Is f ( x ) = { 1 if x ∈ Q 0 if x ∉ Q Lebesgue measurable ?I haven't study rigorous the chapter of measurable function, so I don't have many tools to work on this.My idea was to say that since f − 1 ( 0 ) = { Q } Which is measurable and f − 1 ( 1 ) = { R ∖ Q } which is also measurable then f is measurable, I don't think it's correct. How should I prove this (if f is even measurable)?

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Is   f  (  x  )  =      {                            1                           if           x          ∈                      Q                                                0                           if           x          ∉                      Q                                   Lebesgue measurable ?I haven&#039;t study rigorous the chapter of measurable function, so I don&#039;t have many tools to work on this.My idea was to say that since       f          −      1        (  0  )  =  {      Q    } Which is measurable and       f          −      1        (  1  )  =  {      R    ∖    Q    } which is also measurable then   f is measurable, I don&#039;t think it&#039;s correct. How should I prove this (if   f is even measurable)?

Nola Rivera · Accepted Answer

Your idea, using       f          −      1       is correct, but       f          −      1        (      Q    )  =  {  0  }  ,      f          −      1        (      R    ∖    Q    )  =  {  1  } is not right.For every mesurable set   A,       f          −      1        A must be one of the four cases.      f          −      1        A  =      {                            ∅                          (          0          ,          1          ∉          A          )                                                  Q                                    (          0          ∈          A          ,          1          ∉          A          )                                                  R                    −                      Q                                    (          0          ∉          A          ,          1          ∈          A          )                                                  R                                    (          0          ,          1          ∈          A          )                        And for all cases   ∅  ,      Q    ,      R    −      Q    ,      R   is Lebesgue mesurable.   m  (  ∅  )  =  0  ,  m  (      Q    )  =  0  ,  m  (      R    −      Q    )  =  ∞  ,  m  (      R    )  =  ∞.So   f is Lebesgue mesurable.If you use the same logic as this, then you can get this proposition:For all Lebesgue mesurable set   A  ⊆      R  , characteristic map       χ    A    :      R    →      R   is Lebesgue mesurable function.

Sarai Davenport · Accepted Answer

A function is measurable if the preimage of any measurable set is measurable.The function you gave only has four possible preimages in the first place: if a set contains neither 0, nor 1, then its preimage is   ∅. If it contains 0, but not 1, then the preimage is       R  ∖      Q  . If it contains 1, but not 0, then the preimage is       Q  . And if it contains both, the preimage is       R  .All of these are measurable. So the preimage of any set is measurable - that of a measurable set included. So the function is measurable.

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