Let $(E,\mathcal {A},\mu)$ be a finite measure space and let $ (X, \|\cdot\|)$ be a reflexive Banach space. Let $\{f_n\}\subset \mathcal{L}_{X}^{1}$ be a sequence with : $$ \sup_{n}\int_{E}\|f_n\|<+\infty $$ Suppose that there exists a null set $N$ such that $f_n(T\smallsetminus N)$ is a separable subset of $(X, \|\cdot\|)$, $k = 1, 2 ....$
Let $Y$ be the closed linear subspace of $X$ which is generated by the union of all sets $f_n(T\smallsetminus N)$, $k=1, 2 ....$ ; then $(Y,\|\cdot\|)$ is clearly a separable reflexive Banach space. Hence, the dual space $Y^*$ of $Y$ is certainly separable for the dual norm. Let $\{y_j^*\}$ be a countable dense subset of $Y^*$.
Suppose that there exists the function $\phi_0,\phi_1,\phi_3,....$ in $\mathcal{L}_{\mathbb{R}}^{1}$ such that for every sub-sequence $\{g_{m}\}$ de $\{f_n\}$ :
\begin{equation} \frac{1}{k}\sum_{i=1}^{k}{\|g_{m}\|}\to \phi_{0} ~~~~a.e.~~ \text{sur}~E\setminus A~~~~~~~(1) \end{equation}
\begin{equation} \frac{1}{k}\sum_{i=1}^{k}{\langle y_{j}^{*} ,g_{m}\rangle_{Y^{*},Y}}\to \phi_{j} ,~~~~j=1,2,...~~a.e.~~ \text{sur}~E\setminus A~~~~~~~(2) \end{equation}
Denote by $M$ the union of the exceptional set for $(1)$-$(2)$ and $N$ if $\{f_n\}$ itself acts as the sub-sequence; of course, $M $ is a null set. Define also ${\displaystyle s_n:=\frac{1}{n}\sum_{m=1}^{n}{g_m}}$.
Suppose that for every $t\in E\setminus M$ $s_n(t)$ converges weakly to a point $y_t\in Y$ in $Y$. A fortiori this implies that for every $t$ in $E\setminus M$ $s_n(t)\to y_t$ weakly in X, such that : $$ \langle y_i^*,y_t\rangle_{Y^{*},Y}=\phi_j(t);~~~~j=1,2,...~~~~~~(3) $$
We consider the function $ f:E\to X $ defined by :
$$f(t) = \begin{cases} y_t&\text{if } ~~t\in E\setminus M\\\ 0&\text{if }~~t\in M\;. \end{cases}$$
Show by "the Pettis measurability theorem" that $f$ is strongly measurable.
THEOREM:(Pettis measurability theorem) A function $f: E \to X$ is strongly measurable if and only if :
(i) There exists $M\in \mathcal{A}$ with $\mu (M) = 0$ and such that $f(E\setminus M)$ is a (norm) separable subset of $X$, and
(ii) $f$ is weakly $\mu$-measurable.
Definition : A function $f : E \to X $ is said to be weakly $\mu$-measurable. if for all $x^{*}\in X^{*}$, the application : $$ x \mapsto \langle x^{*},f(x)\rangle_{X^{*},X} $$ is measurable from $E$ to $\mathbb{R}$.
My effort :
$(i)$ we know that : $\mu(M)=0$ and $f(E\setminus M)=\{y_t~:~t\in E\setminus M\}\subset Y$ then it is separable subset of $X$.
$(ii)$ according to (3) and the density of the sequence $\{y_j^*\}$ in $Y^*$, we have : for all $x^{*}\in X^{*}$ then $x_{|Y}^{*}\in Y^{*}$ and the application : $$ t \mapsto \langle x_{|Y}^{*},f(t)\rangle_{Y^{*},Y} $$ is measurable from $E$ to $\mathbb{R}$.
my problem is to show that :
$$ t \mapsto \langle x^{*},f(t)\rangle_{X^{*},X} $$ is measurable from $E$ to $\mathbb{R}$.
An idea please.