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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.

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    @saz Sorry I forgot to put it. – 2017-02-24

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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: $$\lim_{x\to\infty} e^{-\lambda x}\int_0^xT(s)g\,ds -\lim_{x\to\infty}\int_0^x(-\lambda)e^{-\lambda t}\int_0^tT(s)g\,ds\,dt$$

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    How do we know that $\int_0^t T(s)gds$ is almost everywhere differentiable from the given assumptions on $T(\cdot)$ and $g$? Also can you give a reference of the theorem on partial integration? I have not seen this version before. Thanks. – 2017-02-24
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    For your first question see [Theorems 15 and 17 here](https://terrytao.wordpress.com/2010/10/16/245a-notes-5-differentiation-theorems/) or [this question](http://math.stackexchange.com/questions/1731528/differentiable-almost-everywhere-of-antiderivative-function?rq=1]). The partial integration step using a.e. differentiability is basically the same result. The thing you need is $\int_0^x \partial_t f(t)\,dt = f(x)-f(0)$ if $f$ is continuous a.e. differentiable. For Theorem 15 in the link says that the derivative of the left-hand side is $\partial_x f(x)$ and of the right-hand its the same a.e. – 2017-02-25