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#### TheBigBadBen

##### Active member

- May 12, 2013

- 84

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.

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- Thread starter
- #1

- May 12, 2013

- 84

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.

Indeed, for the first question the answer is no: real analysis - Range of function measurable? - Mathematics Stack Exchange

One can change the example to a continuous one.

One can change the example to a continuous one.

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- May 12, 2013

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As it ends up, my intuition for the second problem was totally off. In fact, the counterexample handily provided for my other questionIndeed, 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?

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.