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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?

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  • $\begingroup$ 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. $\endgroup$ – Keshav Aug 14 at 7:58
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    $\begingroup$ 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 $g \circ \phi \in D$, $\int_X{g \circ \phi\,d\mu}=\int_Y{g\,d\nu}$. $\endgroup$ – Mindlack Aug 14 at 8:04
  • $\begingroup$ 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 $Y$. $\endgroup$ – Mindlack Aug 14 at 8:46
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(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.

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Mindlack has given a good answer for the second part.

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 $f$.

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