prove the existence of a measure $\mu$ Suppose $X$ and $Y$ are compact metric spaces and $F : X \rightarrow Y$ is a continuous map from $X$ onto $Y$. If $\nu$ is a finite measure on the Borel sets of $Y$, prove that there exists a measure $\mu$ on the Borel sets of $X$ such that 
$$
\int_{Y} f d\nu = \int_{X}f \circ F d \mu
$$
for all $f$ that are continuous on $Y$.

This is pretty hard to show the existstence of $\mu$ for me (Even for my TA). Currently I am in the chapter of Riesz Representation. Can anybody give me some hints?
 A: Using the fact that $F:X\to Y$ is onto, the map 
$$F^* : C(Y) \to C(X),\ \ \ F^*f(x) := f(F(x))$$
is injective. Thus one can consider $C(Y)$ as a subspace of $C(X)$. By Hahn Banach theorem, the functional $f \mapsto \int_Y f d\nu$ on $C(Y)$ extends to a functional $L$ on $C(X)$. By Riesz representation, there is a Borel measure $\mu$ on $X$ such that 
$$L(f) = \int_X f d\mu , $$
Then 
$$\int_Y fd\nu = L(F^*f) = \int_X f\circ F d\mu\ .$$
A: There is a proof that relies on facts about the topological structure of the compact metrizable space P(X) of all (Borel) probability measures on X, mentioned in the following related thread:
Does the pushforward operator (on measures) preserve surjectiveness?
The proof is this. Let $F_*: P(X) \to P(Y)$ be the pushforward operator. We want to show that $F_*$ is onto. Since the original map $F$ is onto, each Dirac delta measure $\delta_y$ on $Y$ is in the image $F_*(P(X))$. Since $F_*$ is an affine map, each finite convex combination of Diract delta measures on $Y$ is also in the image. Such combinations form a dense subset of $P(Y)$. Since $P(X)$ is compact, the image must also be compact and hence closed. So the image must be the whole set $P(Y)$.
