Characterization of the generator of a measurable contraction semigroup

Let $$E$$ be a $$\mathbb R$$-Banach space and $$(T(t))_{t\ge0}$$ be a contraction$$^1$$ semigroup on $$E$$. Assume $$[0,\infty)\to E\;,\;\;\;t\mapsto T(t)x\tag1$$ is Borel measurable for all $$x\in E$$ and hence $$\mathcal A:=\left\{(x,y)\in E\times E\mid\forall t\ge0:T(t)x-x=\int_0^tT(s)y\:{\rm d}s\right\}$$ is well-defined.

Are we able to show that$$^2$$ $$A:=\left\{(x,y)\in\mathcal A:y\in\overline{\mathcal D(\mathcal A)}\right\}$$ is the generator of $$(T(t))_{t\ge0}$$?

I was able to show the following:

1. $$A$$ is single-valued$$^3$$ and hence the graph of a linear operator on $$E$$
2. If $$t\ge0$$ and $$(x,y)\in\mathcal A$$, then $$(T(t)x,T(t)y)\in\mathcal A$$. In particular, $$T(t)\mathcal D(\mathcal A)\subseteq\mathcal D(\mathcal A)$$ for all $$t\ge0$$.
3. $$T(t)\overline{\mathcal D(\mathcal A)}\subseteq\overline{\mathcal D(\mathcal A)}$$ for all $$t\ge0$$.
4. $$(T(t))_{t\ge0}$$ is strongly continuous on $$\overline{\mathcal D(\mathcal A)}$$
5. $$\mathcal A$$ is closed (with respect to the product topology on $$E\times E$$)

Let $$(\mathcal D(B),B)$$ denote the actual generator of $$(T(t))_{t\ge0}$$. It should be easy to show that $$A\subseteq\left\{(x,Bx):x\in\mathcal D(B)\right\}\tag2.$$ To do so, let $$(x,y)\in A$$. We need to show that $$\left\|\frac{T(t)x-x}t-y\right\|_E\xrightarrow{t\to0+}0.\tag3$$ Since $$(x,y)\in A$$, $$\left\|\frac{T(t)x-x}t-y\right\|_E=\frac1t\left\|\int_0^tT(s)y-y\:{\rm d}s\right\|_E\le\frac1t\int_0^t\left\|T(s)y-y\right\|_E\:{\rm d}s.\tag4$$ Let $$\varepsilon>0$$. Since $$y\in\overline{\mathcal D(\mathcal A)}$$, 4. yields that there is a $$\delta>0$$ with $$\left\|T(s)y-y\right\|<\varepsilon\;\;\;\text{for all }s\in[0,\delta)\tag5$$ and hence $$\frac1t\int_0^t\left\|T(s)y-y\right\|_E<\varepsilon\:{\rm d}s.\tag5$$ Thus, we should obtain $$(3)$$ and hence $$(2)$$.

How can we show the other inclusion?

Since $$\left\{(x,Bx):x\in\mathcal D(B)\right\}\subseteq\mathcal A\tag6$$, we only need to show that $$B\mathcal D(B)\subseteq\overline{\mathcal D(\mathcal A)}$$.

EDIT: Maybe an argument of the following kind is possible: Even when $$(T(t))_{t\ge0}$$ is not strongly continuous (on all of $$E$$) it is always strongly continuous on $$\overline{\mathcal D(B)}$$ (since $$(T(t))_{t\ge0}$$ is locally bounded (even contractive)). This should be enough to show that $$(\mathcal D(B),B)$$ is a closed operator (in the same way as it is usually done for strongly continuous semigroups). This means that $$\overline{\left\{(x,Bx):x\in\mathcal D(B)\right\}}$$ is closed and (since $$\mathcal A$$ is closed) still contained in $$\mathcal A$$.

$$^1$$ i.e. $$\left\|T(t)\right\|_{\mathfrak L(E)}\le1$$ for all $$t\ge0$$.

$$^2$$ As usual, if $$B\subseteq E\times E$$, then $$\mathcal D(B):=\left\{x\in E\mid\exists y\in E:(x,y)\in B\right\}.$$

$$^3$$ i.e. $$\forall y\in E:(0,y)\in A\Rightarrow y=0$$.

Okay, if I'm not missing anything, this is quite trivial: Let $$x\in\mathcal D(B)$$. By $$(6)$$ and 2., $$\frac{T(t)x-x}t\in\mathcal D(A)\tag7\;\;\;\text{for all }t>0$$ and hence $$Bx=\lim_{t\to0+}\frac{T(t)x-x}t\in\overline{\mathcal D(A)}.\tag8$$