1.22

Assume to the contrary that there does exist an $f$, that's continuous everywhere, and $f = \chi_{[0,1]}$ a.e... (From here on out, let's just call it $\chi$.)

We have, $\forall \epsilon > 0,\hspace{0.25cm} \exists \delta > 0$ s.t. $|x - y| < \delta \hspace{0.25cm} \Rightarrow |f(x) - f(y)| < \epsilon$.

Choose $x = 1$ and $\epsilon = \frac{1}{4}$.

We know $\exists \delta > 0$ such that $|f(y) - f(1)| < \frac{1}{4}$ $\forall y \in (1,1+\delta) \hspace{0.25cm} (\dagger$).

Since $f$ is continuous everywhere, we have $\lim_{x \to 1}f(x) = 1 = f(1)$. It follows from $\dagger$ that: $$-\frac{1}{4} < f(y) - 1 < \frac{1}{4}$$ ...thus $$f(y) \in (\frac{3}{4},\frac{5}{4})$$ ...on $(1, 1+\delta)$. This is a contradiction, since $m\big((1, 1+\delta)\big)$ = $\delta$ > 0, and $\chi\big((1, 1+\delta)\big) = \lbrace 0 \rbrace$.