The following is similar to a problem I found on some analysis quals:
$X, Y$ are compact metric spaces and $F: X \to Y$ is surjective and continuous, if there is a finite measure $\nu$ on the Borel subsets of $Y$, then there is a finite measure $\mu$ on the Borel subsets of $X$ such that
$$
\int_{Y} f d \nu = \int_{X} f \circ F d \mu
$$
for all continuous functions on $Y$.
I'm guessing that the Riesz Representation Theorem needs to be used here, but for that I'd need a functional defined on the continuous functions on $X$ although the statement is about continuous functions of $Y$. What is a good approach here?
Edit: Is the following true: If $g: X \to \mathbb{R}$ is continuous, then there is a continuous $f: Y \to \mathbb{R}$ such that $g = f \circ F$? We could simply define $f(F(x)) = g(x)$, but then do we know that $f$ is continuous on $Y$? If we can do this then the claim will follow:
Let $H: X \to \mathbb{R}$ be the functional
$$
H(g) = \int f d \nu
$$
where $f$ the corresponding continuous function on $Y$ as desrcibed above. Then Riesz Representation implies the result.
Best Answer
Consider $\{g \in C(X): g=f\circ F \,\, \text {for some} \, \, f \in C(Y)\}$. Since $F$ is surjective there is only one $f$ for each $g$ in this space.If we show that this is a closed subspace of $C(X)$ then we can use Hahn-Banach Theorem to show that the map $g=f\circ F \to \int f d\nu$ can be extended to an element of the dual of $C(X)$ and the proof is finished by Riesz Theorem.
Suppose $g_n=f_n\circ F \to g$ uniformly. Note that $\sup_{y\in Y} |f_n(y)-f_m(y)| \leq \sup _{x \in X} |g_n(x)-g_m(x)| \to 0$. This implies that $f_n$ converges uniformly to some $f \in C(Y)$ and it is clear that $g=f\circ F$.