E is measurable set , is {$\sqrt x|x∈E$} measurable?

Question

E is Lebesgue measurable subset of [0,1], is {$\sqrt x|x∈E$ } Lebesgue measurable subset of [0,1]?

First,I want to use lipschitz to prove it.But $f(x)=\sqrt x$ is not lipschitz on [0,1] and I can't find the counterexample.

Can you give me some suggestion.Thank you!!!

Is $E$ Borel measurable or Lebesgue measurable? – Jacky Chong Oct 08 '16 at 07:31 — Jacky Chong, Oct 08 '16 at 07:31
special case of this question – Greg Martin Oct 08 '16 at 09:06 — Greg Martin, Oct 08 '16 at 09:06

Greg Martin · Answer 1 · 2016-10-08T18:13:18.543

2

For $n\ge1$ let $E_n = E \cap [\frac1n,1]$. Then, since $f$ is Lipschitz on $[\frac1n,1]$, the set $\sqrt{E_n}$ is measurable. Therefore $\sqrt E \setminus\{0\} = \big( \bigcup_{n=1}^\infty \sqrt{E_n} \big)$ is also measurable (and the single point $0$ won't ruin that if it's in $E$).

edited Oct 08 '16 at 18:13

answered Oct 08 '16 at 09:09

Greg Martin

92,241

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is $\sqrt E = \big( \bigcup_{n=1}^\infty \sqrt{E_n} \big) \setminus{0}$ or $\sqrt E \setminus{0}= \big( \bigcup_{n=1}^\infty \sqrt{E_n} \big) $ ? – Chih Chia Chang Oct 08 '16 at 09:40
You are right about that! – Greg Martin Oct 08 '16 at 18:13

E is measurable set , is {$\sqrt x|x∈E$} measurable?

1 Answers1