I wonder if this is a simpler and correct proof of the Lusin theorem stated below. I was inspired by Stein's proof. However I think the $E_n$ in Stein's proof may be unnecessary. This is the theorem and my proof. Would you help me check it?
Theorem: $E\subseteq \mathbb{R}^d$ is measurable and has finite measure. $f:E\rightarrow \mathbb{R}^d$ is measurable and has finite value. Then $\forall\epsilon>0$ $\exists F\subseteq E$ s.t. $m(E-F)<\epsilon$ and $f$ is continuous viewed as a function on $F$. (In Stein's book $F$ is closed, but whether $F$ is closed or any measurable set is not important. )
(For Stein's proof see this post: Can the proof of Lusin's theorem in Stein-Shakarchi Real Analysis be simplified?)
My proof: First we note that $A\subseteq B$ and both $A$, $B$ are measurable $\Rightarrow m(B-A)=m(B)-m(A)$, since $m(B)=m(A)+m(B-A)$.
Define $$\bar{f}(x)=\begin{cases} f(x)& x\in E\\ 0& x\not\in E\end{cases}$$
Then $\exists \{s_n\}$ a sequence of step functions s.t. $s_n\rightarrow \bar{f}$ a.e $x\in\mathbb{R}^d$, and hence $s_n\rightarrow f$ a.e $x\in E$. If $s_n$ viewed as a function on $E$, then it is continuous at $x_0\in E$ unless $x=x_0$ is a discontinuous point of $s_n$ viewed as a function on $\mathbb{R}^d$, unless it is at the side of a rectangle. Hence $\forall n$, $\{x\in E| s_n \text{ is discontinuous at $x_0$, viewed as a function on $E$ }\}$ has measure zero. Define $E_0=\{x\in E| \exists n,\ s_n \text{ is discontinuous at $x_0$, viewed as a function on $E$ }\}$ $\Rightarrow m(E_0)=m(\{x\in E| \exists n,\ s_n \text{ is discontinuous at $x_0$, viewed as a function on $E$ }\})=0$. Now select $A\subseteq E$ s.t. $s_n\rightarrow f$ uniformly on $A$ and $m(E-A)<\epsilon$, then $f$ is continuous viewd as a function on $E_0\cap A$, and $m(A\cap E_0)=m(A)$ since $m(A\cap E_0)=m(A-(A\cap(E-E_0)))=m(A)-m(A\cap(E-E_0))=m(A)$, and hence $m(E-(E_0\cap A))=m(E)-m(E_0\cap A)=m(E)-m(A)=m(E-A)<\epsilon$, i.e. $E_0\cap A$ is the $F$ we desire.