Since we’ve already discussed the use of simple element in the proof of the ergodic theorem, we will now address the existence of such elements.

Recall that we are working with a measure preserving system as defined previously, where denotes the shift operator. We are looking for an element of such that

for any . First, take a sequence in of elements that are periodic; in other words, each element is of the form

for some integer . We set

where denotes the Dirac measure, as usual. We require that the converge weakly to as .

Here we depart from the sequence of the proof in Kamae in order to first show how to construct such a sequence, before showing that we can use it to obtain a typical element. To show the existence of the sequence, it suffices to show that for any neigbourhood of in the weak topology of Borel measures on , we can find an element such that . The nature of the topology means that we can approximate the measure to an arbitrary degree by considering a finite amount of information. In other words, there are integers , and some positive such that if for any subset of the form

where are integers ranging from to , then .

Now set

We can define a measure on by

for the in the same range as before. We can see as an encoding of the measure on cylinder sets of the form we are considering, and will clearly be used to construct a measure in the neighbourhood of , once we have tweaked it a little.

We are given that is -invariant. From this, we can conclude that

for any . Why? We have that

and the latter term is equal to

From , we want to obtain a measure on which satisfies:

(1) For any , is a positive integer such that , where has been specified earlier.

(2) The relation holds for in the place of .

These do not seem like very great demands on , but we should still confirm that it can satisfy them. The first condition above requires that all the values are rational. This is not much of a challenge, because of the denseness of the rationals. We must just make sure that the adjustment from to does not violate condition (1). We therefore need to make sure, also, that the rationals we pick not only satisfy , but also

(obviously, since must still be a probability). Since there are only a finite number of elements of the measure space, we can ensure first that the latter condition holds, then ensure the validity of by more adjustments, making sure any adjustments upwards are balanced by adjustments downwards to guarantee This brings into question whether there will not be one of these adjustments that will turn the measure of one of these elements negative. But since we’re only dealing with a finite number of points of the measure space, and our adjustments are arbitrarily small, we must just make sure that none of the adjustments are as big as the least measure of any of the elements. If there are any elements of zero measure, we adjust the measure to avoid that, whilst still remaining close to $\latex \nu$.

Now pick a longest sequence of elements of such that

- for any and
- for any ,

where and .

The first condition clearly has to do with the translation operator, and requires that the -th element of is the translate of the -th element, with the last term of the element not dependent on the next element. Notice that there is nothing in the first condition which prevents “looping” the same sequence over and over again. If such were allowed, we would not be able to speak of the longest sequence, and so the second condition is introduced in order to limit the lengths.

Now, suppose that equality does not hold in (2), for some , remembering that is an integer, and also supposing that . In that case, we will be able to extend the sequence to increase the term on the left-hand side (by using closed loops), contradicting the assumption of the longest sequence. Thus, we can conclude that equality holds in (2) for each .

We define a periodic element of with period by

Thus, we have found a periodic element in the neighbourhood of for each . Taking neighbourhoods of such that , denote each corresponding periodic element by . For a sequence of positive integers increasing sufficiently fast, set

for any with , where and (recall that denotes the period of ). This gives us our desired typical element, and completes the argument.

**References: **

Kamae, T., *A simple proof of the ergodic theorem using nonstandard analysis,* Israel Journal of Mathematics, Vol. 42, No. 4, 1982

Hello niice blog

LikeLike