Theorem (Lusin): Let $f$ measurable and finite valued on $E$ a set of finite measure. For all $\varepsilon>0$ there is a closed set $F_\varepsilon\subset E$ s.t. $f|_{F_{\varepsilon}}$ is continuous and $m(E\backslash F_\varepsilon)\leq \varepsilon$.
The proof goes like:
Let $(f_n)_n$ a sequence of step function s.t. $f_n\longrightarrow f$. For all $n$, there is $E_n\subset E$ s.t. $f_n$ is continuous outside of $E_n$ with $m(E_n)<\frac{1}{2^n}$.
my questions
1) If $f_n$ is a step function don't we have that the set of discontinuity is nulle and thus $m(E_n)=0$ ?
2) When we say that $f_n$ is continuous outside of $E_n$, does it mean that $f_n$ is continuous on each point of $E\backslash E_n$ or does it mean that $f_n|_{E\backslash E_n}$ is continuous ? (of course, the first assertion condition implies the second one).
No problem for the rest of the proof.