About measure theoretic interior and boundary Reference:-  Evans-Gariepy, Federer, other books and notes of geometric measure thoery.
I just want to clarify whether these definitions of measure theoretic interior 
and boundary are correct. Given a Borel regular measure $ \mu $ in $\mathbb{R}^n $, given a $\mu$-measurable subset $E \subset \mathbb{R}^n $, let $$ \psi(x,E) = \lim_{r\rightarrow 0}\frac{\mu(E\cap B(x,r))}{\mu(B(x,r))} $$ Here $\psi $ is defined wherever the limit exists. Then do we have interior and exterior as $$ Int_\mu(E) = \{ x \in \mathbb{R}^n\ |\ \psi(x,E) = 1\},\\ Ext_\mu(E) = \{ x \in \mathbb{R}^n\ |\ \psi(x,E) = 0\} $$ and boundary $$\partial_\mu(E) = \mathbb{R}^n\setminus (Int_\mu(E)\cup Ext_\mu(E)) $$ I wold be extremely obliged if someone can tell me whether these are correct and whether they imply $$ \mu(E\setminus Int_\mu(E))= \mu(E^c\setminus Ext_\mu(E)) = \mu(\partial_\mu(E)) = 0 $$ from Lebesgue differentiation theorem. Also how exactly the topological and measure theoretic boundaries are related ? There is also another question, pardon me if it's too lame. Given a $\mu $ can we define a topology on $\mathbb{R}^n $ such that it's topological interior and boundary are the measure theoretic interior and boundary ?
 A: Definitions are not statements; they cannot be correct or incorrect. They may be inconsistent with literature, or perhaps nonsensical. The definition of $\partial_\mu(x,E)$ that you have now is consistent with how the measure-theoretic boundary $\partial_* E$ is defined (when $\mu$ is the Lebesgue measure). 
However, it has a problem: if $x$ is outside of the support of $\mu$, you have $0/0$ inthe definition of $\psi$, no matter what $E$ is. Since $\psi(x,E)$ is undefined, do you put such points into $\partial_\mu(E)$? Then you'll have some strange notion of boundary. For example, if $\mu$ is a Dirac mass at $a$, then $\partial_\mu(E)=\mathbb R^n\setminus \{a\}$, no matter what $E$ is (even if it's empty). This is not going to be consistent with any topological concepts. 
The Lebesgue differentiation theorem  applies to absolutely continuous measures, that is, measures of the form $\mu(E)=\int_E w$. If $\mu$ is of this form (and $w>0$ almost everywhere), then you should get the desired conclusion  $\mu(\partial_\mu E)=0$ by writing 
$$\frac{\mu(E\cap B(x,r))}{\mu(B(x,r))} = \frac{\mu(E\cap B(x,r))}{\lambda(B(x,r))}\cdot \frac{\lambda(B(x,r))}{\mu(B(x,r))}$$
and passing to the limit in each fraction separately. The first will converge to $w\chi_E$, the second to $w^{-1}$ at "most" points $x$, making the ratio $1$ or $0$. 
