1. ## Measurable map

Hello,

I have the following set $\displaystyle X=\mathbb{Z}=\left\{ 0,\pm1,\pm2,\pm3,... \right\}$

I have shown that $\displaystyle \mathcal{A}= \left\{ A\subset \mathbb{Z} \vert \forall n>0: 2n\in A \Leftrightarrow 2n+1 \in A \right\}$ is a $\displaystyle \sigma$-algebra.

I now want to show that the map $\displaystyle T:\mathbb{Z}\rightarrow \mathbb{Z}$ (which is bijective) given as $\displaystyle T(n)=n+2$ is $\displaystyle \mathcal{A}/\mathcal{A}$-measurable.

I am using the following definition: If $\displaystyle (X,\mathcal{A})$ and $\displaystyle (X',\mathcal{A'})$ are measurable spaces then the map $\displaystyle T:X\rightarrow X'$ is $\displaystyle \mathcal{A}/\mathcal{A'}$-measurable if $\displaystyle T^{-1}(A')\in \mathcal{A}$ for all $\displaystyle A'\in \mathcal{A'}$.

According to the definition we must show that: $\displaystyle 2n \in T^{-1}(A) \Leftrightarrow 2n+1 \in T^{-1}(A)$ for $\displaystyle A \in \mathcal{A}$. The we could conclude that $\displaystyle T$ is $\displaystyle \mathcal{A}/\mathcal{A'}$-measurable.

I want to assume that $\displaystyle 2n \in T^{-1}(A)$ and based on this assumption reach that $\displaystyle 2n+1=T^{-1}(2n+3) \in T^{-1}(A)$. How can I use the assumption?

Thanks.

2. ## Re: Measurable map

Never mind, I got it.

We have that $\displaystyle T^{-1}(n)=n-2$.

Assume that $\displaystyle 2n \in T^{1}(A)$. Since $\displaystyle T$ is bijective we have that $\displaystyle T^{-1}(n')=2n$ which means that $\displaystyle n'=2n+2=2(n+1) \in A$.

Since $\displaystyle A \in \mathcal{A}$ we have that $\displaystyle 2(n+1)+1 = 2n+3 \in A$.

Therefore $\displaystyle T^{-1}(2n+3)=2n+1 \in T^{-1}(A)$