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## fa.useful evaluation – Can Birkhoff’s ergodic theorem for integrable features simply breathe deduced from Birkhoff’s ergodic theorem for bounded features?

It appears to me {that a} mighty less complicated proof of Birkhoff’s Ergodic theorem can breathe given for bounded observables than for extra common ones $ L ^ 1 $ observable. So I cerebrate it could breathe higher to show Birkhoff’s ergodic theorem by beginning with the restricted illustration after which extending to the common illustration than to method the common illustration “directly”. But can this extension of $ L ^ infty $ to $ L ^ 1 $ breathe made in some “simple” elementary route?

Let me give one workable model of how one can define this query:

Permit $ (X, mathcal {X}, mu) $ breathe a chance area. A *Markov operator* At $ L ^ 1 ( mu) $ is a linear, monotonic, unit-preserving, integer-preserving duty $ P colon L ^ 1 ( mu) to L ^ 1 ( mu) $.

Suppose we’ve an circumstance $ (P_n) _ {n geq 1} $ of the Markov operators $ L ^ 1 ( mu) $ in order that **(1)** for every $ f in L ^ infty ( mu) $ the next statements apply:

- $ P_n (f) overset { mu textrm {-as}} { to} int_X f , d mu $ how $ n after infty $;
- for all $ m, n geq 1 $, $ | nP_n (f) – mP_m (f) | _ {L ^ infty ( mu)} leq | mn | | f | _ {L ^ infty ( mu)} $;

and **(2)** for every $ f in L ^ 1 ( mu) $ there’s a worth $ P_ infty[f] in overline { mathbb {R}} $ in order that $ , liminf_ {n to infty} P_n (f) overset { mu textrm {-as}} {=} P_ infty[f]$.

(Of passage, if $ f in L ^ infty ( mu) $ then $ P_ infty[f]= int_Xf , d mu $. We will too behold from the proof of the outcome within the point to under that the identical is undoubted for all $ f in L ^ 1 ( mu) $ with $ f geq 0 $.)

We undoubtedly have it for everybody $ f in L ^ 1 ( mu) $, $ P_n (f) overset { mu textrm {-as}} { to} int_X f , d mu $ how $ n after infty $?

(Since $ P_n (-f) = – P_n (f) $, that’s tantamount to maxim this for everybody $ f in L ^ 1 ( mu) $, $ P_ infty[f]= int_Xf , d mu $.)

**Annotation.** I anticipate that when the respond is *Yes sir*, then not all the above circumstances are required to show it.

**Note.** As under, it isn’t troublesome to display [under considerably weaker conditions than those given above] that for everybody $ f in L ^ 1 ( mu) $, $ P_n (f) overset {L ^ 1 ( mu)} { to} int_X f , d mu $ how $ n after infty $.

*Proof*: Take with out limiting the common public $ f overset { mu textrm {-as}} { geq} 0 $. For every $ ok> 0 $ we’ve $ P_n (f wedge ok) overset { mu textrm {-as}} { to} int_X f wedge ok ; d mu $ how $ n after infty $. Thus, because of the monotony of the Markov operators, it’s limpid that

$$ liminf_ {n to infty} P_n (f) overset { mu textrm {-as}} { geq} int_X f , d mu. $$

But due to the integral conservation of the Markov operators, we too have that $ int_X P_n (f) , d mu = int_X f , d mu $ for all $ n $; and so since $ P_n (f) overset { mu textrm {-as}} { geq} 0 $ (by means of the monotony of the Markov operators) Fatou’s lemma supplies that

$$ int_X liminf_ {n to infty} P_n (f) , d mu leq int_X f , d mu. $$

It follows that

$$ liminf_ {n to infty} P_n (f) overset { mu textrm {-as}} {=} int_X f , d mu. $$

The outcome is then a direct consequence of the next Scheffé-like lemma.

**Lemma.** *Given a consequence $ (Y_n) _ {n in mathbb {N}} $ of integrable random variables $ J_n $ that’s uniformly constrained downwards, although $ Y !: = ! underset {n to infty} { liminf} Y_n $ can breathe built-in and $ mathbb {E}[Y_n] to mathbb {E}[Y]$ how $ n after infty $, then $ Y_n overset {L ^ 1} { to} Y $.*

*Proof of Lemma.* Take with out restriction of the common public $ Y = 0 $. We have that $ Y_n wedge 0 $ converges pointwise to $ 0 $ how $ n after infty $, and so since $ (Y_n) $ is uniformly constrained under, the theory of dominated convergence can breathe utilized to secure that $ mathbb {E}[Y_n wedge 0] to $ 0 how $ n after infty $. Since $ | Y_n | = Y_n-2 (Y_n wedge 0) $ and $ mathbb {E}[Y_n] to $ 0 how $ n after infty $, it follows that $ mathbb {E}[|Y_n|] to $ 0 how $ n after infty $. So we’re completed.

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