I am trying to understand the proof of the Dunford-Pettis theorem. While proving relative weak compactness of a uniformly integrable subset, say $K$, of $L_1(\mathscr F, P)$, we start with an arbitrary sequence $(f_n)$ in $K$. Next, it is stated that "the $\sigma$-field generated by $(f_n)$ is separable". I am not able to give the reasoning for this.
In some texts, I found that "there is a countable field $\mathscr G$ such that each $f_n$ is measurable with respect to the $\sigma$-field generated by $\mathscr G$". But I was not able to write down proof of this statement too. Are the two statements equivalent? I am looking for proof of either of these two statements.
Further, is this true for any sequence $(f_n)$ in $L_1(\mathscr F, P)$ without needing uniform integrability?
By a separable $\sigma$-field, I mean a $\sigma$-field generated by a countable collection of sets in $\mathscr F$.
Let $\mathfrak{X}$ be the $\sigma$-field generated by $(f_n)$. This is defined by $$ \mathfrak{X}=\sigma(\{f^{-1}_n(B):B\in \mathscr B\wedge n\in\mathbb{N}\}) $$ where $\mathscr{B}$ is the collection of Borel sets (of reals or whatever space you're looking at).
Put $$ \mathfrak{Y}=\sigma(\{f^{-1}_n((p,q)):p,q\in\mathbb{Q}\wedge n\in\mathbb{N}\}) $$ Clearly, $\mathfrak{Y}$ is separable.
Claim: $\mathfrak{X}=\mathfrak{Y}$.
$\supseteq)$ Trivial.
$\subseteq$) Let $$ \mathscr{S}:=\bigcap_{n\in\mathbb{N}}\{B\in\mathscr{B}:f_n^{-1}(B)\in \mathfrak{Y}\} $$ One can easily show that $\mathscr{S}$ is a $\sigma$-algebra on $\mathbb{R}$. Moreover, it contains every interval with rational endpoints, thus every open subset of $\mathbb{R}$. But then $\mathscr{S}=\mathscr{B}$. This in turn implies that $$ \{f_n^{-1}(B):B\in\mathscr{B}\wedge n\in\mathbb{N}\}\subseteq\mathfrak{Y} $$ and thus $\mathfrak{X}\subseteq\mathfrak{Y}$