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It doesn't really provide you with durations.

The handwavy proof goes as follows(The proof of this theorem is extremely simple in measure theory language, it's actually a quite common exam question in bachelor's level measure theory courses):

You take the set of items that you think never return to their original position.Call this set X. Now take the infinite sequence of pre-images of this set (i.e look at the items that map onto some part of X after one "evolution" of the volume preserving flow, then look at the items that map onto some part of X after 2 flows, etc), for natural numbers.

Note that a key part of the measure theoretic version is that you can assign a finite "volume" to the overall system and some subsets of the system.

This sequence (of pre-images under the volume preserving flow) has items that are pairwise disjoint (i.e an item that flows onto X after 1 evolution cannot also be an item that flows onto X after n evolutions). This is obvious from definition of X(as otherwise that item would recur).

Now here is the tricky part(but basic result in measure theory): The sum of the "volumes" of these sets (of pre-images) is equal to the "volume" of set Y, where Y is this infinite union of pre-image sets together(with union). Because this big set Y is some subset of your system, clearly it has a finite "volume"

This means the "volume" of your non-recurrent points has to be 0. Why? You assume the flow is volume preserving, which means that all the terms in your pre-image sequence have the same volume. Thus each volume has to be 0 (as you have an infinite sequence summing to a finite number).



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