Stein 1.22

Prove that there is no continuous function $f$ on $\mathbb{R}$ s.t. $f= \chi_{[0,1]}$ a.e.

$\textit{Proof} :$
Suppose to the contrary that $f\in{C(\mathbb{R})}$ and $f = \chi_{[0,1]}$ a.e. With $f$ continuous, $\forall \epsilon > 0 $, $\exists \delta = \delta(x,\epsilon) > 0 $ s.t. $|x-z| \leq \delta \Longrightarrow |f(x) - f(z)| \leq \epsilon$. Fix $\epsilon = \frac{1}{2}$, let $x\in{[0,1]}$ and $z <0$. Then, $$\chi_{[0,1]}(x) - \chi_{[0,1]}(z) = 1-0 = 1 \quad \forall x\in{[0,1]}, z<0 $$ and so $$f(x) - f(z) = 1 \quad \text{a.e. } x\in{[0,1]}, z<0 $$ Let $\delta > 0$ and let $x\in{[0,1]},z<0$ s.t. $0 < m(|x-z|)=x-z \leq \delta$.
However, $f(x)-f(z) = 1$ a.e. In particular, $f(x_0)-f(z_0) = 1 $ for some $x_0\in{[0,1]}$, $z_0 < 0$, $|x_0-z_0| \leq \delta$...contradiction. (If we could not find such an $x_0$ and $z_0$ as above, that would contradict the assumption that $f = \chi_{[0,1]}$ a.e.) $\blacksquare$