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[SOLVED] Measurable Function (Another Question)

TheBigBadBen

Active member
May 12, 2013
84
Is it true that if \(\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}\) is a measurable function and \(\displaystyle E\subset\mathbb{R}\) is measurable, then \(\displaystyle f(E)\) is measurable? What if f is assumed to be continuous?

I think that the answer is no for the first and yes for the second, but I have no idea how to prove/disprove either.
 

TheBigBadBen

Active member
May 12, 2013
84
Indeed, for the first question the answer is no: real analysis - Range of function measurable? - Mathematics Stack Exchange

For the second one, the range of $\Bbb R$ of a continuous function is connected. What are the connected subsets of the real line? Are they measurable?
As it ends up, my intuition for the second problem was totally off. In fact, the counterexample handily provided for my other question
http://www.mathhelpboards.com/f50/measurable-function-4789/
does the job here.

We have the following argument:
Let \(\displaystyle f(x)\) be the Cantor function, and let C be the cantor set. Note that \(\displaystyle f(C) = [0,1]\), since f is non-increasing on all points outside of C.

Define \(\displaystyle g:[0,1]\rightarrow [0,2]\) by \(\displaystyle g(x) = x + f(x)\). Since g maps every interval outside of C to an interval of the same length, we can deduce that \(\displaystyle m(g(C))=1\). By Vitali's theorem, there is a non-measurable set \(\displaystyle A\subset g(C)\). Note that \(\displaystyle B:=g^{-1}(A)\) is a subset of C. Because B is a subset of a null set, B is null and hence measurable.

Thus, we have \(\displaystyle g(B) = A\). g is a continuous (and hence measurable) function that takes a measurable set, B, to a non-measurable set, A. Thus, the answer to both questions is no.
 

girdav

Member
Feb 1, 2012
96
You are right. For the second question I misread the question, I believed you asked about $f(\Bbb R)$. I've now edited.

Your counter-example seems correct.