Lebesgue point of density on $[0,1]$ and Dynkin's theorem

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.

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 $C$.
• 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. – PhoemueX Jan 17 '15 at 23:03