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 $0$?

Let $f$ be such a function. Since continuity is local, we may say that $f$ is continuous on $(0,1)$. Let 
$$U = \{x \in (0,1): f(x) \neq \chi_{(0,1)}.\}$$
Then $U$ has measure zero because $f$ is assumed to be equal to $\chi_{[0,1]}$ almost everywhere. Then $U$ is open in $(0,1)$ because $f$ and $\chi_{(0,1)}$ are continuous functions on $(0,1)$. Since $(0,1)$ is open in $\Bbb{R}$, it follows $U$ is open in $\Bbb{R}$. But now no open set in $\Bbb{R}$ has measure zero which is a contradiction.
