"Let μ 0 ( A ) = μ ( A ∪ A 0 ) , A ∈ F be a measure. Show that if ∫ f d μexists, then ∫ A 0 f d μ = ∫ f d μ 0 "However, I think there's a typo in this exercise. μ 0 isn't even a measure when μ ( A 0 ) ≠ 0, since μ 0 ( ∅ ) = μ ( A 0 ). If it were μ 0 ( A ) = μ ( A ∩ A 0 ), which I think that's what the author meant, I could prove it in the following way:For simple functions, ∫ A 0 f d μ = ∫ f 1 A 0 d μ = ∑ i = 1 n f i 1 A 0 μ ( A i ) = ∑ i = 1 n f i μ ( A i ∩ A 0 ) = ∑ i = 1 n f i μ 0 ( A i ) = ∫ f d μ 0 For a non-negative measurable function f, there's a non-decreasing sequence of simple functions ( s n ) n ∈ N such that s n → f, then ∫ A 0 f d μ = ∫ f 1 A 0 d μ = lim n → ∞ ∫ s n 1 A 0 d μ = lim n → ∞ ∫ s n d μ 0 = ∫ f d μ 0 And finally, for any measurable function, we just use f = f + − f − . So, which one is it, μ ( A ∪ A 0 ) or μ ( A ∩ A 0 )?

Question

&quot;Let       μ    0    (  A  )  =  μ  (  A  ∪      A    0    )  ,    A  ∈      F   be a measure. Show that if  ∫  f    d  μexists, then      ∫                  A        0              f    d  μ  =  ∫  f        d                  μ        0            &quot;However, I think there&#039;s a typo in this exercise.       μ    0   isn&#039;t even a measure when   μ  (      A    0    )  ≠  0, since       μ    0    (  ∅  )  =  μ  (      A    0    ). If it were       μ    0    (  A  )  =  μ  (  A  ∩      A    0    ), which I think that&#039;s what the author meant, I could prove it in the following way:For simple functions,      ∫                  A        0              f    d  μ  =  ∫  f            1                      A        0                d  μ  =      ∑          i      =      1              n            f    i              1                      A        0              μ  (      A    i    )  =      ∑          i      =      1              n            f    i    μ  (      A    i    ∩      A    0    )  =      ∑          i      =      1              n            f    i        μ    0    (      A    i    )  =  ∫  f    d      μ    0  For a non-negative measurable function   f, there&#039;s a non-decreasing sequence of simple functions   (      s    n        )          n      ∈              N             such that       s    n    →  f, then      ∫                  A        0              f    d  μ  =  ∫  f            1                      A        0                d  μ  =      lim          n      →      ∞        ∫      s    n              1                      A        0                d  μ  =      lim          n      →      ∞        ∫      s    n      d      μ    0    =  ∫  f    d      μ    0  And finally, for any measurable function, we just use   f  =      f          +        −      f          −      . So, which one is it,   μ  (  A  ∪      A    0    ) or   μ  (  A  ∩      A    0    )?

Ashley Parks · Accepted Answer

Your evidence is strong (with some minor typo). You are correct, in my opinion, that it has toμ&amp;nbsp;0&amp;nbsp;(&amp;nbsp;A&amp;nbsp;)&amp;nbsp;=&amp;nbsp;μ&amp;nbsp;(&amp;nbsp;A&amp;nbsp;∩&amp;nbsp;A&amp;nbsp;0&amp;nbsp;).

"Let μ 0 </msub> ( A ) = μ (

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