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Let $f: \mathbb{R} \to \mathbb{R}$ continuously differentiable. Show that if $E \subset \mathbb{R}$ has measure $0$ then $f(E)$ has measure $0$.

what i have so far: Since $E$ has measure 0 there is a coverage of open intervals $I_n$ such that $\sum \mu(I_n)$< $\varepsilon$.Then from each $I_n$ we create the intervals $I_n\cap (-\infty, 0)$ and $I_n\cap (0,\infty)$ lets call the collection of these intervals $J_n$, they are a countable collection covering $E$ and then $$\sum \mu(J_n)=\sum \mu(I_n)$$.

I'm not sure I'm taking a useful path and I can't think of how to continue, I'd appreciate any help :)

Fermax
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    $C^1$ implies locally Lipschitz (by the mean-value theorem and the fact that $f'$ is bounded on compact sets). Locally Lipschitz implies that we can control how much intervals expand when we map via $f$. This will allow you to prove $f$ sends measure zero sets to measure zero sets (if you search on this site, you'll find more detailed answers). – peek-a-boo Aug 29 '21 at 09:29
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    What about constant functions? – Zuy Aug 29 '21 at 09:58

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