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So, I have recently stumbled upon the following problem while studying measure theory.

Given a Vitali set $V\subset [0,1]$ and a positive measure set $A \subseteq [0,1]$, prove that their intersection is not measurable.

While solving it, I have found an easy counterexample (just take $V$ a Vitali set constructed from $[0,0.1]$ and $A=[0.9,1]$). So, I am quite puzzled now, how can it be false since it is also from the book Mathematics++: Selected topics beyond the basic courses by Kantor, Matousek and Samal? So, have I misunderstood something there? Or is there really a mistake in the book? And also, how can the statement be salvaged? I found one approach which consists of adding the condition that the outer measure of $V$ is greater than $1-m(A)$. If I did not make any mistake, then it is relatively easy to prove the statement assuming that. However, what are alternative salvages?

Note: I know that there is a similar question here Intersection with Vitali Set not measurable But, there they don't mention anything about the counterexample, so I am really puzzled since it is really basic.

frgt
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    I agree with you that the claim you write is false (more easily, take a Vitali set $V$ such that $\frac12\in V\subseteq \left[0,\frac12\right]$ and $A=\left[\frac12,1\right]$ ). Maybe (just a guess, really) it works if you add the condition that $A\cap V$ has positive outer measure, but really it should be the author's job to publish an errata. – Sassatelli Giulio Dec 23 '22 at 14:59
  • @SassatelliGiulio Thanks! That seems to be a nice idea, I will try it! – frgt Dec 23 '22 at 15:01
  • @SassatelliGiulio In fact your salvage should be easy to prove using the fact that every measurable subset of a Vitali set is of measure 0. The proof of the fact can be found here https://math.stackexchange.com/questions/2628721/measurable-subset-of-vitaly-set-has-measure-zero-proof – frgt Dec 25 '22 at 13:33
  • It's a re-statement of the thing I suggested, so yes it is. – Sassatelli Giulio Dec 26 '22 at 13:57

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