How to show a continuous cannot differ a characteristic function by measure $0$? How to show that there is no continuous function on $\mathbb R$ such that it differs from $\chi_{[0,1]}$, the characteristic function of $[0,1]$, by a measure (Lebesgue measure) of $0$?

Hint: Let $f$ be such function. For any $\delta > 0$, $(-\delta, 0)$ must contain an element $x$ such that $f(x) = 0$. Similarly, $(0, \delta)$ must contain an element $x$ such that $f(x) = 1$. 
