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gvaldytist

gvaldytist

Answered question

2022-07-01

Let ( X , M , μ ) be a measueable space, and let f : R R be nonnegative f ( x ) 0 and measurable. Let E n = { x R : f ( x ) 1 n }. Show the following result
R f d m = lim n [ n , n ] f d m = lim n E n f d m .
I was thinking maybe we can assume f χ E n f, monotone increasing sequence of nonnegative measurable functions, and then apply monotone convergence theorem, but I am not sure that we can find such increasing sequence, as 1 n is decreasing.

Answer & Explanation

Zayden Andrade

Zayden Andrade

Beginner2022-07-02Added 22 answers

Both sequences
f χ E n , f χ [ n , n ]
are increasing because f is non-negative, E n E n + 1 , and [ n , n ] [ ( n + 1 ) , n + 1 ]. Now using monotone
convergence theorem we have
lim n E n f d m = lim n R f χ E n d m = R lim n ( f χ E n ) d m = R f d m
and similarly
lim n [ n , n ] f d m = lim n R f χ [ n , n ] d m = R lim n ( f χ [ n , n ] ) d m = R f d m   .
Misael Matthews

Misael Matthews

Beginner2022-07-03Added 5 answers

Thank you, that's helpful!

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