Consider a Cantor like set $C$ with measure $1>\epsilon>0$ on the interval $[0,1]$. Is it possible to find a measurable set $F \subset [0,1]$ with $m(F)=1$ such that $\displaystyle \chi$$_c|_F$ is continuous.
Here $\displaystyle \chi$$_c$ is the characteristic function on the Cantor like set and $\displaystyle \chi$$_c|_F$ is the restriction of the function to $F$ . I have a feeling that it is not possible but I am not sure how to approach this. If somebody could help it would be great. Thanks.
Both $F$ and $[0,1]\setminus C$ must be dense in $[0,1]$ (for different reasons: the former because it has measure 1 and the latter because $C$ is nowhere dense). Moreover, since $m(F) = 1$ and $m(C) > 0$, we know that $F \cap C \not= \emptyset$. Take $x \in F \cap C$. Any interval around $x$ contains points of $F \setminus C$, which shows the function in question cannot be continuous at $x$. Obviously I left out some details, but is the gist. There may be a better proof, I didn't think about it for too long.