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The problem defines a density point $x\in[0,1]$ for a Borel set $A\subset [0,1]$ if $$ \lim_{\varepsilon \rightarrow 0^+} \frac{\mu([x-\varepsilon,x+\varepsilon]\cap A)}{2\varepsilon}=1.$$Denote all the density point of $A$ to be $A^*$. The problem asks to show $$A^*\text{ is Borel}, \mu(A\,\Delta\, A^*)=0, \forall \text{Borel } A\subset[0,1]$$ where $\mu$ is the Lebesgue measure and $\Delta$ means symmetric difference.

The point here is that the professor asks to use Dynkin's $\pi-\lambda$ theorem to prove and he note that interval $I$ would have $\mu(I\,\Delta\, I^*)=0$.

I manage to show $A^*$ is Borel but do not know how to show the second part. I let $\pi$ system be all the intervals in $[0,1]$ and the $\lambda$ system to be $S=\{A\text{ is Borel}:\mu(A\,\Delta\, A^*)=0\}$. I got trouble in checking that if $A\in S$ then $A^c\in S$ and if $A_1,\ldots,A_n,\ldots\in S$, $\bigcup A_n\in S$.

Indeed, I am not sure that whether Dynkin's theorem could be applied here or make the problem easier. Or we still need to go through the proof using Vitali covering lemma.

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Let $C$ be the the collection of intervals. We know that for any $I \in C$ the Theorem holds.

Use that the $\sigma$-algebra generated by $C$ (the Borel sets) coinsides with the Dynkin class generated by $C$.

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  • $\begingroup$ This does not really help. The problem is that it is hard (impossible?) to show (without using Lebesgue's differentiation theorem) that the class of "good sets" (i.e. those for which $\mu(A \Delta A^\ast) = 0$) form a Dynkin class. $\endgroup$ – PhoemueX Jan 17 '15 at 23:03
  • $\begingroup$ Use the regularity of Lebesgue measure. $\endgroup$ – Tomas Persson Jan 18 '15 at 5:54
  • $\begingroup$ Could you please give some more details? I know what regularity of the Lebesgue measure is, but I do not see how this would imply the claim. $\endgroup$ – PhoemueX Jan 18 '15 at 10:10
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    $\begingroup$ Indeed, I even got trouble in checking open and closed set are in the Dynkin class. For open set, its point density set is as large as the open set and is contained in the closure of the open set but then how to show the difference between the open set and the density point set has measure 0. The boundary of the open set might not be zero so I can not proceed. $\endgroup$ – Brian Ding Jan 18 '15 at 16:47
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    $\begingroup$ Well, no. You still have to show that the elements of the generated Dynkin class satisfy the same density condition as the intervals - this is equivalent to the OP's question. What you have written is correct, but it doesn't answer the question. To emphasise, just because something is true for the intervals does not make it automatically true for the generated Dynkin class. (Otherwise if something was true for the intervals it would automatically be true for the Borel sets.) $\endgroup$ – copper.hat Jan 20 '15 at 23:46

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