Proving that if $\lambda (\{ f^\prime = 0\}) = 0$, then $f$ is increasing

Question

I am trying to show that

Let $f:[a,b]\to R$ be a differentiable map, with $f'(x) \geq 0$, for all $x \in [a,b].$ If the set $U = A = \left\{x \in [a,b] : f'(x)=0 \right\}$ has null content, prove that $f$ is an increasing map.

I tried to start this proof by proving the thesis by contradiction. Let $f$ be a non-increasing map. Then, if we take $[x,y] \subset [a,b]$, there exists $y>0$ such that $f(y)=f(x)$ right? I got the tip that for all $z \in [x,y]$, $f(x)+f(z)=f(y)$ but I can't understand why does that happen. Can anybody give me an idea if why is this valid? Thank you so much!

Answer 1

It's obvious that $x\leq y$ implies $f(x)\leq f(y)$. The challenge is to replace $\leq$ with $<$.

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Suppose you had $x<y$. Then, because $(x,y)$ has positive Lebesgue measure, there must be some $c\in(x,y)$ such that $f'(c)>0$.

Given that $f'(c)>0$, there exists $r>0$ such that $f(c+z)>f(c)$ for all $z\in(0,r)$ and also $c+r<y$.

Thus, $f(y)\geq f(c+r)>f(c)\geq f(x)$.

Answer 2

Let $x < y$. By the mean value theorem, $f(y) = f(x) + (y-x)f'(c)$ for some $c$. Since $(y-x)f'(c) \geq 0$, it follows that $f(x) \leq f(y)$.

Towards a contradiction, suppose $f(x) = f(y)$.

For any $z \in (x,y)$, by the mean value theorem again, we have $f(x) \leq f(z) \leq f(y) = f(x)$, which implies $f(z) = f(x)$. Differentiating with respect to $z$, we find $f'(z) = 0$ for all $z \in (x,y)$, so that $$\lambda(\{f' = 0\}) \geq \lambda((x,y)) = y-x > 0,$$ which is the contradiction.

It follows that $f(x)\neq f(y)$, so $f(x) < f(y)$.