I'm reading Pettis's measurability theorem from Diestel-Uhl's Vector Measures.
The basis for this material is a measure space $(\Omega, \Sigma, \mu)$ and a Banach space $X$.
Definition 1. A function $f: \Omega \rightarrow X$ is called $\mu$-simple if there exist $x_1, \ldots, x_n \in X$ and $E_1, \ldots, E_n \in \sum$ such that $\mu(E_i) < \infty$ for all $i=1, \ldots, n$, and $f=\sum_{i=1}^n x_i \chi_{E_i}$, where $1_{E_i}(\omega)=1$ if $\omega \in E_i$ and $1_{E_i}(\omega)=0$ if $\omega \notin E_i$. A function $f: \Omega \rightarrow X$ is called $\mu$-measurable if there exists a sequence of simple functions $\left(f_n\right)$ with $\lim _n\left\|f_n-f\right\|=0$ $\mu$-almost everywhere. A function $f: \Omega \rightarrow X$ is called weakly $\mu$-measurable if for each $x^* \in X^*$ the numerical function $x^* \circ f$ is $\mu$-measurable.
Theorem 2 (Pettis's measurability theorem) Let $(\Omega, \Sigma, \mu)$ be finite. A function $f: \Omega \rightarrow X$ is $\mu$-measurable if and only if
- (i) $f$ is $\mu$-essentially separably valued, i.e., there exists $E \in \Sigma$ with $\mu(E)=0$ and such that $f(\Omega \backslash E)$ is a (norm) separable subset of $X$, and
- (ii) $f$ is weakly $\mu$-measurable.
The authors assume that the measure space is finite. I'm trying to see if this theorem holds for $\sigma$-finite measure spaces, i.e.,
Conjecture: Let $(\Omega, \Sigma, \mu)$ be $\sigma$-finite. A function $f: \Omega \rightarrow X$ is $\mu$-measurable if and only if
- (i) $f$ is $\mu$-essentially separably valued, i.e., there exists $E \in \Sigma$ with $\mu(E)=0$ and such that $f(\Omega \backslash E)$ is a (norm) separable subset of $X$, and
- (ii) $f$ is weakly $\mu$-measurable.
Could you confirm if the conjecture is true and if my below proof is valid?
My attempt: We have 2 lemmas, i.e.,
Lemma 1: Assume $(\Omega, \Sigma, \mu)$ is $\sigma$-finite and not necessarily complete. Let $(f_n)$ be sequence of $\mu$-measurable functions, and $f \in X^\Omega$ the $\mu$-a.e. limit of $(f_n)$. Then $f$ is also $\mu$-measurable.
Lemma 2: A function $f:\Omega\to X$ is $\mu$-measurable if and only if $1_Ef$ is $\mu$-measurable for all $E\in \Sigma$ such that $\mu(E) < \infty$.
Notice that the restriction of a $\mu$-measurable function to a measurable subset is also $\mu$-measurable.
- Assume $f$ is $\mu$-measurable.
Because $\mu$ is $\sigma$-finite, there is a measurable partition $(E_n) \subset \Sigma$ of $\Omega$ such that $\mu(E_n) < \infty$ for all $n$.
By Lemma 2, $1_{E_n} f$ is $\mu$-measurable for all $n$. Notice that $1_{E_n} f$ is supported on $E_n$ which has finite $\mu$-measure. By Theorem 2, $1_{E_n} f$ is $\mu$-essentially separably valued and weakly $\mu$-measurable. Clearly, $f$ is $\mu$-essentially separably valued. Let $x^* \in X^*$. We have the point-wise convergence $$ x^* \circ f = x^* \circ \left(\lim_{n \to \infty} \sum_{i=1}^n 1_{E_i} f\right) = \lim_{n \to \infty} \sum_{i=1}^n x^* \circ (1_{E_i} f). $$
Because $x^* \circ (1_{E_n} f)$ is $\mu$-measurable for all $i \ge 1$, we have $\sum_{i=1}^n x^* \circ (1_{E_i} f)$ is also $\mu$-measurable. By Lemma 1, $x^* \circ f$ is $\mu$-measurable and thus $f$ is weakly $\mu$-measurable.
- Assume $f$ is $\mu$-essentially separably valued and weakly $\mu$-measurable.
Let $E \in \Sigma$ such that $\mu(E) < \infty$. Then $1_E f$ is $\mu$-essentially separably valued and weakly $\mu$-measurable. Notice that $1_{E} f$ is supported on $E$ which has finite $\mu$-measure. Then by Theorem 2, $1_E f$ is $\mu$-measurable. Then by Lemma 2, $f$ is $\mu$-measurable. This completes the proof.
