# Good examples of random variables whose image is not a measurable set?

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...

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Such an analytic set is a continuous image of $[0,1] \setminus \mathbb Q$, and thus a Borel image of $[0,1]$.
*From: e...@math.ohio-state.edu (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: <29114m$e1b@math.mps.ohio-state.edu> References: In$\lt$CEIvIr....@undergrad.math.uwaterloo.ca> emlap...@undergrad.math.uwaterloo.ca (eli lapell) wrote:$\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]  - the link is broken – Nikita Evseev Aug 3 at 11:21 Fixed,${}$thanks. – Gerald Edgar Aug 3 at 20:51 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$. - (now I see you wrote "Borel measurable". Here$W$is Lebesgue measurable, but of course, not a Borel set) – Pietro Majer Jan 29 '13 at 18:50 I wrote$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$. – Pietro Majer Jan 30 '13 at 11:44 I suppose that$|\cdot|\$ denotes the measure of a set, rather than its cardinality... –  Asaf Karagila Feb 1 '13 at 18:35