trying to prove: If $f$ is continious and is lebesgue-almost-everywhere constant, then it is constant

Question

I was wondering if this claim is true, and if it is then how would one try to prove it: If $f\in C[0,1]$ (and thus is continuous) and is Lebesgue-almost-everywhere constant, then it is constant.

It looks very logical to me, since in order to be continuous and travel between two different values, $f$ would need to change over at least a small interval, which has a positive measure.

Therefore, I figured out that I would need to use the Intermediate Value Property of continuous functions. Yet I didn't really manage to use it properly in that case.

Any help would be blessed, thanks!

Answer 1

If $\;f\;$ is a continuous function in $\;[0,1]\;$ and $\;f(x_1)\neq f(x_2)$ for some $x_1,x_2\in [0,1]\;$, suppose $\;f(x_1)> f(x_2)$, then by continuity, there exists $\;\epsilon_1,\epsilon_2 >0$, such that $f(x)>\frac{f(x_1)+f(x_2)}{2}\forall x\in(x_1-\epsilon_1,x_1+\epsilon_1),$ and $f(x)<\frac{f(x_1)+f(x_2)}{2}\forall x\in(x_2-\epsilon_2,x_2+\epsilon_2)$. From this, you can find a contradiction to "constant a.e" (For example, if $f=c$ a.e with $c<\frac{f(x_1)+f(x_2)}{2}$, can you find a contradiction?).

Answer 2

I think Devil's staircase is a counterexample. It is constant on the complement of the Cantor-set and continous, but the function itself is non-constant.

Answer 3

To get a pretty straightforward contradiction prove the following lemma:

Lemma: If $\;f\;$ is a continuous function in $\;[0,1]\;$ and $\;f(w)>0\;$ for some $\;w\in [0,1]\;$ , then there exists $\;\epsilon >0\;$ s.t. $\;f(x)>0\;\;\forall\,x\in (w-\epsilon\,,\,w+\epsilon)\;$ (take half neighborhoods if $\;w=0\;\;or\;\;w=1\;$ )