Ezekiel Yoder

2022-06-28

Let $u\in {\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mu \mathcal{)}$, $T$ A measure preserving map s.t.: $\int (u)d\mu =\int u\circ Td\mu $. ${A}_{n,\u03f5}=\{\frac{|u({T}^{n}(x))|}{{n}^{2}}>\u03f5\}$. ${T}^{n}=T\circ T\circ T\dots $ n times.

Prove $\sum _{n\ge 1}\mu ({A}_{n,\u03f5})$ is finite.

My attempt:

By Markov inequality we get: $\mu ({A}_{n,\u03f5})\le \frac{1}{\u03f5}\int \frac{|u({T}^{n}(x))|}{{n}^{2}}d\mu $, since $T$ is measure preserving we get: $|u({T}^{n}(x))|\in {\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mu \mathcal{)}$. Summing over n gives:

$\sum _{n\ge 1}\mu ({A}_{n,\u03f5)}\le \sum _{n\ge 1}{\u03f5}^{-1}{n}^{-2}\int |u({T}^{n}(x))|d\mu .$

This is the part where I am stuck. I know that all the integrals are finite but how do I ensure that summing over all the integrals stays finite? Can I make the integral independent of n? Or can I just state by assumption: $\int |u({T}^{n}(x))|d\mu =\int |u|d\mu $?

Prove $\sum _{n\ge 1}\mu ({A}_{n,\u03f5})$ is finite.

My attempt:

By Markov inequality we get: $\mu ({A}_{n,\u03f5})\le \frac{1}{\u03f5}\int \frac{|u({T}^{n}(x))|}{{n}^{2}}d\mu $, since $T$ is measure preserving we get: $|u({T}^{n}(x))|\in {\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mu \mathcal{)}$. Summing over n gives:

$\sum _{n\ge 1}\mu ({A}_{n,\u03f5)}\le \sum _{n\ge 1}{\u03f5}^{-1}{n}^{-2}\int |u({T}^{n}(x))|d\mu .$

This is the part where I am stuck. I know that all the integrals are finite but how do I ensure that summing over all the integrals stays finite? Can I make the integral independent of n? Or can I just state by assumption: $\int |u({T}^{n}(x))|d\mu =\int |u|d\mu $?

Daniel Valdez

Beginner2022-06-29Added 19 answers

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Since $T$ is a measure preserving map, then for $n=1$ and $f$ measurable we have

$$

Now suppose that for $n>1$ it holds that

$$

We would have

$$

Therefore by induction

$$

Now set $f(x)=|u(x)|$ and you're done: $u$ is integrable so $|u|$ is too and

$$

$$

Now suppose that for $n>1$ it holds that

$$

We would have

$$

Therefore by induction

$$

Now set $f(x)=|u(x)|$ and you're done: $u$ is integrable so $|u|$ is too and

$$

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