A particular measure in the Cantor space $2^\infty$ / How to prove it also defines a $\sigma$-algebra?

Question

Consider the following measure $\mu$ for the Cantor set (seen as the space of infinite sequences of 0's and 1's): $$ \mu\left(E\right) = \lambda \left(g\left(E\right) \right) \tag{1}$$ where $g:2^\infty \to \left[0,1\right]$ is the function with the rule $(x_i)_{i=1}^\infty \mapsto \sum_{i=1}^\infty x_i/2^i$ and $\lambda$ is the Lebesgue measure. This rule suggests that a set must be called measurable in $2^\infty$ iff its image under $g$ its Lebesgue-measurable in $[0,1]$. This function is continuous and surjective but not injective (e.g., $(01000...)$ and $(00111..)$ have the same image).

This was taken from J. Oxtoby's "Measure and Category" [2nd ed. p. 84]. One thing that is not proven in the book is that this actually defines a $\sigma$-algebra in the space $2^\infty$. One condition is very easy to check, for instance, if the sets $(E_i)_{i=1}^\infty$ are such that $g(E_i)$ is measurable for every $i$ then $\bigcup_i E_i$ is measurable since $g(\bigcup_i E_i)= \bigcup_i g(E_i)$.

I am having some trouble proving that if $E\subseteq 2^\infty$ is such that $g(E)$ is measurable, then $g(2^\infty\setminus E)$ is measurable.

I think this condition is necessary for (1) to define a $\sigma$-algebra and a measure in $2^\infty$. Am I missing some elementary set-theoretic fact here that makes this evident?

Answer 1

Hint: A couple of observations and a question:

1. Note that the image of $g$, call is $A$, is Lebesgue measurable.
2. Note that if $g$ were one-to-one, the result would be trivial.
3. How big is the set $\{ \mathbf{x} \in 2^\infty : ( \exists \mathbf y \in 2^\infty ) ( \mathbf{y} \neq \mathbf{x} , g ( \mathbf{y} ) = g (\mathbf{x}) ) \}$? (This will give you an approximation of the "error" between $g [ 2^\infty \setminus E ]$ and $A \setminus g [ E ]$, and you should be able to show that the Lebesgue measurability of one implies that of the other.)