 # Let u &#x2208;<!-- ∈ --> <mrow class="MJX-TeXAtom-ORD"> <mi class="MJX-tex-caligra Ezekiel Yoder 2022-06-28 Answered
Let $u\in {\mathcal{L}}^{\mathcal{1}}\mathcal{\left(}\mu \mathcal{\right)}$, $T$ A measure preserving map s.t.: $\int \left(u\right)d\mu =\int u\circ Td\mu$. ${A}_{n,ϵ}=\left\{\frac{|u\left({T}^{n}\left(x\right)\right)|}{{n}^{2}}>ϵ\right\}$. ${T}^{n}=T\circ T\circ T\dots$ n times.
Prove $\sum _{n\ge 1}\mu \left({A}_{n,ϵ}\right)$ is finite.
My attempt:
By Markov inequality we get: $\mu \left({A}_{n,ϵ}\right)\le \frac{1}{ϵ}\int \frac{|u\left({T}^{n}\left(x\right)\right)|}{{n}^{2}}d\mu$, since $T$ is measure preserving we get: $|u\left({T}^{n}\left(x\right)\right)|\in {\mathcal{L}}^{\mathcal{1}}\mathcal{\left(}\mu \mathcal{\right)}$. Summing over n gives:
$\sum _{n\ge 1}\mu \left({A}_{n,ϵ\right)}\le \sum _{n\ge 1}{ϵ}^{-1}{n}^{-2}\int |u\left({T}^{n}\left(x\right)\right)|d\mu .$
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: $\int |u\left({T}^{n}\left(x\right)\right)|d\mu =\int |u|d\mu$?
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MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document 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 $f\left(x\right)=|u\left(x\right)|$ and you're done: $u$ is integrable so $|u|$ is too and