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


*

*for each $x'\in X'$, the composition map
$$x'\circ f:T\to \mathbb{R}$$
is Lebesgue 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
$$x_E=\int_E fd\mu.$$
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...

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, $m_f\ll\mu$
