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 $\mathcal{L}^2(E\setminus A)<\varepsilon$?