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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. – Quinn Culver Dec 23 '12 at 20:42
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 $\lambda(g[B(\varphi)])=\mu(B(\varphi))=2^{-|F|}$. – Brian M. Scott Dec 23 '12 at 20:48
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. – Forever Mozart Dec 23 '12 at 21:56
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 $2^{-|F|}$. – Brian M. Scott Dec 23 '12 at 22:11
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? – Forever Mozart Dec 24 '12 at 2:48

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