Pettis integrability condition Let $B$ be a Banach Space, $(X,\mu)$ a measure space and $f:X\longrightarrow B$ a weak measurable function, that is a function such that:
$$ \phi\circ f:X \longrightarrow \mathbb C $$
is measurable forall $\phi\in B^*$ (the topological dual).
If $||f(x)||\leq M(x)$ with $M\in L^1$, can we deduce that $f$ is Pettis integrable?
 A: $\newcommand{\bf}[1]{\mathbb #1}\newcommand{\sc}[1]{\mathscr #1}$The answer is affirmative for separable Banach spaces.
In order to prove it, consider the linear functional $\Lambda $ defined on the (topological) dual space $B'$  by
$$
  \Lambda (\varphi )=\int_X \varphi \big (f(x)\big )\, d\mu (x), \quad\forall \varphi \in B'.
  $$
Notice that
the integrand satisfies
$$
  |\varphi \big (f(x)\big )| \leq  \|\varphi \| \|f(x)\| \leq  \|\varphi \| M(x),
  $$
so
$$
  |\Lambda (\varphi )|\leq \|\varphi \|\|M\|_1,\quad\forall \varphi \in B',
  $$
and hence  $\Lambda $ is seen to be continuous.  We next claim that $\Lambda $ is also continuous relative to  the
weak-star topology on $B'$.
Using
V.12.8 in Conway's "A Course in Functional Analysis", a Corollary of Krein-Smulian's Theorem,  it is enough to show that $\Lambda $ is weak-star sequentially
continuous.   So let us suppose that $\{\varphi _n\}_n$ is a sequence in $B'$ converging to some $\varphi \in B'$ relatively to  weak-star topology.
If tollows that
$$
  \varphi _n\circ f \to  \varphi \circ f
  $$ pointwise everywhere on $X$.  Moreover, by the Banach-Steinhauss Theorem, we have that the $\varphi _n$ are uniformly
bounded, whence
$$
  K := \sup_{n\in {\bf N}}\|\varphi _n\| <\infty ,
  $$
and hence
$$
  |\varphi _n(f(x))| \leq  \|\varphi _n\| \|f(x)\| \leq   K M(x),\quad \forall x \in X.
  $$
We may then  use Lebesgue's Dominated Convergence Theorem to conclude that
$$
  \lim_{n\to \infty } \Lambda (\varphi _n) =
  \lim_{n\to \infty } \int_X \varphi _n(f(x))\, d\mu (x) = $$$$ =
  \int_X \varphi (f(x))\, d\mu (x) = \Lambda (\varphi ),
  $$
hence concluding the proof of the weak-star continuity of $\Lambda $.
As it is well known, every weak-star continuous linear functional on $B'$ must be a point-evaluation, so there exists
$b\in B$ such that
$$
  \Lambda (\varphi )= \varphi (b),\quad\forall \varphi \in B'.
  $$
Consequently
$$
  \int_X \varphi (f(x))\, d\mu (x) = \varphi (b),
  $$
so $f$ is Pettis integrable!

Thanks to MathOverflow user @Mikael de la Sale for pointing me towards Krein-Smulian's Theorem!
