Does equality of antiderivatives imply equality almost everywhere?If two Lebesgue integrable functions f , g : [ a , b ] → R satisfy ∫ a x f ( s ) d s = ∫ a x g ( s ) d s ,for every x ∈ [ a , b ] , , then is is it true that f ( x ) = g ( x ) almost everywhere?

Question

Does equality of antiderivatives imply equality almost everywhere?If two Lebesgue integrable functions     f  ,  g  :  [  a  ,  b  ]  →      R     satisfy      ∫    a    x    f  (  s  )    d  s  =      ∫    a    x    g  (  s  )    d  s  ,for every     x  ∈  [  a  ,  b  ]  ,  , then is is it true that   f  (  x  )  =  g  (  x  ) almost everywhere?

Gallichi5mtwt · Accepted Answer

Step 1It suffices to show that:- If       f  ∈      L    1    [  a  ,  b  ]     and                 ∫      a      x        f    (    s    )        d    s    =    0    ,          , for all     x  ∈  [  a  ,  b  ]  ,  , then     f  =  0   a.e.Set   μ  (  E  )  =      ∫    E    f  (  s  )    d  s  .  . Then   μ defines a signed measure in [a,b]. We know that   μ  (  F  )  =  0, for all subintervals F of [a,b]. In particular   μ  (  U  )  =  0, for all U open in [a, b], as they are countable unions of intervals.Let now   λ  ,  ν be two signed-measures on       B    (  [  a  ,  b  ]  ), the Borel subsets of [a,b]. Clearly, if   λ  (  [  a  ,  b  ]  )  =  μ  (  [  a  ,  b  ]  ), then the collection of sets       S    =  {  E  ∈      B    (  [  a  ,  b  ]  )  :  λ  (  E  )  =  ν  (  E  )  }  ,, is a   σ-algebra.Step 2In our case   μ agrees with the identically zero measure on the open sets of [a,b], and hence on the   σ-algebra they generate, which is       B    (  [  a  ,  b  ]  ). Thus   0  =  μ  (  E  )  =      ∫    E    f  (  x  )    d  x  ,, for all     E  ∈      B    (  [  a  ,  b  ]  ). Let   A  =  {  x  ∈  [  a  ,  b  ]  :  f  (  x  )  ≥  0  } and   B  =  {  x  ∈  [  a  ,  b  ]  :  f  (  x  )  &amp;lt;  0  }. Then A and B are Borel sets and       ∫    A    f  (  x  )    d  x  =      ∫    B    f  (  x  )    d  x  =  0..But as   f  (  x  )  ≥  0 and       ∫    A    f  (  x  )    d  x  =  0, then   f  (  x  )  =  0 a.e. in A, and so happens in B. Thus   f  (  x  )  =  0 a.e. in [a,b].

et1wdwk4o · Accepted Answer

Step 1We prove if f be Lebsegue integrable on [a,b] which satisfies       ∫    a    c    f  d            L        1    =  0  ,, for every c. Show that f is equal to 0 a.e..Step 2We can define two measures       m    +    ,      m    −   on [a,b] by       m    +    (  E  )  =      ∫    E        f    +    d            L        1    ,      m    −    (  E  )  =      ∫    E        f    −    d            L        1  Using       ∫    a    c    f  d            L        1    =      m    +    (  (  a  ,  c  )  )  −      m    −    (  (  a  ,  c  )  )  =  0, we can see that       m    +    (  I  )  =      m    −    (  I  ), for every open interval   I  ⊂  [  a  ,  b  ]. Since every open set can be represented as a countable union of disjointed intervals, one has that       m    +    (  O  )  =      m    −    (  O  ), for every open set   O  ⊂  [  a  ,  b  ]. Since Borel sets are generated by open sets, this holds for every Borel, and hence measurable sets E. This shows that       ∫    E    f  d            L        1    =      m    +    (  E  )  −      m    −    (  E  )  =  0.Setting   E  =  {  x  ∈  [  a  ,  b  ]      |    f  (  x  )  ≥  0  }, we have       f    +    =  0. Similarly       f    −    =  0. Hence   f  =  0 a.e.Step 3If they are not Lebesgue integrable, the result is not true. Consider   [  a  ,  b  ]  =  [  0  ,  1  ], where   f  (  x  )  =      1    x    ,  g  (  x  )  =      1          2      x        ∀  x  ∈  (  0  ,  1  ]  ,  f  (  0  )  =  g  (  0  )  =  0, the integral is always   ∞, but f, g are not equal a.e..

Does equality of antiderivatives imply equality almost everywhere? If two Lebesgue integrable funct

Answered question

Answer & Explanation

New Questions in Calculus 1