I am actually studying integration theory for vector-valued functions in a general Banach space, defining the integral with Riemann's sums.
Everything seems to work exactly as in the finite dimensional case:
Let X be a Banach space, $f,g \colon I = [a,b] \to X$, $\alpha$, $\beta \in \mathbb{R}$ then:
$\int_I \alpha f + \beta g = \alpha \int_i f + \beta \int_i g$, $\|\int_I f\| \le \int_I \|f\|$, etc…
The fundamental theorem of calculus holds.
If $f_n$ are continuous and uniformly convergent to $f$ it is also true that $\lim_n \int_I f_n = \int_I f$.
My question is: is there any property that hold only in the finite dimensional case? Is it possible to generalize the construction of the integral as Lebesgue did? If so, does it make sense?
Thank you for your help and suggestions
Best Answer
You can define the integral of $f:A \to X$ implicitly by requiring that $\phi( \int_A f) = \int_A \phi (f)$ for every $\phi \in X^\ast$. If it exists, it must be unique by the Hahn-Banach theorem. You may now use your favorite theory for integrating real valued functions to obtain a version for functions with values in a Banach space (or more generally a locally convex TVS). The difficult part is now to find appropriate conditions under which the integral will exist.
Using this implicit definition, we may easily generalize a lot of properties to the infinite dimensional case. An important property which does not generalize is that the integral operator $\int_A: \mathcal L^1(A;X) \to X$ will no longer be compact.