While studying concepts of measurable functions, there was a theorem suggesting that if any measurable function is altered on a null set, its measurability still remains.If f : E → R is measurable, E ∈ M , g : E → R is such that the set { x : f ( x ) ≠ g ( x ) } is null, then g is measurable.They showed the following proof.Consider the difference d ( x ) = g ( x ) − f ( x ). It is zero except on a null set, so { x : d ( x ) &gt; a } = { a null set a ≥ 0 a full set a &lt; 0 Here, a full set is the complement of a null set. Since both null and full sets are measurable, d is a measurable function. g = f + d is thus measurable.Now, I was curious whether the statement remains true if I change the set { x : f ( x ) ≠ g ( x ) } into { x : f ( x ) = g ( x ) } , i.e., differ at points in a full set, because there was no doubt if I alter the proof as the following.The difference d ( x ) is still measurable since { x : d ( x ) &gt; a } = { a full set a ≥ 0 a full set a &lt; 0 However, the later statement is actually not plausible at all. Is there any contradictory logic among here?Appreicate as always.

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While studying concepts of measurable functions, there was a theorem suggesting that if any measurable function is altered on a null set, its measurability still remains.If   f  :  E  →      R   is measurable,   E  ∈      M    ,  g  :  E  →      R   is such that the set       {    x    :    f    (    x    )    ≠    g    (    x    )    }   is null, then   g is measurable.They showed the following proof.Consider the difference   d  (  x  )  =  g  (  x  )  −  f  (  x  ). It is zero except on a null set, so      {    x    :    d    (    x    )    &amp;gt;    a    }    =      {                            a null set                          a          ≥          0                                      a full set                          a          &amp;lt;          0                        Here, a full set is the complement of a null set. Since both null and full sets are measurable,   d is a measurable function.   g  =  f  +  d is thus measurable.Now, I was curious whether the statement remains true if I change the set       {    x    :    f    (    x    )    ≠    g    (    x    )    }   into       {    x    :    f    (    x    )    =    g    (    x    )    }   , i.e., differ at points in a full set, because there was no doubt if I alter the proof as the following.The difference   d  (  x  ) is still measurable since      {    x    :    d    (    x    )    &amp;gt;    a    }    =      {                            a full set                          a          ≥          0                                      a full set                          a          &amp;lt;          0                        However, the later statement is actually not plausible at all. Is there any contradictory logic among here?Appreicate as always.

robegarj · Accepted Answer

Let   f  =  0  ,  A  ⊆  E be a non-measurable subset and   N is a null set with   N  ∩  A  =  ∅  .Then we let   g  (  x  )  =      {                            1          ,                          x          ∈          A                                      0          ,                          x          ∈          N                                      −          1          ,                          x          ∈          E          ∖          (          A          ∪          N          )                        Thus,  {  x  :  d  (  x  )  &amp;gt;  0  }  =  A which is non-measurable. So here   d  (  x  ) is not a measurable function.

While studying concepts of measurable functions, there was a theorem suggesting that if any measurab

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