Let $m(\partial A)=0$, show that $A$ is Lebesgue measurable

Question

Let $A\in\mathbb{R}^d$ such that the boundary, $\partial A$, has Lebesgue measure $0$. Show that $A$ is Lebesgue measurable.

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I got a bit confused on this exercise since I'm not sure which approach to use to prove that a set is Lebesgue measurable. This is my attempt:

$$A = int(A) \cup (\partial A \cap A) \quad\text{so we don't have to worry if $A$ is open}$$ $$m(\partial A) = 0 \implies m(\partial A \cap A) = 0 $$

So then $m(A) = m(int(A))$ but $int(A)$ is an open set and hence Lebesgue measurable.

Tangential question: is it possible to write proofs showing that a set is measurable because it is an element of the Lebesgue sigma-algebra? What is the generator of the Lebesgue sigma-algebra? Just all the close rectangles in $\mathbb{R}^d$?

Answer 1

Let $B \subseteq \partial A$ be a set such that $A=A^{\circ} \cup B$. Notice that $A^{\circ}$ is measurable since it is open, and $m(B)=0$ (since is a subset of a set with measure zero). Therefore $A$ is measurable.

NOTE: $A^{\circ}$ denotates the interior of $A$.