Complement of a set of measure zero is dense in $[a,b]$?

Question

Assume we have a subset of measure zero $N$ in the Borel-$\sigma$-field $\mathcal{B}([a,b])$.

I think that $[a,b]\setminus N$ is a dense set. But how to prove that?

My attempt:

let $x \in [a,b]$ and $\varepsilon>0$ arbitrary then $\lbrace z \in [a,b]: |x-z|<\varepsilon\rbrace \cap ([a,b]\setminus N)\neq \emptyset$ so there are infinitely many in any neighbourdhood. Hence it is dense, right?

The reason why the intersection is non-empty is if it would be empty, then $N\cup \lbrace z \in [a,b]: |x-z|\geq \varepsilon\rbrace = [a,b]$ so $\lambda(N\cup \lbrace z \in [a,b]: |x-z|\geq \varepsilon\rbrace )=b-a$ which is wrong for $\varepsilon>0$ hence we have a contradiction.

Answer 1

Your proof is correct but the way you obtain a contradiction strikes me as somewhat awkward: I would just note that if $(x - \epsilon, x + \epsilon) \cap ([a, b] \setminus N) = \emptyset$, then $(x - \epsilon, x + \epsilon) \subseteq N$ (by definition of complement), and $\lambda((x - \epsilon, x + \epsilon)) = 2\epsilon$, contradicting the fact that $\lambda(N) = 0$.