By Martin Rütter
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Additional info for Eine Couch für alle Felle. Probleme im Hundealltag?
Example text
X \ (A ∪ A )) + δ + 1 − µ(X \ A′ ) µ(A′ ) √ √ Hence µ{x : |E(11A |Bm ) − 11A | ≥ 3δ} ≤ 3δ. 18. We end this Section with the ergodic decomposition theorem and the adequate entropy formula. 13. Let T be a measure preserving endomorphism of a Lebesgue space. A measurable partition A is said to be T -invariant if T (A) ⊂ A for almost every A ∈ A. The induced map TA = T |A : A → A is a measurable endomorphism of the Lebesgue space (A, FA , µA ). One calls TA a component of T . 11. (a) There exists a finest measurable partition A (mod 0) into T -invariant sets (called the ergodic decomposition).
X \ (A ∪ A )) + δ + 1 − µ(X \ A′ ) µ(A′ ) √ √ Hence µ{x : |E(11A |Bm ) − 11A | ≥ 3δ} ≤ 3δ. 18. We end this Section with the ergodic decomposition theorem and the adequate entropy formula. 13. Let T be a measure preserving endomorphism of a Lebesgue space. A measurable partition A is said to be T -invariant if T (A) ⊂ A for almost every A ∈ A. The induced map TA = T |A : A → A is a measurable endomorphism of the Lebesgue space (A, FA , µA ). One calls TA a component of T . 11. (a) There exists a finest measurable partition A (mod 0) into T -invariant sets (called the ergodic decomposition).
Let (X, F , µ) be a probability space. If A, B and C are countable partitions of X then: H(A ∨ B|C) = H(A|C) + H(B|A ∨ C) H(A ∨ B) = H(A) + H(B|A) A ≤ B ⇒ H(A|C) ≤ H(B|C) B ≤ C ⇒ H(A|B) ≥ H(A|C) (a) (b) (c) (d) H(A ∨ B|C) ≤ H(A|C) + H(B|C) H(A|C) ≤ H(A|B) + H(B|C) (e) (f) Proof. Let A = {An : n ≥ 1}, B = {Bm : m ≥ 1}, and C = {Cl : l ≥ 1}. Without loss of generality we can assume that all these sets are of positive measure. 4) we have H(A ∨ B|C) = − But i,j,k µ(Ai ∩ Bj ∩ Ck ) log µ(Ai ∩ Bj ∩ Ck ) µ(Ck ) µ(Ai ∩ Bj ∩ Ck ) µ(Ai ∩ Bj ∩ Ck ) µ(Ai ∩ Ck ) = µ(Ck ) µ(Ai ∩ Ck ) µ(Ck ) unless µ(Ai ∩ Ck ) = 0.
