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I saw an example to show how to calculate the Measure-theoretic entropy, and here is the example:

Let $T:X\to X$ be the doubling map $T(x)=2x\;(mod\;1)$ and there is a partition $\alpha=\{[0,\frac{1}{2}),[\frac{1}{2},1)\}$

$\lor_{i=0}^{n-1}T^{-i}\alpha=\{[\frac{i}{2^n},\frac{i+1}{2^n})\,:i=0,...2^n-1\}$ and the example just say $\alpha$ is the generator, then calculate that the result is log2.

What I don't unstand is that why $\alpha$ is a generator for T ?

According to Kolmogorov-Sinai Theorem, is this $\lor_{i=0}^{\infty}T^{-i}\alpha$ generating $\sigma$ algebra of [0,1)? I think it's not, so I don't unstand what kind of $\sigma$ algebra it generates and why $\alpha$ can be a generator.

I suspect the problem is my thought about X, maybe X is not [0,1)?

Zane
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1 Answers1

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$$\lor_{i=0}^{n-1}T^i\alpha=\{[\frac{i}{2^n},\frac{i+1}{2^n})\,:i=0,...2^n-1\}$$ and the union of these over $n$ generates the Borel $\sigma$ algebra.

Yuval Peres
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  • So the equation above is not sigma algebra , it just can generate Borel σ algebra, so alpha is called a generator, right? – Zane Apr 19 '22 at 11:15
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    @Zane One can equivalently consider partitions of $X$ or sub-$\sigma$-algebras of $X$. In this case $\vee_{i=1}^\infty \alpha_i=\mathcal{B}$ means that if a sub-$\sigma$-algebra of $\mathcal{B}$ contains $\vee_{i=1}^n\alpha_i$ for any $n$ (up to negligible sets), then that sub-$\sigma$-algebra has to be $\mathcal{B}$ (up to negligible sets). – Alp Uzman Apr 19 '22 at 13:45
  • Without converting partitions to sub-$\sigma$-algebras, $\alpha$ being generating is equivalent to $\vee_{i=1}^\infty\alpha_i$ being the partition into points. – Alp Uzman Apr 19 '22 at 13:45
  • @Alp Uzman Do you mean that the "=" in $,\lor_{i=0}^{\infty} T^{-i}\alpha=\mathcal B,$ isn't like the symbol "=" in "1+1=2", but shows the specific relation between $ \mathcal B$ and $\lor_{i=0}^{\infty}T^{-i}\alpha$ ? And if I don't consider the partitions to sub-σ-algebras, when $,\lor_{i=0}^{\infty}T^{-i}\alpha$ can contain every point of the full set X, $\alpha$ could also be seen a generator. Did I understand correctly? – Zane Apr 19 '22 at 16:05
  • @Zane Yes (more precisely the two $=$'s are alike, but not the same; the one used in this context implicitly is a combination of the equivalence of the filtered categories of a.e. measurable partitions and sub-$\sigma$-algebras of the measure algebra, and notions of isomorphism in these two categories), and yes (again a.e.; there are countably many points missing). – Alp Uzman Apr 19 '22 at 16:12
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    @Alp Uzman OK，I understand much more about this than before, Thank you for your answer. Have a nice day! – Zane Apr 19 '22 at 16:20