Suppose we have a measurable, surjective function $f: (X, \Sigma) \to (Y, \Omega)$ and $X,Y$ are two locally compact metric spaces (can even assume $X,Y$ are subsets of $\mathbb{R}^n$). Given a probability measure $\nu$ on $Y$, I would like to construct a "pullback measure", i.e. a probability measure $\mu$ on $X$ such that the pushforward $f_*\mu = \nu$. I can already prove the existence of such a measure under these conditions for $f, X, Y$ using a Riesz argument, but I'm wondering if an explicit expression for $\mu$ can be derived.
I've been looking into the disintegration of measure, i.e. defining a kernel $K: Y \times \Sigma \to [0,1]$ such that $K(y, \cdot)$ is a probability measure and $K(\cdot, B)$ is a measurable function and defining for all $A \in \Sigma$ $$\mu(A) = \int_Y K(y, A)d\nu(y).$$ As a (perhaps naive?) first attempt, I considered taking $K(y, \cdot)$ to be the uniform measure over $f^{-1}(\{y\})$, i.e. $$K(y, A) = \frac{\lambda(A \cap f^{-1}(\{y\}))}{\lambda(f^{-1}(\{y\}))}$$ where $\lambda$ is the Lebesgue measure over the space $f^{-1}(\{y\})$. Then $K(y, f^{-1}(B)) = \mathbf{1}_{B}(y)$ for every $B \in \Omega$ so defining $\mu$ as above we get $f_*\mu = \nu$. The problem is that $f^{-1}(\{y\})$ can have measure 0. I'm wondering if this approach can be modified to work, if alternate constructions exist, or if more conditions on $f$ needs to be enforced (for instance $\mu(\{y: \mu(f^{-1}(\{y\}) = 0\}) = 0$).
Thanks!
There is a standard way to prove this when $X$ and $Y$ are analytic subsets of Polish (separable and completely metrizable) spaces. However, it is not constructive and uses a measurable selection theorem. The original result (which has been frequently rediscovered) is from
where it is Lemma 2.2.