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

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.

• I'm probably missing something, but if $\mu$ is symmetric under rotation of $\mathbb R^d$ and $s, t$ are two different rotations of $\mathbb R^d$, doesn't then $(\textrm{id}\times s)_\#\mu=(\textrm{id}\times t)_\#\mu$ hold although $s(\mathbf{x})=t(\mathbf{x})$ at only a single point ($\mu$-almost nowhere)? – md2perpe Jul 14 '18 at 15:02
• I don't think so: the definition brings $(\textrm{id}\times s)_\#\mu(A)=\mu(A\cap s^{-1}(A))$ for every Borel set A. – RIccardo Plati Jul 14 '18 at 15:11
• Okay, then it's not the ordinary push-forward. That's obviously what was missing. – md2perpe Jul 14 '18 at 15:19
• That's the ordinary push-forward, but through the function $(\textrm{id}\times s):\mathbb{R}^d\to\mathbb{R}^d\times\mathbb{R}^d$, defined as $\mathbf{x}\mapsto (\mathbf{x},s(\mathbf{x}))$. – RIccardo Plati Jul 14 '18 at 15:31
• Is really $(\textrm{id}\times s)_\#\mu(A)=\mu(A\cap s^{-1}(A))$ correct? Shouldn't $A \subseteq \mathbb R^d \times \mathbb R^d$? But then $s^{-1}(A)$ isn't valid, since $s : \mathbb R^d \to \mathbb R^d$ so for $s^{-1}(A)$ to make sense we should have $A \subseteq \mathbb R^d.$ – md2perpe Jul 23 '18 at 14:58

## 1 Answer

### 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.