Antiderivative of lebesgue integrable functionLet f : [ 0 , b ] → R be Lebesgue integrable. We define g ( x ) = ∫ x b f ( t ) t d t , 0 &lt; x ≤ b .Show that g(x) is Lebesgue integrable in [0,b]. Also show that ∫ 0 b g ( x ) d x = ∫ 0 b f ( t ) d t .If we show the asked equality then it is obvious that g is Lesbegue integrable, since f is. Now, I have try to show the equality using the fact that, ∫ 0 b [ g ′ ( x ) − f ( x ) x ] d x = 0, since g is the antiderivative of f but I had no success. I am pleased to know other ideas to approach this problem.

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Antiderivative of lebesgue integrable functionLet   f  :  [  0  ,  b  ]  →      R   be Lebesgue integrable. We define  g  (  x  )  =      ∫    x    b              f      (      t      )        t    d  t  ,    0  &amp;lt;  x  ≤  b  .Show that g(x) is Lebesgue integrable in [0,b]. Also show that      ∫    0    b    g  (  x  )  d  x  =      ∫    0    b    f  (  t  )  d  t  .If we show the asked equality then it is obvious that g is Lesbegue integrable, since f is. Now, I have try to show the equality using the fact that,      ∫    0    b              [            g    ′    (  x  )  −            f      (      x      )        x              ]        d  x  =  0, since g is the antiderivative of f but I had no success. I am pleased to know other ideas to approach this problem.

thatuglygirlyu · Accepted Answer

Explanation:Let   Δ  =  {  (  x  ,  t  )  ∣  0  ≤  x  ≤  t  ≤  b  }. Define h(x,t) on   [  0  ,  b  ]  ×  [  0  ,  b  ] by   h  (  x  ,  t  )  =      χ          Δ        (  x  ,  t  )            f      (      t      )        t  . Apply Fubini&#039;s theorem to h to integrate it in two different ways. (There will be an initial step involving |h| to prove integrability of h.)Edit:   Δ would more accurately be changed in order to be a subset of   (  0  ,  b  ]  ×  (  0  ,  b  ]

Antiderivative of lebesgue integrable function Let f : [ 0 , b ] →<

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