# About the induced vector measure of a Pettis integrable function(part 2)

Notations: 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 $\mathcal{M}$.

Definition. A function $f:T\to X$ is said to be Pettis-integrable if

1. for each $x'\in X'$, the composition map $$x'\circ f:T\to \mathbb{R}$$ is Lebesgue integrable and

2. 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 $$x_E=\int_E fd\mu.$$

Remark. Let $f:T\to X$ be Pettis-integrable. Define $$m_f:\mathcal{M}\to X$$ by $$m_f(E)=\int_E fd\mu$$ for any $E\in \mathcal{M}.$ Hahn-Banach Theorem ensures that $x_E$ in the above definition is necessarily unique and so $m_f$ is a well-defined mapping. Moreover, Orlicz-Pettis Theorem imply that the induced vector measure $m_f$ is countably additive, see for instance this.

Question. With the above discussions, how do we show that $m_f$ is $\mu$-continuous?

I would be thankful to someone who can help me...

• What is definition of $\mu$-continuiuty? Commented May 18, 2013 at 3:28
• I get the entire discussions above in an article, entitled "Pettis Integration in Locally Convex Spaces", by Ali and Chakraborty. They did not put the definition of $\mu$-continuous, I guess they are referring $\mu$-continuous to the absolute continuity of the vector measure $m_f$ with respect to $\mu$. Commented May 18, 2013 at 3:57

Let $\mu(E)=0$. Take arbitrary $x'\in X'$, then $$x'(m_f(E))=\int_E (x'\circ f)d\mu=0$$ then. Since $x'\in X'$ is arbitrary by corollary of Hahn-Banach theorem $m_f(E)=0$. Thus, $m_f\ll\mu$
• I got the whole idea of your answer:) In a locally convex Hausdorff space, the topological dual $X'$ separates the points of $X$ because of the Geometric form of the Hahn-Banach Theorem. Am I right? Commented May 18, 2013 at 4:57