exact duplicate of Lebesgue measurable but not Borel measurable

BUT! can you please translate Miguel's answer and expand it with a formal proof? I'm totally stuck...

In short: Is there a Lebesgue measurable set that is not Borel measurable?

They are an order of magnitude apart so there should be plenty examples, but all I can find is "add a Lebesgue-zero measure set to a Borel measurable set such that it becomes non-Borel-measurable". But what kind of zero measure set fulfills such a property?

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    $\begingroup$ See also this thread $\endgroup$
    – t.b.
    May 4, 2012 at 17:47
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    $\begingroup$ @TheChaz because nobody would ever see that comment (except for people randomly looking for this...) $\endgroup$
    – example
    May 4, 2012 at 17:48
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    $\begingroup$ It would notify Miguel and Jonas... $\endgroup$ May 4, 2012 at 17:50
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    $\begingroup$ Prove: there are $\mathfrak{c} = \#\mathbb{R}$ Borel measurable sets. Every subset of the Cantor set is Lebesgue measurable. There are $2^\mathfrak{c} = \#P(\mathbb{R})$ subsets of the Cantor set. $\endgroup$
    – t.b.
    May 4, 2012 at 17:52
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    $\begingroup$ Also relevant: math.stackexchange.com/questions/70880/… $\endgroup$
    – user83827
    May 4, 2012 at 18:52

1 Answer 1


Let $\phi(x)$ be the Cantor function, which is non-decreasing continuous function on the unit interval $\mathcal{U}_{(0,1)}$. Define $\psi(x) = x + \phi(x)$, which is an increasing continuous function $\psi: [0,1] \to [0,2]$, and hence for every $y \in [0,2]$, there exists a unique $x \in [0,1]$, such that $y = \psi(x)$. Thus $\psi$ and $\psi^{-1}$ maps Borel sets into Borel sets.

Now choose a non Borel subset $S \subseteq \psi(C)$. Its preimage $\psi^{-1}(S)$ must be Lebesgue measurable, as a subset of Cantor set, but it is not Borel measurable, as a topological mapping of a non-Borel subset.

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    $\begingroup$ This is. ${}{}{}{}$ $\endgroup$
    – leo
    May 5, 2012 at 0:36
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    $\begingroup$ Why is it possible to choose a non Borel subset $S\subset \psi(C)$ ? $C$ is the Cantor set, right ? $\endgroup$
    – Hua
    Mar 13, 2017 at 10:15
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    $\begingroup$ @bridger because $\psi(C)$ has positive measure and every set with positive Lebesgue measure contains a non-measurable set. Since Borel sets are measurable, the non-measurable set contained in $\psi(C)$ must be non-Borel. Now consider its preimage under $\psi$, you get a null set. So, it's Lebesgue measurable but it is not Borel because $\psi$ and $\psi^{-1}$ map Borel sets into Borel sets. $\endgroup$ Feb 10, 2019 at 11:38
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    $\begingroup$ Let $(a_n,b_n)$ be an open interval we removed in the Cantor construction (note$[0,1]\setminus C=\cup_{n\in\mathbb{N}}(a_n,b_n)$ and also note that $\phi$ is constant on these intervals). Then since $\psi$ is increasing $\mathcal{L}(\phi(a_n,b_n))=\psi(b_n)-\psi(a_n)=b_n+\phi(b_n)-(a_n+\phi(a_n))=b_n-a_n=\mathcal{L}(a_n,b_n)$. Then $\mathcal{L}(\psi([0,1]\setminus C))=\mathcal{L}(\cup_{n\in\mathbb{N}}\psi((a_n,b_n)))=\sum_n \mathcal{L}(\psi(a_n,b_n))=\sum_n\mathcal{L}((a_n,b_n))=\mathcal{L}([0,1]\setminus C)=1$ But $\mathcal{L}(\psi(C)\cup \psi([0,1]\setminus C))=2$ so $\mathcal{L}(\psi(C))=1$ $\endgroup$
    – edamondo
    Nov 18, 2021 at 15:00

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