I recently learn that every open set and closed set is Lebesgue Measurable. However, from what I understand, every set $S$ can be decomposed into the union of its interior, $\text{int}(S)$, which is open, and its boundary, $\partial{S}$, which is closed.
So ${S}=\text{int}(S)\cup \partial{S}$ is a finite union of measurable sets. Then shouldn't any set S should be measurable too? I know this is incorrect by the existence of non-measurable sets, so what goes wrong?
Edit: by correction in comments: $\overline{S}=\text{int}(S)\cup \partial{S}$
The problem is about the distinction between the boundary of a set in itself and the boundary in the ambient space. For a set considered as a subspace of itself, you have this decomposition, but the topological properties of Lebesgue measure are defined relative to the ambient topology, not subspace topologies of individual sets. In the ambient topology the boundary of $S$ is contained in $S$ if and only if $S$ is closed.
These kinds of subspace topology issues come up elsewhere. For example you can look into Lusin's theorem. In looking at Lusin's theorem we find that a function with a single jump discontinuity must become continuous in the subspace topology when an arbitrarily small neighborhood of the jump is removed from the domain...which is indeed the case, even though it feels a bit weird at first.