I can`r do this, need help Given a locally integrable function &#x03B3;<!-- γ --> :

Mohamed Mooney 2022-06-13 Answered
I can`r do this, need help
Given a locally integrable function γ : R 0 R , we define the absolutely continuous function Γ ( t ) := max u t u t γ d λ.
I want to show, that γ ( t ) 0 holds for almost all t with Γ ( t ) = 0. In other words, I want to show that the set
A := { t R 0 | Γ ( t ) = 0   and   γ ( t ) > 0 }
is a Lebesgue-null set, i.e. λ ( A ) = 0.
All my attempts have failed. Nevertheless, I was able to show, that if Γ vanishes on a (proper) interval [ a , b ] with a < b, then λ ( A [ a , b ] ) = 0. This however does not lead to a proof of the more general claim λ ( A ) = 0.
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Answers (2)

Marlee Norman
Answered 2022-06-14 Author has 18 answers
Let B be the following measurable set.
B = { t ( 0 , ) lim u t 1 t u u t γ ( s ) d s = γ ( t ) } .
By the differentiation theorem, R 0 B is a Lebesgue null set. I leave it to you to show that Γ > 0 holds in the measurable set B { γ > 0 }. Hence Γ > 0 holds Lebesgue almost everywhere in { γ > 0 }, or { Γ = 0 , γ > 0 } is Lebesgue null.
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minwaardekn
Answered 2022-06-15 Author has 6 answers
We first define two measurable sets:
B := { t R 0 | x 0 x γ d λ  is differentiable in  t  with derivative  γ ( t ) }
C := { t R 0 | Γ  is differentiable in  t  with  Γ ( t ) = { γ ( t ) , for  Γ ( t ) > 0 , 0 , otherwise. }
By Lebesgue's differentiation theorem, we know that λ ( B c ) = 0.
Now let t B C with Γ ( t ) = 0. Then we have
d d x t x γ d λ | x = t = d d x 0 x γ d λ 0 t γ d λ | x = t = γ ( t )
and thus it holds that
0 = Γ ( t ) = lim x t Γ ( x ) Γ ( t ) x t = lim x t Γ ( x ) t t γ d λ x t lim x t t x γ d λ t t γ d λ x t = d d x t x γ d λ | x = t = γ ( t ) .
This means A B C = and thus A ( B C ) c . Moreover, we have
λ ( A ) λ ( ( B C ) c ) = λ ( B c C c ) λ ( B c ) + λ ( C c ) = 0
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