Let
- $E$ be a $\mathbb R$-Banach space;
- $(T(t))_{t\ge0}$ be a semigroup on $E$ and $$E_0:=\left\{x\in E:\left\|T(t)x-x\right\|_E\xrightarrow{t\to0+}0\right\};$$
- $A$ denote the generator of $(T(t))_{t\ge0}$.
Note that
- $T(t)\mathcal D(A)\subseteq\mathcal D(A)$ and $$AT(t)x=T(t)Ax\tag0$$ for all $t\ge0$;
- $\mathcal D(A)\subseteq E_0$.
Assume that $(T(t))_{t\ge0}$ is locally bounded$^1$. Then,
- $T(t)\overline{\mathcal D(A)}\subseteq\overline{\mathcal D(A)}$ for all $t\ge0$;
- If $x\in E_0$, then $$\int_0^tT(s)x\:{\rm d}s\in\mathcal D(A)\tag2$$ and $$A\int_0^tT(s)x\:{\rm d}s=T(t)x-x\tag3$$ for all $t\ge0$;
- $\mathcal D(A)$ is a dense subset of $E_0$.
Question: Let $x\in\mathcal D(A)$. Are we able to show that $$A\int_0^tT(s)x\:{\rm d}s=\int_0^tT(s)Ax\:{\rm d}s\tag4$$ for all $t\ge0$?
Formally, it is easy to derive $(4)$ given the present assumptions. The only technical issue which I don't see how I can resolve it is the Borel measurability of $$[0,\infty)\to E\;,\;\;\;s\mapsto T(s)Ax\tag5$$ which is needed for the integral of the right-hand side of $(4)$ to be well-defined.
$^1$ i.e. $$\exists t_0>0:\sup_{t\in[0,\:t_0)}\left\|T(t)\right\|_{\mathfrak L(E)}<\infty\tag1.$$
It's a Riemann integral that's a limit of Riemann sums where the corresponding Riemann sum identity holds. Then you can used the fact that $A$ is closed on order to take the limit of the sums.