# Pull back of a finite measure

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 continuous surjective map. Let $$D= \{f\in C(X): f(x)=f(x’)\ \text{whenever}\ \phi(x)=\phi(x’)\}.$$

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 subsets $$F$$ 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?

• I don't know the answer, I also came to the same issues as you, but I was also reminded of this point-set topology fact: A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Maybe this will help, I am not sure. It seemed related because you are dealing with X up to points that are not injectively mapped by phi. – Keshav Aug 14 at 7:58
• Consider the bounded linear form $L$ on $D$, such that $L(g \circ \phi)=\int{g\,d\nu}$. By Hahn-Banach and Riesz-Markov, there is a non-negative Borel measure $\mu$ with $\mu(X)=\|L\|=\nu(Y)$ such that for all $g \circ \phi \in D$, $\int_X{g \circ \phi\,d\mu}=\int_Y{g\,d\nu}$. – Mindlack Aug 14 at 8:04
• For the first part, bitesizebo’s answer seems to work, but here is my suggestion: since $\phi$ is surjective, for any closed (thus compact) $F \subset Y$, $T=(g \circ \phi)^{-1}(F)$ is a closed (thus compact) subset of $X$ and $g^{-1}(F)=\phi(\phi^{-1}(g^{-1}(F))=\phi(T)$ is a compact, hence closed ($Y$ is metric) subset of $Y$. – Mindlack Aug 14 at 8:46

(a) I believe you know that \begin{align*} g(y)=f(x), \end{align*} where $$x$$ is any point satisfying $$\phi(x)=y$$. (Note that you should write $$g(y)=something$$ instead of $$g(\phi(x))=something$$ to give a definition of a function on $$Y$$. This is very important. )
To show $$g$$ is continuous, we could show that $$g^{-1}(A)$$ is closed in $$Y$$ for every closed set $$A$$ in $$\mathbb{R}$$. Observe that \begin{align*} g^{-1}(A)=\phi(f^{-1}(A)) \end{align*} and use the compactness of $$X$$, we could conclude that $$g^{-1}(A)$$ is closed in $$Y$$. The verification of the above formula is left to you.
(b) As the remark in my first paragraph, $$\mu(\phi^{-1}(F))=something$$ is not a definition of a measure on the Borel algebra (in $$X$$). (you already pointed it out in the question) The correct definition should be \begin{align*} \mu(A)=\nu(\phi(A)), \end{align*} where $$A$$ is $$any$$ Borel set in $$X$$. To check this is well-defined, we have to show that $$\phi(A)$$ is a Borel set in $$Y$$ for any Borel set $$A$$ in $$X$$. This uses the surjectivity of $$\phi$$ and the fact that Borel algebra is the smallest $$\sigma$$-algebra containing all closed sets.
Of course, checking $$\mu(\phi^{-1}(F))=\nu(F)$$ is required (and is left to you). During the verification, you would need $$\phi(\phi^{-1}(F))=F$$, which is true when $$\phi$$ is surjective.
To show continuity of $$g$$ you use the closed map lemma. In particular $$X$$ is compact, $$Y$$ is Hausdorff and $$\phi$$ is continuous therefore $$\phi$$ is a closed map. Hence as a continuous, closed surjection, $$\phi$$ is a quotient map. Thus by the defining property of the quotient topology any $$g: Y \to Z$$ (where $$Z$$ is any topological space) is continuous if and only if $$g \circ \phi$$ is. $$g \circ \phi$$ is clearly continuous as it is equal to $$f$$.