Prove that, there exists no continuous function $f:\mathbb R\rightarrow\mathbb R$ with $f=\chi_{[0,1]}$ almost everywhere.

Question

Prove that, there exists no continuous function $f:\mathbb R\rightarrow\mathbb R$ with $f=\chi_{[0,1]}$ almost everywhere.$\textbf(Make\ sure\ that\ your\ proof\ is\ completely\ rigorous)$.

I don't know, which property to use. (It is not allowed to show it with $\epsilon-\delta-criterion$, our last topics were: Lp-Spaces, Radon-Nikodym Theorem, Riesz Representation Theorem,Lipschitz-Functions, Product measures, Fubini Theorem)

but i can't find anything to do with continuity, which of them could be useful ?

Answer 1

Hint1: if a subset of $\mathbb{R}$ has a complement with Lebesgue measure $0$ then it is dense.

Hint2: continuous functions are determined by their values on a dense set.

Answer 2

This is an alternative way to see it:

A function is continuous iff the preimage of every open set is open. Look for example at

$$f^{-1}((\frac{1}{3},\frac{2}{3}))=U$$ Why is $U \neq \emptyset$? What can you say about the measure of $U$?

Answer 3

When $f(x)=\chi_{[0,1]}(x)$ almost everywhere then any neighborhood $U:=\ ]{-h},h[\ $ of $0$ contains points $x<0$ with $f(x)=0$ as well as points $x>0$ with $f(x)=1$. Such a function cannot be continuous at $0$.