Given a measurable set $A$, if we square every element $B=\{a^2: a\in A\}$, is $B$ still measurable? I have no idea about this question, can someone help me?
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I think you want to specify A to be a lebesgue measurable subset of the real numbers – Dionel Jaime Aug 02 '17 at 17:28
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Yes, assume that A is lebesgue measurable. – user335468 Aug 02 '17 at 17:30
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Hmm, the collection of $A \subseteq [0, \infty)$ such that $f(A)$ is Borel (where $f(x) = x^2$) is a $\sigma$-algebra and it contains all the open intervals, so the collection contains all Borel subsets of $[0, \infty)$. – Daniel Schepler Aug 02 '17 at 17:46
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Let $B_+=\bigl\{a^2\,|\,a\in A\cap[0,+\infty)\bigr\}$ and let $B_-=\bigl\{a^2\,|\,a\in A\cap(-\infty,0]\bigr\}$. Since $B=B_+\cup B_-$, if you prove that $B_+$ and $B_-$ are both measurable, then $B$ is measurable.
Now, note that if $s(x)=\sqrt x$, then $B_+=s^{-1}\bigl(A\cap[0,+\infty)\bigr)$. Since $s$ is continuous and $A\cap[0,+\infty)$ is measurable, $B_+$ is measurable. A similar argument shows that $B_-$ is measurable.
José Carlos Santos
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1I think this works if "measurable" means "Borel measurable", but not "Lebesgue measurable". There are continuous functions such that the inverse image of a Lebesgue measurable set is not Lebesgue measurable. – Alex Zorn Aug 02 '17 at 17:53
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