$\exists A\in \Sigma$ with $\mu (A)=0$ such as , $\forall n\ge 1$ : $f_n(E\smallsetminus A)$ is a separable subset of $

Question

Let $(\Omega,\Sigma,\mu)$ be a $\sigma$-finite measure and $X$ be a banach space:

Theorem (Pettis Measurability Theorem) :
$f:\Omega\to X$ is $\mu$-measurable if and only if :

$(i)$ $\exists A\in \Sigma$ with $\mu (A)=0$ such as : $f(E\smallsetminus A)$ is a separable subset of $X$.

$(ii)$ $f$ is weakly $\mu$-measurable.

let $(f_n)_n$ be a sequence of $\mu$-measurable. Show that :

$\exists A\in \Sigma$ with $\mu (A)=0$ such as , $\forall n\ge 1$ : $f_n(E\smallsetminus A)$ is a separable subset of $X$

Answer 1

For each $n$ there exists $A_n$ such that $\mu (A_n)=0$ and $f_n(E\setminus A_n)$ is separable. Take $A =\cup_n A_n$. Then $\mu (A)=0$ and $f_n(E\setminus A) \subseteq f_n(E\setminus A_n)$ so $f_n(E\setminus A)$ is separable for each $n$.