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Lena Bell 2022-07-16 Answered
If f : I R R is a Lebesgue-measurable function ( I is a closed interval), μ is the Lebesgue measure, my question is whether the limit
lim u u 0 μ ( f 1 ( ( u , u 0 ] ) ) = 0.
I don't really know how to justify the value of this limit, if it is 0 or possibly μ ( f 1 ( { u 0 } ) ). I would appreciate your help.
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Answers (1)

Brendan Bush
Answered 2022-07-17 Author has 14 answers
wlog you can use a sequence u n u 0 . We have μ ( f 1 ( R ) ) = μ ( I ) = λ ( I ) < . Then you can see that
f 1 ( ( u n , u 0 ] ) = { x : f ( x ) ( u n , u 0 ] } { x : f ( x ) ( u n + 1 , u 0 ] } = f 1 ( ( u n + 1 , u 0 ] )
and therefore, since inverse images preserve set operations:
n N f 1 ( ( u n , u 0 ] ) = f 1 ( n N ( u n , u 0 ] ) = f 1 ( { u 0 } )
By continuity of measures (we have μ ( f 1 ( ( u 1 , u 0 ] ) ) λ ( I ) we then have
lim n μ ( f 1 ( ( u n , u 0 ] ) ) = μ ( f 1 ( { u 0 } ) )
To see that it is not always 0, consider I = [ 1 , 1 ], f ( x ) = 1 [ 0 , 1 ] ( x ) + ( 1 / 2 ) 1 [ 1 , 0 ) ( x ) and u 0 = 1. Then
lim n μ ( f 1 ( ( u n , 1 ] ) ) = μ ( f 1 ( { 1 } ) ) = μ ( [ 0 , 1 ] ) = 1
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