Let $X$ be a Banach space. Suppose I have a 2 parameter family of bounded operators on $X$: $V(t,s)$, $0\leq s\leq t \leq T$, such that $V(t,s)x=U(t,s)x+\int_s^t V(t,r)H(r)U(r,s)x\,dr$ and $(t,s)\mapsto V(t,s)x$ continuous (both for every $x\in X$). The only relevant information about $H(t), U(t,s)$ are that they're strongly continuous bounded operators and have a uniform bound on their norms (say by $C$).
I want to show uniqueness of the solution. This is how the book (Pazy's Semigroups) does it: Let $V(t,s), V_1(t,s)$ be two solutions of the integral equation. Set $W(t,s)=V(t,S)-V_1(t,s)$. Then $W(t,s)x=\int_s^tW(t,r)H(r)U(r,s)x\,dr$. Fix $t$; we have $\|W(t,s)x\|\leq C^2\int_s^t\|W(t,r)x\|dr$, so by Gronwall's inequality $W(t,s)x=0$.
The problem is these operators don't commute so I can't "get $x$ to stick with $W(t,s)$". I tried to resolve this by just using Gronwall's inequality on $\|W(t,s)\|\leq C^2\int_s^t\|W(t,r)\|dr$, but to do so I need that $s\mapsto\|W(t,s)\|$ is measurable. I'm not sure if this is implied by the fact that $s\mapsto W(t,s)$ is strongly continuous.