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In what follows, $X$ stands for a Hausdorff LCTVS and $X'$ its topological dual.

Let $(T,\mathcal{M},\mu)$ be a finite measure space, i.e., $T$ is a nonempty set, $\mathcal{M}$ a $\sigma$-algebra of subsets of $T$ and $\mu$ is a nonnegative finite measure on $\mathcal{M}$.

A function $f:T\to X$ is called scalarly integrable iff for each $x'\in X'$, the composition map $$x'\circ f:T\to \mathbb{R}$$ is Lebesgue integrable.

A function $f:T\to X$ is said to be Pettis-integrable iff it is scalarly integrable and for each $E\in \mathcal{M}$, there exists $x_E\in X$ such that $$x'(x_E)=\int_E(x'\circ f)d\mu$$ for all $x'\in X'$. In this case, $x_E$ is called the Pettis integral of $f$ over $E$ and is denoted by $$x_E=\int_E fd\mu.$$

Question. How can we prove that if $f:T\to X$ is Pettis integrable then the map $$m_f:\mathcal{M}\to X$$ defined by $$m_f(E)=\int_E fd\mu, \quad E\in \mathcal{M}$$ is a vector measure on $\mathcal{M}$ which is countably additive?

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Let $\{E_n:n\in\mathbb{N}\}\subset\mathcal{M}$ be a sequence of disjoint sets. By $E$ we denote their union. Fix $x'\in X^*$, then $$ x'(m_f(E)) =x'\left(\int_E fd\mu\right) =\int_E(x'\circ f)d\mu =\sum\limits_{n=1}^\infty\int_{E_n}(x'\circ f)d\mu =\sum\limits_{n=1}^\infty x'\left(\int_{E_n} fd\mu\right) =\sum\limits_{n=1}^\infty x'\left(m_f(E_n)\right) =x'\left(\sum\limits_{n=1}^\infty m_f(E_n)\right) $$ Using Orlicz-Pettis theorem for locally convex spaces we can be sure that the series $\sum\limits_{n=1}^\infty m_f(E_n)$ is convergent. Since $x'\in X^*$ is arbitrary by corollary of Hahn-Banach theorem $$ m_f(E)=\sum\limits_{n=1}^\infty m_f(E_n) $$

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    $\begingroup$ Maybe I'm missing something, but how do you know $\sum_{n=1}^\infty m_f(E_n)$ converges? $\endgroup$ Commented May 14, 2013 at 15:37
  • $\begingroup$ One can use the Orlicz-Pettis Theorem to deduce this. $\endgroup$ Commented May 14, 2013 at 15:45
  • $\begingroup$ @David Mitra, Actually, that is what I want to see. Kluvanek and Knowles in their book "Vector Measures and Control Systems" at p.22, said (without a proof) that the result follows from Orlicz-Pettis Theorem, of which I do not know how... $\endgroup$ Commented May 14, 2013 at 16:02
  • $\begingroup$ @DavidMitra, thank you, now the proof seems to be correct $\endgroup$
    – Norbert
    Commented May 14, 2013 at 16:14
  • $\begingroup$ @juniven Given any subseries $\sum_{k=1}^\infty m_f(E_{n_k})$, using the argument above, it follows that for any $x'$, $x'(m_f(\cup_k E_{n_k} )) = \sum_{k=1}^\infty x'(m_f(E_{n_k}))$. That is, $\sum_{n=1}^\infty m_f(E_n)$ is weakly subseries convergent. O-P then tells you $\sum_{n=1}^\infty m_f(E_n)$ is norm convergent (unconditional even). $\endgroup$ Commented May 14, 2013 at 16:15

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