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We have $f:\mathbb{R}\rightarrow\mathbb{R}$ which is continunous and its derivative is also continuous. We define set $D=\{x:f'(x)=0\}$. Prove that set $f(D)$ is of 0 (Lebesgue) measure.

If we assume that $f$ is monotonic then every maximal interval on which $f$ is constant is landing on single point. Between such interval there is always a gap, so there are no intervals in $f(D)$. However, I believe that $f(D)$ can have "more" that countably infinite elements. If that's true, then I do not know how to proceed.

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    are you sure this is how the question was stated? Couldn't we just take $f$ to be a constant function and this would not hold? – 2017-01-17
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    @DanielXiang In fact, the statement in the body is correct, while the one in the title is blatantly false. – 2017-01-17
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    i don't understand, the two look the same. Do we mean to show that the set $D = \{x : f(x) = \alpha, f'(x) \neq 0\}$ for any $\alpha \in \mathbb{R}$ has 0 measure? – 2017-01-17
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    I hope that title is ok now. – 2017-01-17
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    It's true that $f(D)$ can have uncountably many values, e.g. if $D$ is the Cantor set, $g$ is a nonnegative continuous function zero exactly on $D$, and $f(x)=\int_0^x g(t)\,dt$, then $f$ will increase between any 2 points in $D$. – 2017-01-17

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