Showing $\chi_E$ restricted to $F$ is not continuous under certain conditions. Here is a problem from a practice qualifying exam that I am having some trouble with.
Let $E$ be a closed subset of $[0,1]$ with positive measure and dense complement.  Show that if $F \subseteq [0,1]$ has measure one, then the restriction on $\chi_E$ to $F$ is not continuous.
Here is what I have so far:
$F$ is dense in $[0,1]$ because if it weren't, then it couldn't possibly have measure 1.
If $x,y \in E$ with $x<y$, then there is a $z \in E^c$ such that $x<z<y$.
Suppose $\chi_E$ restricted to $F$ is continuous and let $(x_n)_{n=1}^\infty$ be a sequence in $F$ converging to some point $a \in F$.  If $a \in E$, then there is an $N$ such that $x_n \in E$ for all $n \geq N$.  If $a \in E^c$, then there is an $N$ such that $x_n \in E^c$ for all $n \geq N$.
 A: Let us prove by contradiction. 
Suppose that the restriction on $\chi_E$ to $F$ is continuous. Then, 
$$\textrm{for all $x\in E\cap F$, there is $\delta>0$ such that $B(x,\delta) \cap F \subseteq E$.}\tag{1}$$
Since $\mu(E)>0$ and $\mu(F)=1$, we have that $E\cap F\neq \emptyset$. Let $x_0 \in E\cap F$.  So, by $(1)$, we have that there  there is $\delta_0>0$ such that $$B(x_0,\delta_0) \cap F \subseteq E \tag{2}$$
Note that $E^c$ is dense in $[0,1]$, so $B(x_0,\delta_0)\cap E^c\neq \emptyset$. Since $E$ is closed, $E^c$ is open, so we have that  $B(x_0,\delta_0)\cap E^c$ is a non-empty open set. So $\mu(B(x_0,\delta_0)\cap E^c)>0$.  From $(2)$, we have that
$$(B(x_0,\delta_0)\cap E^c)\cap F= (B(x_0,\delta_0)\cap F)\cap E^c =\emptyset$$ 
So $(B(x_0,\delta_0)\cap E^c) \subseteq F^c$, but since $\mu(F)=1$, we have 
$$ 0 < \mu(B(x_0,\delta_0)\cap E^c) \leqslant \mu(F^c) =0 $$
Contradiction.
Remark: In fact, the above proof proves that the restriction on $\chi_E$ to $F$ is not continuous at any point $x\in E\cap F$.
