# Pull back of a finite measure

by WhoKnowsWho   Last Updated August 14, 2019 07:20 AM

I came across the following problem asked in a prelim exam.

Let X and Y be compact metric spaces. Suppose, $$\phi:X\to Y$$ is a continuos surjective map. Let $$D= \{f\in C(X): f(x)=f(x’) whenever \phi(x)=\phi(x’)\}.$$

a) Show that D is a closed subspace of $$C(X)$$ and that $$D=\{g\circ \phi : g\in C(Y)\}.$$

b) Let $$\nu$$ be a finite positive Borel measure on Y. Prove that there is a finite positive Borel measure $$\mu$$ on X such that $$\mu(\phi^{-1}(F))=\nu(F)$$ for all Borel sunsets $$F$$ of Y.

As far as part a) is concerned, it is easy to prove that D is closed subspace and also that a function $$f$$ in D looks like $$g\circ \phi$$ for some g. But, I am not able to argue why g should be continuous.

My main problem is regarding the part b). While it is easy to show that $$\{\phi^{-1}(F)\}$$, where F runs over Borel subsets of Y, is a sigma algebra. And we can define a finite measure $$\mu$$ on this sigma algebra by $$\mu(\phi^{-1}(F))=\nu(F)$$. I can show that this is well defined. My trouble is that this sigma algebra on $$X$$ can be much smaller than the Borel sigma algebra on X. So, do I need to extend this measure to whole Borel sigma algebra? If yes, how can I do that? Also, I do not see the use of surjectivity of $$\phi$$ in part b). Am I correct in my understanding?

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