I was working with one of my friend about this problem and I have said 2 things
$1 \text{ $f$ continuous a e } \rightarrow \exists N \text{ null set such that } f\vert_{\mathbb R \setminus N} \text{ is continuous} $
$2$ $\forall A $ open set $f^{-1}(A)=(f^{-1}(A)\cap N) \cup f^{-1}(A)\cap (\mathbb R\setminus N)$ the first one is measurable because subset of a set of null maeasure and the other also because $f^{-1}(A)\cap (\mathbb R\setminus N)= f\vert_{\mathbb R \setminus N}^{-1}(A)$ and $f\vert_{\mathbb R \setminus N} \text{ is continuous}$ so the preimmage will be $U \cap \mathbb (R\setminus N)$ where $U$ is an open set
my friend said no because if a function is continuous a e does not mean it is the restriction of a continuous function I said it is right but now I have said that the restriction is continuous and it's all a different story . There is some mistake ?
