I was trying to solve the next question: let X be a finite measure space ( &#x03BC;<!--

London Ware 2022-05-11 Answered
I was trying to solve the next question:
let X be a finite measure space ( μ ( X ) < ) and let f L 1 ( X , μ ), f ( x ) 0 almost everywhere.
Show that for each measurable subset E X:
lim n E | f | 1 n d μ = μ ( E )
My idea for a solution is to use Fatou's lemma:
In one direction:
E lim inf n | f | 1 n d μ lim inf n E | f | 1 n d μ
and
E lim inf n | f | 1 n = E 1 d μ = μ ( E )
So we get:
μ ( E ) lim inf n E | f | 1 n d μ
In the other direction, I thought of maybe saying that we know there is an ε > 0
And an N N so for all n > N we get that 1 + ε > | f | 1 n which means 1 + ε | f | 1 n > 0
and use Fatou's lemma again on the expression above to get the lim sup smaller or equal to μ ( E )

Is it valid? Am I missing something?
If I do, what can I do to prove the other direction?
Thank you!
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Answers (1)

Mathias Patrick
Answered 2022-05-12 Author has 22 answers
Lebesgue dominated convergence theorem will do:
Note that | f | 1 / n | f | + 1 and | f | + 1 L 1 ( μ ) since μ is of finite measure. Then interchange of integral with limit is fine.
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