"Let μ 0 ( A ) = μ ( A ∪ A 0 ) ,...

kolutastmr

kolutastmr

Answered

2022-07-12

"Let μ 0 ( A ) = μ ( A A 0 ) , A F be a measure. Show that if
f d μ
exists, 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 )?

Answer & Explanation

Ashley Parks

Ashley Parks

Expert

2022-07-13Added 11 answers

Your proof looks good (with some minor typo). I think you are right that it should be μ 0 ( A ) = μ ( A A 0 ).

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