Let $\lambda_d$ be the Lebesgue measure on $\mathbb{R}^d$. Let $A \subset \mathbb{R}^d, \ddot{A}$ be the interior of $A$ and $\bar{A}$ be the closure of $A$. Show: if $\lambda_d(\bar{A})=\lambda_d(\ddot{A})$, then $A$ is Lebesgue-measurable and $\lambda_d(A)=\lambda_d(\bar{A})=\lambda_d(\ddot{A})$.
Proof: $\bar{A}$ is Lebesgue-measurable: Since $\bar{A}$ is closed, it is Borel-measurable (every closed set is Borel-measurable). Since Borel-measurable sets are Lebesgue-measurable, $\bar{A}$ is also Lebesgue-measurable.
$\dot{A}$ is Lebesgue-measurable: $\dot{A}$ is the interior of $A$. The interior of a set is open, and open sets are also Borel-measurable and hence Lebesgue-measurable.
$A$ is Lebesgue-measurable: $A$ is the difference between the closure $\bar{A}$ and the edge $\partial A$ of $A$:
$A = \bar{A} \setminus \partial A.$
Since closed sets (like $\bar{A}$) and edge sets (like $\partial A$) are Borel-measurable, the difference $A$ is also Borel-measurable and therefore Lebesgue-measurable.
Now that we have shown that $A$ is Lebesgue-measurable, we can show the equality of the Lebesgue measures:
$\lambda_d(A) = \lambda_d(\bar{A} \setminus \partial A) = \lambda_d(\bar{A}) - \lambda_d(\partial A).$
Since $\bar{A}$ is Lebesgue-measurable, $\lambda_d(\bar{A})$ is well-defined. The edge $\partial A$ is a null set since $\lambda_d(\partial A) = 0$ (since $\partial A$ is a null set). Therefore, we get:
$\lambda_d(A) = \lambda_d(\bar{A}) - \lambda_d(\partial A) = \lambda_d(\bar{A}) - 0 = \lambda_d(\bar{A}).$
Is it right?
The following proof works in the case that $\lambda(\dot{A})=\lambda(\bar{A})<\infty$:
Note that $A\setminus\dot{A}\subseteq\bar{A}\setminus\dot{A}$ and $\lambda(\bar{A}\setminus\dot{A})=\lambda(\bar{A})-\lambda(\dot{A})=0$. Since the Lebesgue measure is complete, we obtain that $A\setminus\dot{A}$ is measurable and thus $A=\dot{A}\cup\left(A\setminus\dot{A}\right)$ is, too. Moreover, we get that $\lambda(A)=\lambda(\dot{A})+\lambda(A\setminus\dot{A})=\lambda(\dot{A})$.