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Is this a true claim or false? Recall that $m$ is Lebesgue measure and $m^\ast$ is Lebesgue outer measure. Also, $\mathcal{B}(\mathbb{R})$ in this case represents the Borel $\sigma$-algebra on $\mathbb{R}$.

I know that since $m^\ast(A) = 0$, then it can be shown that $A$ is Lebesgue measurable and hence $m(A) = 0$. I also know that since $B,C$ are Borel measurable, then they are Lebesgue measurable. My current strategy is to find a $B$ and $C$ that are Borel measurable and hence Lebesgue measurable such that $A = B \setminus C$ since $A$ is Lebesgue measurable. I'm not sure how to go about this.

Perhaps the claim isn't true?

Thanks for any help.

GNUSupporter 8964民主女神 地下教會
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Darkdub
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    If $B, C$ are in the Borel $\sigma-$algebra then so is $B \setminus C$. So it boils down to whether you think every measure zero set is Borel. –  Feb 18 '18 at 20:05

1 Answers1

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As the comment suggests, for all Borel sets $B,C$, $B \setminus C$ is a Borel set.

Since there exists a non Borel Lebesgue measure zero set $A$, $A$ can't be a Borel set, so the claim is not true.

GNUSupporter 8964民主女神 地下教會
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  • Perfect. Do we have a clear/simple example of a set $A$ with $m^\ast(A) = 0$ such that $A$ is Lebesgue measurable but not Borel measurable? Or only that we can guarantee the existence of one via the link you gave? Thanks. – Darkdub Feb 19 '18 at 00:08
  • @Darkdub, I can't think of any example simpler than the pre-image of a non-measurable subset of a Cantor set under $x \mapsto x + \phi(x)$, where $\phi$ is the Cantor Lebesgue function. The simplicity of this example can be subjective. When I first heard that in a tutorial, I was lost. In fact, I've enhanced my memory of this through answering your question. – GNUSupporter 8964民主女神 地下教會 Feb 19 '18 at 08:42