Remark 1

Suppose that G and H are Banach spaces, and let U be a mapping of T into L(G; H) such that (t, x) ↦ U(t) · x is continuous on T × G. Note that this implies that the finite function t ↦ \|U(t)\| is bounded on every compact subset of T and lower semi-continuous on T, being the upper envelope of the continuous functions $t \mapsto |U(t) \cdot x|$ as $x$ runs over the ball $|x| \leq 1$ in G. Set $h(t) = \| U(t) \|$. Then, for every positive measure $\mu$ on T such that $h$ is $\mu$-integrable, we again have $\int U d\mu \in \mathscr{L}(G; H)$. For, the measure $\nu = h \cdot \mu$ is bounded by hypothesis; there therefore exists a partition of T formed by a $\nu$-negligible set N and a sequence $(K_n)$ of compact subsets. The argument made at the beginning of this No., applied to the measure $\varphi_{K_n} \cdot \mu$, shows that

and, moreover (No. 2, Prop. 6), $\| A_n \| \leq \int \varphi_{K_n} \| U \| \, d\mu \leq \nu(K_n)$. The series with general term $A_n$ is therefore absolutely convergent in the Banach space $\mathscr{L}(G; H)$, and it is immediate that its sum is $\int U \, d\mu$ and that $\| \int U \, d\mu \| \leq \int \| U \| \, d\mu$.