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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...

edited Feb 1 at 6:50

asked Jan 29 at 18:06

Uwe Franz
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Gerald Edgar · Accepted Answer · 2013-01-29 19:25:13Z UTC

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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 $[0,1]$.

answered Jan 29 at 19:25

Gerald Edgar
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Pietro Majer · Answer 2 · 2013-01-30 11:35:24Z UTC

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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 $C$.

edited Jan 30 at 11:35

answered Jan 29 at 18:45

Pietro Majer
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(now I see you wrote "Borel measurable". Here $W$ is Lebesgue measurable, but of course, not a Borel set) – Pietro Majer Jan 29 at 18:50

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Yes, in probability one takes usually the Borel σ-algebra on the range (when this is a topological space, in standard courses it is R or Rd anyway) and an arbitrary (complete) σ-algebra on the domain (see, e.g., mathoverflow.net/questions/31603/…). Therefore it is natural that the map f:R→R in Dynkin's lemma should be Borel, too – Uwe Franz Jan 30 at 10:08

I think that is the right setting for the question. – Uwe Franz Jan 30 at 10:43

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 at 11:44

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I suppose that $|\cdot|$ denotes the measure of a set, rather than its cardinality... – Asaf Karagila Feb 1 at 18:35

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

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