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Consider $T: [0,1)\to [0,1), x\mapsto 10x-\lfloor 10x\rfloor$.

Is that the same as $$ 10x (mod 1)? $$ or in which sense is that multiplication with 10 mod 1?

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    It's true in general that $g(x)-[g(x)]$ is identically $g(x) \text{ mod } 1$. This is just because $x - [x] =x \text{ mod } 1$ (compose this with $g$). – MPW Feb 25 '14 at 19:46

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It is exactly the same. It is also just move the decimal one position right and erase what comes to the left.

Ross Millikan
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  • What is f.e. $T^{-1}([0.5,0.75))$? Is that $\bigcup_{k=0}^{9}[0.05,0.075]\pm k$? –  Feb 25 '14 at 20:08
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    No, $k$ is $0,0.1,0.2\dots0.9$ and no minus sign, so $[0.05,0.075)\cup [0.15,0.175)\dots \cup[0.95,0.975)$ – Ross Millikan Feb 25 '14 at 20:16
  • Then my script is wrong? It says: $T^{-1}(B)=\left{x\in [0,1): Tx\in B\right}=\bigcup_{k=0}^{9}\frac{1}{10}B-k$. –  Feb 25 '14 at 20:19
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    Yes, it is wrong. The domain of $T$ is given as $[0,1)$, so you can't add (for example) $5$. Try applying $T$ to some of the points in my set, you can see how it works. I think you lost parentheses around $B-k$-the $k$ should be divided by $10$, but the sign should be $+$. – Ross Millikan Feb 25 '14 at 20:58