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This question already has an answer here:

In my module notes, if $A$ is a Borel and $m(A)=0$, then it is not necessarily true that any subset $B$ of $A$ (with $m(B=0)$) is Borel.

So I am wondering if there is a null set that is not a Borel set?

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marked as duplicate by user228113, Daniel W. Farlow, Silvia Ghinassi, hardmath, Shailesh Mar 6 '16 at 1:34

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  • $\begingroup$ Examples would be better $\endgroup$ – 王向皓 Mar 5 '16 at 13:15
  • $\begingroup$ There must exist non-Borel null sets by a simple cardinality argument. The set of all Borel sets has the cardinality of the continuum, $c$. But, any subset of Cantor's ternary set $C$ has measure zero and the cardinality of ${\cal P}(C)=2^c$, which is strictly larger than the cardinality of the contiuum. $\endgroup$ – Mark McClure Mar 5 '16 at 13:29
  • $\begingroup$ @G.Sassatelli: I believe that the OP is asking about a different thing: a measure is called "complete" if from $m(A) = 0$ and $B \subset A$ it results that $B$ is measurable and $m(B) = 0$. Well, It seems that the OP asks for an example of a Borel measure that is not complete. $\endgroup$ – Alex M. Mar 5 '16 at 13:41
  • $\begingroup$ @AlexM. I did not think of that generality. It can be as you say, but I honestly see no element excluding that the OP refers to the real case only. In the end, the proof of what I linked is "$(\Bbb R,\mathcal B,\lambda^1)$ is not a complete measure space, while $(\Bbb R,\mathcal M,\lambda^1)$ is". I thought that by "examples would be better" he meant "explicit construction of non-Borel null sets". $\endgroup$ – user228113 Mar 5 '16 at 14:05
  • $\begingroup$ @G.Sassatelli Yes, that's what I mean. Thank you~ $\endgroup$ – 王向皓 Mar 5 '16 at 14:59
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We can guarantee the existence of one, but I do not know if one can be found in a direct way.

Let $\phi$ be the Cantor Lebesgue function and define $\psi(x)=\phi(x)+x$. Then $\psi$ is a strictly increasing continuous function mapping $[0,1]$ to $[0,2]$, and moreover, maps the Cantor set onto a set of positive measure.

Let $C$ be the Cantor set. Then since every set that has positive outer measure contains a non-measurable set, $\psi(C)$ contains an non-measurable set, $A$. Then $\psi^{-1}(A)$ is a subset of $C$ so it is measurable with measure zero, and it is not Borel, because the image of a Borel set through a continuous function is measurable.

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See this enlightening Wikipedia discussion.

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  • $\begingroup$ Cantor set is Borel measurable, of Lebesgue measure zero and of the cardinality of the continuum $c$ , The cardinality of the class of all Borel measurable subsets of the real axis $R$ is equal to $c$. The cardinality of the class of all subsets of the Cantor set is equal to $2^c$. Hence there is a subset of the Cantor set which is not Borel measurable. $\endgroup$ – George Mar 6 '16 at 6:11

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