29.1 Weak Integrals*
Definition 29.1.1 (Weakly Measurable).label Let $(X, \cm)$ be a measurable space, $E$ be a locally convex space over $K \in \RC$, and $f: X \to E$, then $f$ is weakly measurable if for each $\phi \in E^{*}$, $\phi \circ f: X \to K$ is Borel measurable.
As I know so little about weak integrals, I will Dunning-Kruger myself right now, give an opinion, and laugh about it later. My gripe with seeing the definition comes from the need to test against every continuous linear functional. To me, this seems quite inflexible: consider integrating a distribution-valued function. Surely it is wiser to only test this function against test functions rather than the dual of $\mathcal{D}'$ (dual with respect to $\mathcal{D}'$ with the bounded convergence topology). As such, it may be more productive to consider a more flexible form of testing, such as using duality.
Definition 29.1.2 (Weak Integrability*).label Let $(X, \cm, \mu)$ be a measure space, $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, and $f: X \to E$, then $f$ is Dunford $\lambda$-integrable* if:
- (I1)
$f$ is weakly measurable.
- (I2)
For each $\phi \in F$, $\phi \circ f \in L^{1}(X; K)$.
- (I3)
For each $A \in \cm$, the mapping
\[F \to K \quad \phi \mapsto \int_{A} \dpn{f(x), \phi}{\lambda}d\mu\]is a continuous linear functional on $F$.
For each $A \in \cm$, the element $\phi \mapsto \int_{A} \dpn{f(x), \phi}{\lambda}d\mu$ of $F^{*}$ is the Dunford $\lambda$-integral of $f$ over $A$, denoted $\int_{A}^{*} f d\mu$.
The function $f$ is Pettis $\lambda$-integrable* if it satisfies (I1), (I2), and
- (I3+)
For each $A \in \cm$, the mapping
\[F \to K \quad \phi \mapsto \int_{A} \dpn{f(x), \phi}{\lambda}d\mu\]is a $\sigma(F, E)$-continuous linear functional on $F$.
In which case, for each $A \in \cm$, $\int_{A}^{*} f d\mu$ is the Pettis $\lambda$-integral of $f$ over $A$.
It is at this point that I start to understand why the bidual setup is useful: existence.
Proposition 29.1.3.label Let $(X, \cm, \mu)$ be a measure space, $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, and $f: X \to E$. If
- (I1)
$f$ is weakly measurable.
- (I2)
For each $\phi \in F$, $\phi \circ f \in L^{1}(X; K)$.
- (C)
$F$ is a Fréchet space.
then $f$ is Dunford $\lambda$-integrable.
Proof, [Section 3.3, Rya02]. Let $T: F \to L^{1}(X; K)$ be defined by $T\phi = \phi \circ f$. To see that $T$ is continuous, it is sufficient to apply the Closed Graph Theorem.
Let $\seq{\phi_n}\subset F$, $\phi \in F$, $\seq{g_n}\subset L^{1}(X; K)$, and $g \in L^{1}(X; K)$ such that
- (i)
$g_{n} = \phi_{n} \circ f$ almost everywhere for all $n \in \natp$.
- (ii)
$\phi_{n} \to \phi$ and $g_{n} \to g$ as $n \to \infty$.
By passing through a subsequence using Theorem 26.7.4, assume further without loss of generality that $g_{n} \to g$ almost everywhere. In which case, $\phi_{n} \circ f \to g$ almost everywhere as well, and $\phi \circ f = g$ almost everywhere. Thus $T$ is continuous.
Now, let $T^{*}: L^{1}(X; K)^{*} \to F^{*}$ be the adjoint of $T$. For each $A \in \cm$, the mapping $\Phi_{A}: L^{1}(X; K) \to K$ defined by $g \mapsto \int_{A} g d\mu$ is a continuous linear functional on $L^{1}(X; K)$. As such, for each $\phi \in F$,
Therefore $T^{*}\Phi_{A} = \int_{A}^{*} f d\mu$ is the desired Dunford $\lambda$-integral of $f$ over $A$.$\square$
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