$(P_t)_{t\geq 0}$ is the semigroup of $d$-dimensional Brownian motion, i.e. $$P_tu(x) = E^x[u(B_t)]$$ where $u$ is a real-valued and bounded function on $\mathbb{R}^d$. Then, it is true that $P_tu$ is continuous and bounded.
Furthermore, whenever $u$ is continuous and vanishing, $x \mapsto \int_0^tP_su(x)\,ds$ belongs to the domain of the generator of Brownian motion, denoted by $\mathfrak{D}(\Delta)$.
The book that I am following (Schilling and Partzsch) gives proofs of these facts but also claims the following.
Let $g\in\mathfrak{D}(\Delta)$. Then the mapping $$(t,x) \mapsto \int_0^tP_sg(x)\,ds$$ is once continuously differentiable in $t$ on $(0,\infty)$ and twice continuously differentiable in $x$ on $\mathbb{R}^d$.
There is no proof of this latter claim but it is mentioned that it follows from the proof of the first fact and the statement of the second fact above. I don't see how. I reason as follows. $$g\in\mathfrak{D}(\Delta)\implies g \in C_{\infty}(\mathbb{R}^d)\implies x \mapsto \int_0^tP_sg(x)\,ds \in \mathfrak{D}(\Delta)$$
If $d >1$, $\mathfrak{D}(\Delta) = \{u \in C_{\infty}(\mathbb{R}^d): \text{ in the weak sense of derivative }\Delta u \in C_{\infty}(\mathbb{R}^d)\}$. So clearly, my reasoning above does not lead to the conclusion that $x \mapsto \int_0^tP_sg(x)\,ds$ is twice continuously differentiable. Then there is still the question of why it is once continuously differentiable in $t$ but something like
$$\frac{d}{dt}\int_0^tP_sg\,ds = P_tg$$
must be true even though I did not check this.