Push-forward of a measure from $\mathbb{R}^d$ to $\mathbb{R}^d\times\mathbb{R}^d$

Question

I'm having some problems in showing that, given a probability measure $\mu$ on $\mathbb{R}^d$, if $s,t:\mathbb{R}^d\to\mathbb{R}^d$ are such that $(\textrm{id}\times s)_\#\mu=(\textrm{id}\times t)_\#\mu$, then $s(\mathbf{x})=t(\mathbf{x})$ holds $\mu$-almost everywhere. Could someone help me? Thank you.

Answer 1

Proof by contradiction

First assume that $\mu((\mathrm{id} \times s)^{-1}(A)) = \mu((\mathrm{id} \times t)^{-1}(A))$ for all measurable $A \subseteq \mathbb R^d \times \mathbb R^d.$

Then assume that there is a Borel set $E \subseteq \mathbb R^d$ such that $\mu(E) > 0$ and $s \neq t$ on $E.$

Let $A = (\mathrm{id} \times s)(E).$ Then $(\mathrm{id} \times s)^{-1}(A) = E,$ but $(\mathrm{id} \times t)^{-1}(A) = \emptyset.$ Thus $\mu((\mathrm{id} \times s)^{-1}(A)) \neq \mu((\mathrm{id} \times t)^{-1}(A))$ since $LHS>0$ while $RHS=0.$

We conclude that if the first assumption is true, then the second assumption must be false, i.e. $s=t$ must be valid $\mu$-a.e.