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cyfwelestoi 2022-05-25 Answered
Let ( X , A , μ ) be a measure-space and f : X [ 0 , ] a unsigned measurable map. Show that μ ( { f ( x ) > 0 } ) = 0 μ ( { f ( x ) > t } ) = 0 For every t > 0.
Suppose that μ ( { f ( x ) > 0 } ) = 0. Since f = 0 a.e we have that X f = 0. Thus μ ( { f ( x ) > t } ) 1 t X f = 0 and since the measure is positive we have that μ ( { f ( x ) > t } ) = 0.
Conversely suppose that μ ( { f ( x ) > t } ) = 0. Then what can be done for this case? It seems that I could write X f = X { f ( x ) > t } f + { f ( x ) > t } f and the latter integral would be zero, but I don’t really get anywhere from here?
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Answers (1)

grindweg1v
Answered 2022-05-26 Author has 12 answers
Note that { f ( x ) > 0 } = n 1 { f ( x ) > 1 / n }, it holds
μ ( { f ( x ) > 0 } ) = lim n μ ( { f ( x ) > 1 / n } ) = 0
since the sequence of set A n = { f ( x ) > 1 / n } is increasing.
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