Lusin's theorem states that if $f:[a,b]\rightarrow\mathbb{R}$ is a Lebesgue measurable function, then for any $\epsilon>0$ there exists a compact subset $E$ of $[a,b]$ whose complement has Lebesgue measure less than $\epsilon$ and a function $g$ continuous relative to $E$ such that $f=g$ on $E$. Now the way Lusin's theorem is usually proven is by using the approximation theorem for $L_1$ functions, which states that if $f$ is a Lebesgue integrable function, there exists a continuous function (with compact support) $g$ such that the Lebesgue integral of $f-g$ is less than $\epsilon$.
My question is, is it possible to prove Lusin's theorem without using this approximation theorem?