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Aug 25, 2015 at 14:18 vote accept Uchiha
Aug 23, 2015 at 20:37 answer added user78465 timeline score: 1
Aug 21, 2015 at 12:55 history edited Uchiha CC BY-SA 3.0
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Aug 21, 2015 at 12:36 history edited Uchiha CC BY-SA 3.0
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Aug 21, 2015 at 12:26 comment added Uchiha Thanks. I have seen a related concept in infinite ergodic theory which is called pointwise dual ergodicity. But this is a stronger requirement than that, roughly speaking.
Aug 21, 2015 at 9:46 comment added Asaf @Ray, when most people say ergodic theory, they mean dynamics on finite measure space, hence the measure limitation is not very interesting either. Moreover, try to mimic the proof of ergodicity out of mixing, and it fails in your definition. Perhaps you should tell us what notation exactly do you have in mind, and do you know a reasonable interesting system which have this property (most favorably, homogeneous one, on which such a statement can be analyzed in algebraic tools).
Aug 21, 2015 at 6:57 history edited Uchiha CC BY-SA 3.0
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Aug 21, 2015 at 6:56 comment added Uchiha Probably I should not say for any A and B, but any A and B in some finite measure set.
Aug 20, 2015 at 23:10 comment added Anthony Quas Then taking $A=B=E$ would make this not make sense.
Aug 20, 2015 at 21:57 comment added Uchiha Ah sorry. There was a typo in my previous question. It is $\mu(A)\mu(B)$, not $\mu(A\cap B)$.
Aug 20, 2015 at 21:54 history edited Uchiha CC BY-SA 3.0
Corrected a typo.
Aug 20, 2015 at 20:00 comment added Anthony Quas Probably not very interesting as this would say that if $A$ and $B$ don't intersect initially, they never will. (The identity transformation has this property, but not much else).
Aug 20, 2015 at 18:08 history asked Uchiha CC BY-SA 3.0