Let $a , b\in \Bbb R$ , $a>0$
$T_{a,b}: \Bbb R\to \Bbb R$ defined as:
$T_{a,b}(x)=ax+b$.
Show that
- $T_{a,b}(B_R)=B_R$
2.$\forall A\in{B_R}$ $m(T_{a,b}(A))=am(A)$
Hints:
Notice that $T_{a,b}$ is bijection and its inverse is $T_{\frac{1}{a},\frac{-b}{a}}$.
First prove that $T_{a,b}$ is a sigma-algebra that contains all the open finite intervals and then look at:
$s: B_R\to [0,\infty]$
$s(A)=am(T^{-1}_{a,b}(A))$
$m$ is Lebesgue measure.
I tries to write:
$T_{a,b}(B_R)={aI+b : I\in B_R}$.
We know that the borel sigma algebra is generated by the collection of open intervals, closed intervals..
It's obvious that $T_{a,b}(B_R)\subset B_R$ as $I\in T_{a,b}(B_R)$ and $I=T_{1,0}$.
Then I have to show that $T_{a,b}(B_R)$ is a sigma-algebra that contain all of the finite segments so by the fact that $B_R$ is the minimal sigma algebra that contains all the finite segments, we get that $B_R\subset T_{a,b}(R)$.
So, we must prove two axioms, T is closed under complements and countable unions.
I tried to use the hint mentioned, but it's not obvious for me how to do it.
Thanks for any help.
Best Answer
In fact, if $T:X\to X$ is any homeomorphism, then $T$ carries Borel sets bijectively to Borel sets. Prove this in the usual way, letting $B=\{T(E):E\in \mathscr B(X)\}$ and showing that $B$ is a $\sigma$-algebra that contains the open sets in $X$:
If $T(E_i)\in B$ then $\bigcup T(E_i)=T\left(\bigcup E_i\right)\in B$ because $\bigcup E_i\in \mathscr B(X).$ A similar argument works for complements.
Now, if $U$ is open in $X$, then $T^{-1}(X)$ is open in $X$ and therefore $T^{-1}(U)\in \mathscr B(X).$ But then $U=T(T^{-1}(U))\in B$ by defnition.
So, $B$ is a $\sigma$-algebra on $X$ that contains the open sets. That is, $\mathscr B(X)\subseteq B$. Replacing $T$ by $T^{-1}$ gives the reverse inclusion, so $\mathscr B(X)=B.$