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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