Consider the binary expansions of $[0,1]$: $ f: \{0,1\}^{\mathbb{N}} \to [0,1]; \quad (x_n)_n \mapsto \sum\limits_{n=1}^{\infty}\: \frac{1}{2^n}\:x_n.$
I have already prooved that $f$ is continuous and that $f$ restricted to $B := \{\:(x_n)_n \in \{0,1\}^{\mathbb{N}} \mid x_n=0\: \text{for infinitely many}\: n \in \mathbb{N}\:\}\: \cup \: \{1\}^{\mathbb{N}} $
is bijective to [0,1] and that $B$ is a Borel-subset. I'm now want to check that $f^{-1}: [0,1] \to B \subset\{0,1\}^{\mathbb{N}} $ is Borel-measureable. My Proof: We only have to consider the components: $\pi_n \circ f^{-1}: [0,1] \to \{0,1\}$. Then
$(f^{-1})^{-1}(\pi_{n}^{-1}(\{0,1\})=[0,1]$ ;
$(f^{-1})^{-1}(\pi_{n}^{-1}(\{0\})= (f^{-1})^{-1}(\{0,1\} \times\{0,1\}\times \dots \times\{0\}\times\{0,1\}\times \dots)= [0,1-\frac{1}{2^n})$
$(f^{-1})^{-1}(\pi_{n}^{-1}(\{1\})= (f^{-1})^{-1}(\{0,1\} \times\{0,1\}\times \dots \times\{1\}\times\{0,1\}\times \dots)= [\frac{1}{2^n},1]$
Hence $f^{-1}$ is Borel-measureable.
Is this proof correctly?
No. For example $(f^{-1})^{-1}\pi_2^{-1}(\{0\}) = [0,1/4) \cup [1/2,3/4)$.