A really strange question here in my opinion:Let A , B be two measurable subsets of R 1 . Define f ( x ) = | ( A − x ) ∩ B | . Evaluate ∫ R 1 f d x. Here | ⋅ | refers to the measure.Here f is clearly non-negative, so I tried using the definition of Lebesgue integral: ∫ R 1 f d x = sup { ∫ R 1 s d x : 0 ≤ s ≤ f } ,where s are simple functions. But I realised that I couldn't come up with any simple functions.I tried breaking up R 1 into two parts: E 1 = { y ∈ R 1 : y ∈ A ∩ B } , E 2 = { y ∈ R 1 : y ∉ A ∩ B }and then considering the sum ∫ E 1 f d x + ∫ E 2 f d xand then I'm clueless as to how to proceed. I'm guessing the answer should be something intuitive like | A ∩ B | . Any help would be appreciated.

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A really strange question here in my opinion:Let   A  ,  B be two measurable subsets of             R        1  . Define   f  (  x  )  =      |    (  A  −  x  )  ∩  B      |  . Evaluate       ∫                            R                1              f  d  x. Here       |    ⋅      |   refers to the measure.Here   f is clearly non-negative, so I tried using the definition of Lebesgue integral:      ∫                            R                1              f  d  x  =  sup      {          ∫                                    R                    1                      s    d    x    :    0    ≤    s    ≤    f    }    ,where s are simple functions. But I realised that I couldn&#039;t come up with any simple functions.I tried breaking up             R        1   into two parts:      E    1    =  {  y  ∈            R        1    :  y  ∈  A  ∩  B  }  ,      E    2    =  {  y  ∈            R        1    :  y  ∉  A  ∩  B  }and then considering the sum      ∫                  E        1              f  d  x  +      ∫                  E        2              f  d  xand then I&#039;m clueless as to how to proceed. I&#039;m guessing the answer should be something intuitive like       |    A  ∩  B      |  . Any help would be appreciated.

Buckokasg · Accepted Answer

One could proceed as follows. Note that we can rewrite   f  (  x  )  =      |    (  A  −  x  )  ∩  B      |    =      ∫                  R                  1          B        (  y  )      1    A    (  x  +  y  )  d  y. Then the question becomes evaluating the expression      ∫                  R                  ∫                  R                  1          B        (  y  )      1    A    (  x  +  y  )  d  y  d  x  .By translation invariance of the Lesbeque measure on       R  , it follows that      ∫                  R                  ∫                  R                  1          B        (  y  )      1    A    (  x  +  y  )  d  y  d  x  =      ∫                  R                  ∫                  R                  1          B        (  y  )      1    A    (  y  )  d  y  d  x  =      ∫                  R                  1    B    (  y  )  d  y      ∫                  R                  1    A    (  x  )  d  x  =      |    A      |    ⋅      |    B      |    .

A really strange question here in my opinion: Let A , B be two measurable subsets of

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