Define a semigroup $\{T(t)\}$ on $L$ to be measurable if $T(\cdot)f$ is measurable as a function on $[0,\infty),\mathcal{B}[0,\infty))$ for each $f\in L$. We define the full generator $\hat{A}$ of a measurable contraction semigroup $\{T(t)\}$ on $L$ by $$\hat{A}=\{(f,g)\in L\times L: T9t)f-f=\int_0^t T(s)gds, t\ge 0\}.$$
Now let $L=B(\mathbb{R})$ with the sup norm.Below is a part of the proof of Proposition 5.1 from Ethier and Kurtz' Markov Processes. I don't understand the second equality here. I think the authors used integration by parts, but I don't understand how the integration by parts can be used when we don't assume $T(t)g$ to be continuous, so that it has an antiderivative. I would greatly appreciate it if anyone could explain how this works here.
You need that if $T(t)g$ is $L^1_{\mathrm{loc}}$ then $\int_0^t T(s)g\,ds$ is a continuous almost everywhere differentiable function with differential being equal to $T(t)g$ almost everywhere. The differential need not be continuous.
In such a setting you can use partial integration, let $A,B$ be continuous and almost everywhere differentiable then you've got: $$\int_0^x (\partial_t A(t))B(t)\,dt = \int_0^x\partial_t(A(t)B(t))\,dt-\int_0^xA(t)\partial_tB(t)\,dt\\ =(A(x)B(x)-A(0)B(0))-\int_0^xA(t)\partial_tB(t)\,dt$$ Doing this with $\int_0^\infty e^{-\lambda t}T(t)g\,dt$ ($A=\int_0^tT(s)g\,ds$, $B=e^{-\lambda t}$) gives you: $$\lim_{x\to\infty} e^{-\lambda x}\int_0^xT(s)g\,ds -\lim_{x\to\infty}\int_0^x(-\lambda)e^{-\lambda t}\int_0^tT(s)g\,ds\,dt$$