# Show that if $f:\mathbb R\longrightarrow \mathbb R$ is measurable, there is $A_\varepsilon$ s.t. $f|_{A_\varepsilon}$ continuous

Show that if $f:\mathbb R\longrightarrow \mathbb R$ is measurable, then for all $\varepsilon>0$, there is $A_\varepsilon\subset \mathbb R$ s.t. $f|_{A_\varepsilon}$ is continuous and $m(A_\varepsilon^c)<\varepsilon$.

Is my proof correct ?

Let $\mathbb R=\bigcup_{i=0}^\infty \underbrace{]-i-1,-i]\cup[i,i+1[ }_{=:A_i}$. The $A_i$ are disjoint and of finite measure. By lusin, there is $F_i\subset A_i$ s.t. $f|_{F_i}$ is continuous for all $i$ and $m(F_i^c\cap A_i)<\frac{\varepsilon}{2^{i+1}}$. Then, if we set $A_\varepsilon=\bigcup_{i=0}^\infty F_i$, then $f|_{A_\varepsilon}$ is continuous and $$m(A_\varepsilon^c)=m(\bigcup_{i=0}^\infty A_i\cap F_i^c)\underset{A_i\cap F_i^c\ disjoint}{\leq} \sum_{i=0}^\infty \frac{\varepsilon}{2^{i+1}}=\varepsilon$$ what prove the claim.

Is it correct ?

• You have a small typo (you should have $-i-1$, not $i-1$) but otherwise I think it is OK. – Ian Jan 19 '16 at 14:10
• good to know, thank you very much :-) (yes I corrected it) – user301068 Jan 19 '16 at 14:11