Luzin's theorem states that: let $f:[a,b]\rightarrow R$ be an a.e. finite function, $f$ is measurable iff $\forall \epsilon \geq 0: \exists \phi_\epsilon$ continuous on $[a,b]$ and $\mu\{x: f(x)\neq \phi_\epsilon (x)\} \leq \epsilon $
Why doesn't it imply that a measurable function $f$ equals a.e. to a continuous function $\phi_\epsilon$?