A problem in Billingsley's Probability and Measure says that if f : [ 0 , 1 ] → R is finite-valued and Borel-measurable, then f may be assumed integrable, and even bounded.I don't see how this is true though. For example define f as f = 0 on [ 0 , 1 / 2 ) f = 1 on [ 1 / 2 , 3 / 4 ) f = 2 on [ 3 / 4 , 7 / 8 ) ⋮ f = n on [ 2 n − 1 2 n , 2 n + 1 2 n + 1 ) Then f ( x ) &lt; ∞ for all x ∈ [ 0 , 1 ] and f is measurable since it is a step function and takes only discrete values. However f is not bounded on [0,1].What's wrong with this attempted counterexample? How would the statement be proven?

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A problem in Billingsley&#039;s Probability and Measure says that if   f  :  [  0  ,  1  ]  →      R   is finite-valued and Borel-measurable, then   f may be assumed integrable, and even bounded.I don&#039;t see how this is true though. For example define   f as  f  =  0   on   [  0  ,  1      /    2  )  f  =  1   on   [  1      /    2  ,  3      /    4  )  f  =  2   on   [  3      /    4  ,  7      /    8  )  ⋮  f  =  n   on             [                  2      n      −      1              2      n        ,            2      n      +      1              2              n        +        1                        )      Then   f  (  x  )  &amp;lt;  ∞ for all   x  ∈  [  0  ,  1  ] and   f is measurable since it is a step function and takes only discrete values. However   f is not bounded on [0,1].What&#039;s wrong with this attempted counterexample? How would the statement be proven?

Marlee Guerra · Accepted Answer

As I suspected, you left out context. The problem is about proving Lusin&#039;s theorem for functions on [0,1]. When part (a) says &quot;Show that   f may be assumed to be integrable [or bounded]&quot;, what Billingsley means is that&quot;Show that if Lusin&#039;s theorem is true for all integrable [or bounded] functions   f  :  [  0  ,  1  ]  →      R  , then it is true for all Borel functions   f  :  [  0  ,  1  ]  →      R  .&quot;

A problem in Billingsley's Probability and Measure says that if f : [ 0 , 1 ]

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