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I have a set $S$ that is a nonmeasurable subset of $X=\{0,1\}^{\mathbb{N}}$ (with respect to the normed product measure on $X$.

Now let $g:X\to[0,1]$ be defined by $g(x)=\underset{n\in\mathbb{N}}{\sum}\frac{x_{n}}{2^{n+1}}$.

Why is $g[S]$ not Lebesgue measurable?

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    Because the given map is a [measure theoretic isomorphism](http://en.wikipedia.org/wiki/Measure-preserving_dynamical_system#Homomorphisms).2012-12-23
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    For each finite $F\subseteq\Bbb N$ and each function $\varphi:F\to\{0,1\}$ let $B(\varphi)=\{x\in X:x\upharpoonright F=\varphi\}$; then $\lambda(g[B(\varphi)])=\mu(B(\varphi))=2^{-|F|}$.2012-12-23
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    I see why $\mu(B(\varphi))=2^{-|F|}$, and I know that the Cantor space is homeomorphic to the Cantor set, but I am lost when it comes to the measure preserving part.2012-12-23
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    Let $\varphi_i:\{0\}\to\{0,1\}:0\mapsto i$; $g[B(\varphi_0)]=\left[0,\frac12\right]$, and $g[B(\varphi_1)]=\left[\frac12,1\right]$. Every $g[B(\varphi)]$ is a ‘nice’ union of closed intervals whose lengths sum to $2^{-|F|}$.2012-12-23
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    Ok I see why this works. But what about sets other than the B(φ)? These are measurable sets in the algebra on X, but surely different types occur in the sigma algebra. Is it trivial to see that this works for all meas. sets?2012-12-24

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