Measurable function on the Cantor Set with the Product Topology

Question

I am new to measure theory and I have worked out that the cantor set can be represented as $C = \{ 0,1 \}^N$ with the product topology metric. But I don't know how to show that for $\phi : N \to N $ , the map $ \sigma_{\phi} : (C, B_{C}) \to (C, B_{C})$ such that $\sigma_{\phi} = (x_1,x_2,x_3,...)$ = $\{x_{ \phi (1)}, x_{\phi (2)}, x_{\phi (3)}\}$ is measurable. Do I have to make sure it works in the definition of measure? Any help would be appreciated.

Answer 1

For any "index set" $I$ and any collection $\{S_i:i\in I\}$ of spaces, the (Tychonoff) product topology $T$ on $P=\prod_{i\in I}S_i$ is defined as the weakest (i.e. $\subset$-smallest) topology such that every projection $p_j: P\to S_j$ is continuous. [ If $x=(x_i)_{i\in I}\in P$ then $p_j(x)=x_j.$ ]

For each $j\in I$ let $B_j$ be the collection of all sets of the form $\prod_{i\in I}U_i,$ where $U_i=S_i$ if $i\ne j,$ and $U_j$ is some (any) open subset of $S_j.$

Then $B=\cup_{j\in I}B_j$ is a sub-base (sub-basis) for the topology $T$ on $P$.

Therefore any $\sigma: P\to P$ is continuous iff $\sigma^{-1}b\in T$ whenever $b\in B.$

In your Q:

We have $I=\Bbb N$ and every $S_i=\{0,1\}$ has the discrete topology. And $P=\{0,1\}^{\Bbb N}.$

Now for $j\in \Bbb N$ the only members of $B_j$ other than $P$ and $\emptyset$ are $B_j(0)=\{(x_i)_{i\in \Bbb N}: x_j=0\}$ and $B_j(1)=\{(x_i)_{i\in \Bbb N}: x_j=1\}.$

For $x=(x_i)_{i\in\Bbb N}\in P,$ and for $j\in \Bbb N,$ and for $y\in \{0,1\},$ we have $$(\sigma_{\phi})^{-1} B_j(y)=\{ x\in P: x_{\phi (j)}=y\}= B_{\phi(j)}(y)\in T.$$ Therefore $\sigma_{\phi}$ is continuous.

This might suspiciously easy. But that's all there is.