Prove that $(\mu_1 \otimes \mu_2)\circ {\Pi_1}^{-1}=\mu_1$

Question

Suppose $\Pi_1 :(\Omega_1 \times \Omega_2, \mathcal{F_1} \otimes \mathcal{F_2} ,\mu_1 \otimes \mu_2) \rightarrow (\Omega_1, \mathcal{F_1},(\mu_1 \otimes \mu_2) \circ {\Pi_1}^{-1})$ is a projection map which basically maps $(x,y) \rightarrow x$.

We are to prove that $(\mu_1 \otimes \mu_2)\circ {\Pi_1}^{-1}=\mu_1$.

I was thinking of showing that the two measures are equal on a generator of $\mathcal{F_1}$, then we can say that they agree over $\mathcal{F_1}$. Bu how to explicitly choose a generator such that this holds? Can anyone help?

Answer 1

Assuming these are probability measures (otherwise this is not true):

Since these are general $\sigma$-algebras, you will not be able to use your approach. You would only be able to do that if you already knew a nice generator for $\mathcal{F}_1.$ Instead, the solution is quite straightforward if you realize what $\Pi_1^{-1}$ is exactly. Suppose $A\in \mathcal{F}_1$, then $\Pi_1^{-1}(A) = A\times \Omega_2.$ Now fill this in: $$(\mu_1\otimes \mu_2)\circ\Pi_1^{-1}(A) = (\mu_1\otimes \mu_2)(A\times \Omega_2) = \mu_1(A)\cdot \mu_2(\Omega_2) = \mu_1(A).$$

Answer 2

Let $F\in\mathcal F_1$.

Then $\Pi_1^{-1}(F)=F\times\Omega_2$ so that - if $\mu_2$ is a probability measure:$$(\mu_1\otimes\mu_2)\circ\Pi_1^{-1}(F)=(\mu_1\otimes\mu_2)(F\times\Omega_2)=\mu_1(F)\times\mu_2(\Omega_2)=\mu_1(F)\times1=\mu_1(F)$$

So in that case indeed: $$(\mu_1\otimes\mu_2)\circ\Pi_1^{-1}=\mu_1$$

Note however that this is evidently not the case if $\mu_2$ is not a probability measure.