Does the antiderivative of an indicator function?I have an indicator function of the form I { a &lt; t &lt; b } and I need the antiderivative of it with respect to t.I believe I can't use the fundamental theorem of calculus to simply write it as ∫ 0 t I { a &lt; s &lt; b } d s (which would be very convenient) since it's not a continuous function.It's not clear to me whether such a function even exists or how I would write it.

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Does the antiderivative of an indicator function?I have an indicator function of the form   I  {  a  &amp;lt;  t  &amp;lt;  b  } and I need the antiderivative of it with respect to t.I believe I can&#039;t use the fundamental theorem of calculus to simply write it as       ∫    0    t    I  {  a  &amp;lt;  s  &amp;lt;  b  }  d  s (which would be very convenient) since it&#039;s not a continuous function.It&#039;s not clear to me whether such a function even exists or how I would write it.

alisonhleel3 · Accepted Answer

Explanation:There does not exists a function f defined on an open interval properly containing (a,b) and such that       f    ′    (  x  )  =  I  {  a  &amp;lt;  x  &amp;lt;  b  }  (  x  ). The reason is that derivatives have the intermediate value property, but the indicator function doesn&#039;t.

polivijuye · Accepted Answer

Explanation:For   t  ∈  (  a  ,  b  ) we have   I  (  t  )  =  1, so the antiderivative, say F, would be   F  (  x  )  =  x  +      C    1  . For   x  &amp;gt;  b it would be   F  (  x  )  =      C    2   and for   x  &amp;lt;  a,   F  (  x  )  =      C    3  . It is impossible to find       C    1    ,      C    2    ,      C    3   to make F a differentiable function on the whole       R  , or on any interval containing properly (a,b).

Does the antiderivative of an indicator function? I have an indicator function of the form I

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