# Pullback probability measure : how to pullback the finite uniform probability?

push forward:

For a measurable map $f : (\Omega_1, \mathscr{F}_1, \mathbb{P}_1) \to (\Omega_2, \mathscr{F}_2)$ textbooks often define the push forward measure $\mathbb{P}_2$ of $\mathbb{P}_1$ by : $\forall F_2 \in \mathscr{F}_2$, $\mathbb{P}_2(F_2) = \mathbb{P}_1(f^{-1}(F_2))$.

pull back: (for a special case)

Given a set $\Omega$ and a family of finite probability spaces with finite uniform probability measure : $(C_i, \mathscr{C}_i, \mathbb{P}_i)$ and a family of maps $f_i : \Omega \to C_i$ : $$f_i : \Omega \to C_i \quad \text{ where } (C_i, \mathscr{C}_i, \mathbb{P}_i) \text{ is a uniform finite probability space}$$ How can we define a $\sigma$-algebra $\mathscr{F}$ and a probability measure $\mathbb{P}$ on $\Omega$ such that each $f_i$ is a measurable map from $(\Omega, \mathscr{F})$ to $(C_i, \mathscr{C}_i)$ ? $$\mathscr{F}, \mathbb{P} \quad \text{ such that } \quad f_i : \Omega_i \to C_i \quad \text{ is \mathscr{F}/\mathscr{C}_i-measurable }$$

Notes:

• The finite uniform probability is $\mathbb{P}(F) = \mathrm{card}(F) / \mathrm{card}(\Omega) \quad \forall F \in \mathscr{F}$
• I'm interested in probability spaces so we might consider independence assumptions.

• There is a functional analysis mathoverflow question related to pullback measure, but it's too complex for me to grasp.