Question: Suppose there exists an operator $I: C^{\infty}(0,1) \to \mathbb R$ satisfying the following properties:
(1) $I (\chi_{(0,1)})=1$ ;
(2) $I(kf)=kI(f)$, where $k\in \mathbb R$ and $f\in C^{\infty}(0,1)$;
(3) $I(f+g)=I(f) + I(g)$, where $f,g\in C^{\infty}(0,1)$;
(4) $I(f)\le I(g)$, whenever $f\le g$ on $(0,1)$
Can we conclude that $I$ must be the integral operator $\int_0^1 d \mu$ for some measure $\mu$ on $(0,1)$?
Note: What if the property (3) is replaced by:
(3)'$I(f+g)\le I(f) + I(g)$, where $f,g\in C^{\infty}(0,1)$