Stone's Theorem Integral: Avanced Integral Reference
This problem grew out from: Stone's Theorem Integral: Basic Integral
Problem
Given the real line as measure space $\mathbb{R}$ and a Hilbert space $\mathcal{H}$.
Consider a strongly continuous unitary group $U:\mathbb{R}\to\mathcal{B}(\mathcal{H})$.
Take the time evolution $\varphi(t):=U(t)\varphi$.
This time the integral is taken over an infinite measure:
$$\int_0^\infty e^{-\lambda s}\varphi(s) \, \mathrm ds$$
What interpretations are available and how do they agree?
 A: The Riemann integral for vector functions with values in a Banach space $X$ is essentially the same as for scalar functions.
$$
             \int_{a}^{b} F(t)\,dt = \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{\mathcal{P}} F(t_{j}^{\star})\Delta_{j}t,
$$
where $\mathcal{P}$ is a partition with partition points
$$
                           a = t_{0} < t_{1} < t_{2} < \cdots < t_{n} =b
$$
and augmented with evaluation points $t_{j}^{\star}\in [t_{j-1},t_{j}]$. If you're used to the Darboux-Riemann integral using upper and lower integrals, then forget that one. Stick with the classical definition stated above. For scalar functions, these definitions are equivalent. Of course that's not the case here because The Darboux-Riemann integral makes no sense.
For scalar functions $F$, the integral exists iff the set of discontinuities of $F$ is of Lebesgue measure $0$. I believe the 'if' part is still true for Banach-space-valued functions. However, if you're concerned, piecewise continuous is always safe. Your integral over the infinite interval has to be an improper Riemann integral, just as it is for scalar functions, and the integral makes sense if the integrand is absolutely Riemann integrable on the infinite interval. So the following makes sense as an improper Riemann integral:
$$
  \int_{0}^{\infty}e^{-\lambda t}U(t)x\,dt,\;\;\; \mathcal{R}\lambda > 0,\;\; x\in X.
$$
This Laplace transform integral is fundamental to $C^0$ semigroup theory because it gives the resolvent of the generator.
A: Bochner
Since it is separable valued:
$$\varphi\in\mathcal{C}(\mathbb{R},E):\quad\mathbb{R}\text{ separable}\implies (\alpha\varphi)(\mathbb{R})\text{ separable}$$
and weakly measurable:
$$l\in E':\quad(\alpha\varphi)\text{ continuous}\implies l\circ(\alpha\varphi)\text{ measurable}$$
so by Pettis' criterion strongly measurable:
$$\varphi\text{ Bochner measurable}$$
Also it is absolutely integrable:
$$\int\|\varphi(s)\|\alpha(s)\mathrm{d}s=\frac{1}{\lambda}\|\varphi\|<\infty$$
So the Bochner integral exists!
Improper Bochner
This one coincides with the former by dominated convergence.
Improper Riemann
Especially, it is bounded:
$$\lambda(A)<\infty:\quad\|\alpha\varphi\|_A\leq1\|\varphi\|<\infty$$
so for subspaces of finite measure:
$$\lambda(A)<\infty:\quad\alpha\varphi\in\mathcal{L}_\mathfrak{R}(A)\cap\mathcal{L}_\mathfrak{B}(A)$$
But the real line is $\sigma$-finite so one has:
$$\int_A\alpha(s)\varphi(s)\mathrm{d}s\to\int_0^\infty\alpha(s)\varphi(s)\mathrm{d}s$$
So the improper Riemann integral exists and agrees with others!
Induced Bochner
Consider the induced Borel measure:
$$\mu(A):=\int_A\alpha(s)\mathrm{d}s:\quad\mu(\mathbb{R})=\frac{1}{\lambda}<\infty$$
Then one has absolute integrability as:
$$\int\|\varphi(s)\|\mathrm{d}\mu(s)=\int\|\varphi(s)\|\alpha(s)\mathrm{d}s<\infty$$
and again measurability by continuity.
So it is Bochner integrable.
As the function is measurable and bounded one can construct:
$$\|\sigma_n\|_\infty\leq\|\varphi\|_\infty+1:\quad\sigma_n\to\varphi$$
Thus one obtains by dominated convergence:
$$\int\varphi(s)\mathrm{d}\mu(s)\leftarrow\int\sigma_n(s)\mathrm{d}\mu(s)=\int\sigma_n(s)\alpha(s)\mathrm{d}s\to\int\varphi(s)\alpha(s)\mathrm{d}s$$
So the induced Bochner integral exists and agrees with others!
Induced Riemann
One last time by boundedness:
$$\|\varphi\|_\infty<\infty:\quad\varphi\in\mathcal{L}_\mathfrak{R}(\mu)\cap\mathcal{L}_\mathfrak{B}(\mu)$$
So the induced Riemann integral exists and agrees with others!
