whose image is not a Borel set? Something that occurs "naturally"?

I am teaching Doob's lemma (for two real-valued random variables $X$ and $Y$, $X$ is $\sigma(Y)$-measurable iff there exists a Borel-measurable function $f:\mathbb{R}\to\mathbb{R}$ such that $X=f(Y)$) and the main difficulty in the proof comes from the fact that $Y(\Omega)$ is in general not a Borel set. So I am wondering if there is a "natural" example that I can use to convince 4th year students that this "pathology" can naturally come up.

It is easy to construct examples, e.g., choose $A\subseteq \mathbb{R}$ any set that is not a Borel set, and equip it with the $\sigma$-algebra $\mathcal{A}=\{A\cap B; B\in \mathcal{B}(\mathbb{R})\}$, where $\mathcal{B}(\mathbb{R})$ denotes the $\sigma$-algebra of Borel sets in $\mathbb{R}$. Then the inclusion $X:(A,\mathcal{A})\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable and has $A$ as image, so its image is not a Borel set. But this feels like cheating...


2 Answers 2


An analytic set that is not a Borel set...see this post* from long ago.

Such an analytic set is a continuous image of $[0,1] \setminus \mathbb Q$, and thus a Borel image of $[0,1]$.

*From: [email protected] (Gerald Edgar) Newsgroups: sci.math Subject: Re: Real Measurable, non-Borel. Date: 7 Oct 1993 08:13:10 -0400 Organization: The Ohio State University, Dept. of Math. Message-ID: <[email protected]> References:

In $\[email protected]> [email protected] (eli lapell) wrote: $\gt$What is a set of real numbers which is measurable but not Borel? Or just not Borel, period ??

An explicit example of a set of real numbers that is measurable (indeed, analytic) but not Borel [due to Lusin, Fundamenta Math. 10 (1927) p. 77]:

the set of all real numbers x with continued fraction expansion
x = a[0] + 1/(a[1] + 1/(...))
such that, for some positive integers r[1] < r[2] < ..., we have a[r[i]] divides 
a[r[i+1]] for all i.

Other examples of analytic sets that are not Borel can be given in (complete separable) metric spaces other than the line:

In the space K[0,1] of nonempty compact subsets of [0,1] with the Hausdorff
metric:  The subset consists of the uncountable compact subsets.  [Hurewicz, 1930]


In the space C[0,1] of real-valued continuous functions on [0,1] with the
unform metric:  The subset consists of the differentiable functions.
[Mazurkiewicz, 1936]

I like this example, which is as natural as can be an example with sets that are not Lebesgue measurable. Start from the Cantor function $f:[0,1]\rightarrow \mathbb{R}$, and consider $h(x):= x+f(x)$, which is a homeomorphism $[0,1]\rightarrow[0,2]$. On each interval on the complement of the Cantor set $C$ this functions is a translation. Therefore $|h([0,1]\setminus C)|=|[0,1]\setminus C|=1$. Thus $|h(C)|=|[0,2]\setminus h([0,1]\setminus C)|=1$. So there exists a non measurable subset $V$ of $ h(C)$; let $W$ be $h^{-1}(V)\subset C$. Finally, the homeomorphism $h$ maps this Lebesgue measurable set $W$ into the non-measurable set $V$.

Also note that any Lebesgue, non Borel set in $h(C)$ is mapped by the homeomorphism $h^{-1}$ into a Lebesgue, non Borel subset of $C$.

  • $\begingroup$ (now I see you wrote "Borel measurable". Here $W$ is Lebesgue measurable, but of course, not a Borel set) $\endgroup$ Commented Jan 29, 2013 at 18:50
  • $\begingroup$ I wrote $h^{-1}$ for $h$, now fixed. Note that $h^{-1}$ is also interesting, and closer to what you want: since it is a homeomorphism, it bijects Borel subsets. And sends any subset of $h(C)$ (non-Lebesgue, Lebesgue and non-Borel, etc) into a (Lebesgue) subset of $C$. $\endgroup$ Commented Jan 30, 2013 at 11:44
  • 2
    $\begingroup$ I suppose that $|\cdot|$ denotes the measure of a set, rather than its cardinality... $\endgroup$
    – Asaf Karagila
    Commented Feb 1, 2013 at 18:35
  • $\begingroup$ Indeed, the Lebesgue measure $\endgroup$ Commented Feb 2, 2013 at 10:32

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