Fundamental theorem of calculus for semigroups I have a Feller semigroup $(P_t)_{t\geq 0}$. Based on this semigroup I define the linear operator $L = \int_0^tP_s\,ds$ as follows.
$$x \mapsto Lu(x) = \int_{0}^t\int u(y) p_s(x,dy)\,ds$$
where $p_s$ is the unique kernel for $P_s$. What I would like to know is why the following is true.
$$P_tu - u = \int_0^t\frac{d}{ds}(P_su) ds$$
I am self-studying this subject so my apologies if this is too trivial. I wrote the following but I am not happy with it.
\begin{align}
\int_0^t\frac{d}{ds}(P_su) ds &= \int_0^t\frac{d}{ds}\left(\int u(y)p_s(x,dy)\right) ds \\
&= \frac{d}{ds}\int_0^t\int u(y)p_s(x,dy) ds \\
& = \frac{d}{ds}\int_0^tP_su(x) ds \\
& = P_tu(x) - P_0u(x)\\
& = P_tu(x) - u(x)
\end{align}
The reason why I am not convinced of this reasoning is because I am treating these new differentiation and integration operators as if they behave like ordinary differentiation and integration operators. While that may be the case I haven't proven this yet so I am looking for a proof of the fact above in terms of first principles, so to speak.
 A: Let us first the recall the following (standard) definition of the infinitesimal generator

Let $(P_t)_{t \geq 0}$ be a Feller semigroup. Then the infinitesimal generator $A: \mathcal{D}(A) \to C_{\infty}(\mathbb{R}^d)$ is defined by $$Af := \lim_{t \to 0} \frac{P_t f-f}{t}$$ for any $f \in C_{\infty}(\mathbb{R}^d)$ where the limit exists with respect to the uniform norm.

There is the following well-known statement.

Let $(P_t)_{t \geq 0}$ be a Feller semigroup. For any $u \in \mathcal{D}(A)$ and $t>0$ it holds that $P_t u \in \mathcal{D}(A)$ and $$\frac{d}{dt} P_t u = AP_t u = P_t Au.$$

The diffentiability of the mapping $t \mapsto P_t u$ can be shown for a larger class of functions; however, the derivative fails, in general, to be integrable. Consequently, we we cannot expect to prove the identity
$$P_t u-u = \int_0^t \frac{d}{ds} P_s u \, ds \tag{1}$$
for any $u$; we have to restrict ourselves to a smaller class of functions, e.g. $u \in C_{\infty}(\mathbb{R}^d)$ such that $\frac{d}{dt} P_t u(x)$ is Riemann-integrable on $(0,T)$ for any $T>0$ and $x \in \mathbb{R}^d$ (which holds, by the above statement, for instance for any $u \in \mathcal{D}(A)$).
Now, given such a mapping $u$, the identity $(1)$ is a direct consequence of the fundamental theorem of calculus. For fixed $x \in \mathbb{R}^d$ we know that the mapping
$$t \mapsto F(t) := P_t u(x)$$
is differentiable and its derivative is Riemann integrable. By the fundamental theorem of calculus, we find
$$F(t)-F(0)= \int_0^t \frac{d}{ds} F(s) \,ds,$$
i.e.
$$P_t f(x)- \underbrace{P_ 0f(x)}_{=f(x)} = \int_0^t \frac{d}{ds} P_s f(x) \, ds.$$
