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The question is

If $E\subset[0,1]$, $|E|=0$ and $f(x)=x^3$, show $|f(E)|=0$, where |E| denotes Lebesgue measure of $E$.

Can anyone provide a hint on this? Thank you!

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    $\begingroup$ Hint: The Mean Value Theorem or whatever shows that $|f(x)-f(y)|\le3|x-y|$ for $x,y\in[0,1]$. (Now say you've covered $E$ by open intervals of total length less than $\epsilon$...) $\endgroup$ Jul 9, 2015 at 18:51
  • $\begingroup$ Thank you! That is very helpful. $\endgroup$
    – Tony
    Jul 10, 2015 at 0:54

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As the set $E$ is of measure zero, for all $\epsilon > 0$ you can find a countable union of open segments $I_n$ with $$E \subset \cup_{n \ge 0} I_n$$ and $$\sum_{n \ge 0} \ell(I_n) < \epsilon$$ That is the definition of $E$ having a measure equal to zero.

You have to prove the same thing for $f(E)$.

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  • $\begingroup$ I solved it. Thank you! $\endgroup$
    – Tony
    Jul 10, 2015 at 0:54

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