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Ezekiel Yoder 2022-06-28 Answered
Let u L 1 ( μ ), T A measure preserving map s.t.: ( u ) d μ = u T d μ. A n , ϵ = { | u ( T n ( x ) ) | n 2 > ϵ }. T n = T T T n times.
Prove n 1 μ ( A n , ϵ ) is finite.
My attempt:
By Markov inequality we get: μ ( A n , ϵ ) 1 ϵ | u ( T n ( x ) ) | n 2 d μ, since T is measure preserving we get: | u ( T n ( x ) ) | L 1 ( μ ) . Summing over n gives:
n 1 μ ( A n , ϵ ) n 1 ϵ 1 n 2 | u ( T n ( x ) ) | d μ .
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: | u ( T n ( x ) ) | d μ = | u | d μ?
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Answers (1)

Daniel Valdez
Answered 2022-06-29 Author has 19 answers
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
X f ( T ( x ) ) μ ( d x ) = X f ( x ) μ T ( d x ) = X f ( x ) μ ( d x )
Now suppose that for n > 1 it holds that
X f ( T n 1 ( x ) ) μ ( d x ) = X f ( x ) μ ( d x )
We would have
X f ( T n ( x ) ) μ ( d x ) = X f ( T n 1 ( T ( x ) ) ) μ ( d x ) = = X f ( T ( x ) ) μ ( d x ) = = X f ( x ) μ ( d x )
Therefore by induction
X f ( T n ( x ) ) μ ( d x ) = X f ( x ) μ ( d x ) , n N
Now set f ( x ) = | u ( x ) | and you're done: u is integrable so | u | is too and
n N μ ( A n ) 1 ε ( X | u | d μ ) n N 1 n 2 <
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