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.

Question

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&#039;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.

Brendan Bush · Accepted Answer

wlog you can use a sequence       u    n    ↑      u    0  . We have   μ  (      f          −      1        (      R    )  )  =  μ  (  I  )  =  λ  (  I  )  &amp;lt;  ∞. 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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