Let be a measure space with measure ; let be a measure-preserving transformation. Last time we looked at how the averages

behave in . But, now we want pointwise convergence.

**The pointwise ergodic theorem **

We consider the pointwise ergodic theorem of Garrett George Birkhoff:

Theorem 1 (Birkhoff)Let . Then the averages converge almost everywhere to a function with a.e.

There is something of analogy between Birkhoff’s theorem and the well-known fact from real analysis that a function in on a euclidean space can be recovered from its integrals over balls, i.e., for almost all ,

where is Lebesgue measure. The proof of this latter theorem usually proceeds by associating to a locally integrable function on the **Hardy-Littlewood maximal function**

and proving that it defines a bounded sublinear operator from to weak . Then, one uses an approximation argument since the continuous functions with compact support are dense in .

**The maximal ergodic theorem **

In proving the Birkhoff ergodic theorem, we will define the **maximal operator**

(When , that expression is set to be zero, so everywhere.) There is a similar weak-type inequality for this, which we will prove from the **maximal ergodic theorem**:

Theorem 2For ,

To prove this, we use the abbreviation ; then becomes a transformation of onto itself. When , we have

as is easily seen. In particular,

which is at least \( ||M_T f||_1 – ||UM_T f||_1 \geq 0 \) since

and is bounded by 1. This completes the proof.

It isn’t quite clear how the maximal ergodic theorem is a weak-type inequality. To do this, we fix and note that

In particular, by the maximal theorem,

which implies , a weak-type bound. What we will actually use, however, is the boxed statement above, or rather a variant of it. If with , then

which follows by doing all this with replacing .

**Proof of the ergodic theorem **

Given and , consider the sets

and

I will show that when , . Taking the union of these intersections for with , one gets a set of measure zero outside of which the limit of the averages exists. So, it is enough to prove . Now , as is easily seen. Also, at each point of , we have so by the last boxed statement,

Now we can do the exact same thing with , since is the same thing as for , which implies

and putting this all together gives , possible only if .

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