I can`r do this, need helpGiven 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 ) &gt; 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 &lt; 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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I can`r do this, need helpGiven 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    )    &amp;gt;    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  &amp;lt;  b, then   λ  (  A  ∩  [  a  ,  b  ]  )  =  0. This however does not lead to a proof of the more general claim   λ  (  A  )  =  0.

Marlee Norman · Accepted Answer

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  &amp;gt;  0 holds in the measurable set   B  ∩  {  γ  &amp;gt;  0  }. Hence   Γ  &amp;gt;  0 holds Lebesgue almost everywhere in   {  γ  &amp;gt;  0  }, or   {  Γ  =  0  ,      γ  &amp;gt;  0  } is Lebesgue null.

minwaardekn · Accepted Answer

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                    )                    &amp;gt;                    0                                    ,                                                                                    0                ,                                            otherwise.                                                                    }  By Lebesgue&#039;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

I can`r do this, need help Given a locally integrable function γ :

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