Approximation of Borel sets

Let $\nu$ be a finite Radon measure on $\mathbb{R}^2$ and denote the Lebesgue measure on $\mathbb{R}^2$ by $\mathcal{L}^2$. Assume that $\nu<<\mathcal{L}^2$. We denote the boundary of $A\subset\mathbb{R}^2$ by $\partial A$, that is, \begin{equation} \partial A\equiv \{z\in\mathbb{R}^2\ |\ \forall \varepsilon>0, \ B(z, \varepsilon)\cap A\neq\emptyset\text{ and } B(z, \varepsilon)>0\cap\mathbb{R}^2\setminus A\neq\emptyset\} \end{equation} Set \begin{equation} \mathcal{G}\equiv\{A\subset\mathbb{R}^2\ |\ A\text{ is Borel and }\nu(\partial A)=0\}. \end{equation}It is not hard to show that $\mathcal{G}$ is a $\sigma$-algebra on $\mathbb{R}^2$.

Proof

The collection of Borel sets is a $\sigma$-algebra. We have $\emptyset$ and $\mathbb{R}^2$ are Borel. Also $\partial\emptyset=\partial\mathbb{R}^2=\emptyset$, and consequently, \begin{equation} \nu(\partial\emptyset)=\nu(\partial\mathbb{R}^2)=\nu(\emptyset)=0. \end{equation}Thus, $\emptyset, \mathbb{R}^2\in \mathcal{G}$.

If $A\in \mathcal{G}$, then it is Borel, and consequently, $\mathbb{R}^2\setminus A$ is Borel. Furthermore, $\partial A=\partial(\mathbb{R}^2\setminus A)$, and therefore, \begin{equation} \nu(\partial(\mathbb{R}^2\setminus A))=\nu(\partial A)=0. \end{equation}In particular, $\mathbb{R}^2\setminus A\in\mathcal{G}$.

Lastly, suppose that $\{A_i\}_{i=1}^{\infty}\subset\mathcal{G}$. Then each $A_i$ is Borel, and therefore, the union $\cup_{i\in\mathbb{N}}A_i$ is Borel. We have that $\nu(\partial A_i)=0$ for each $i\in \mathbb{N}$. Since \begin{equation} \partial\left(\bigcup_{i\in\mathbb{N}}A_i\right)\subset\bigcup_{i\in\mathbb{N}}\partial A_i, \end{equation}it follows that \begin{equation} \nu\left(\partial\left(\bigcup_{i\in\mathbb{N}}A_i\right)\right)\leq\sum_{i=1}^{\infty}\nu(\partial A_i)=0, \end{equation}that is, $\nu(\partial(\cup_{i\in\mathbb{N}}A_i))=0$. In particular, $\mathcal{G}$ is closed under countable unions.

Question. Suppose that $\nu<<\mathcal{L}^2$ on $\mathcal{G}$ in the following sense: for all $\varepsilon>0$ there is a $\delta>0$ such that if $A\in \mathcal{G}$ and $|A|<\delta$, then $\nu(A)<\varepsilon$.

If $A$ is Borel but $\nu(\partial A)>0$, can $A$ be approximated by sets in $\mathcal{G}$? In particular, given $\varepsilon>0$, does there exist an $E\in\mathcal{G}$ such that $A\subset E$ and $\mathcal{L}^2(E\setminus A)<\varepsilon$?

• It is not hard to show that $G$ is a $\sigma$-algebra on $\mathbb R^2$. Really? Have you tried it? (I assume that $\partial A$ is the topological boundary of $A$, if it is not, you'd better tell us what it is before more sarcastic questions come your way). – fedja Dec 13 '17 at 3:53
• The boundary of a countable union is not contained in the union of the boundaries. Write $\mathbb Q^2$ as a union of singletons. – Jochen Wengenroth Dec 13 '17 at 7:48
• @JochenWengenroth: And indeed, if you take some finite weighting of Lebesgue measure, the boundary of a singleton has measure zero, but the boundary of $\mathbb{Q}^2$ doesn't. So the premise of this question is false. – Nate Eldredge Dec 13 '17 at 7:49
• @JochenWengenroth, Ah I see. Thank you. – Nirav Dec 13 '17 at 8:01