Does the antiderivative of an indicator function?

Question

I have an indicator function of the form $I\{a<t<b\}$ 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 $\int_0^t I\{a<s<b\} ds$ (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.

Answer 1

There does not exists a function $f$ defined on an open interval properly containing $(a,b)$ and such that $f'(x)=I\{a<x<b\}(x)$. The reason is that derivatives have the intermediate value property, but the indicator function doesn't.

Answer 2

For $t\in(a,b)$ we have $I(t)=1$, so the antiderivative, say $F$, would be $F(x)=x+C_1$. For $x>b$ it would be $F(x)=C_2$ and for $x<a$, $F(x)=C_3$. It is impossible to find $C_1, C_2, C_3$ to make $F$ a differentiable function on the whole $\Bbb R$, or on any interval containing properly $(a,b)$.