Let $(X, \mathcal F, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Here we use the Bochner integral. I'm trying to generalize Theorem 1.40 in Rudin's Real and Complex Analysis, i.e.,
Theorem: Let $f \in L_1(X, \mu, E)$ and $F$ be a non-empty closed subset of $E$. If $$ \varphi(A) :=\frac{1}{\mu(A)} \int_A f \mathrm d \mu \in F \quad \forall A \in \mathcal F \text{ s.t. } \mu(A) \in (0, \infty), $$ then $f(x) \in F$ for almost all $x \in X$.
Could you have a check on my attempt?
My attempt: By Pettis's theorem, $f$ is $\mu$-essentially separably valued. By restricting to a separable subspace of $E$, we can assume that $E$ is separable.
- $\mu$ is finite.
We have $F^c := E \setminus F$ is open. For each $e \in F^c$, there is $r_e>0$ such that $B(e, r_e) \subset F^c$. Clearly, $\{B(e, r_e/2) ; e \in F^c\}$ is an open cover of $F^c$. Separable metric space is Lindelöf, so there is a sequence $(e_n, r_n/2)_n \subset F^c \times \mathbb R_{>0}$ such that $\{B(e_n, r_n/2) ; n \in \mathbb N\}$ covers $F^c$. Let $B_n := f^{-1}[B(e_n, r_n/2)] \in \mathcal F$. We want to prove that $\mu (B_n) = 0$. Assume the contrary that $\mu(B_n)>0$. Then $$ |\varphi(B_n) - e_n| = \left |\frac{1}{\mu(B_n)} \int_{B_n} f \mathrm d \mu - e_n \right | \le \frac{1}{\mu(B_n)} \int_{B_n} |f-e_n| \mathrm d \mu \le \frac{r_n}{2}. $$
This implies $\varphi(B_n) \in \overline{B (e_n, r_n/2)} \subset F^c$, which is a contradiction.
- $\mu$ is $\sigma$-finite.
There is a partition $(X_n) \subset \mathcal F$ of $X$ such that $\mu(X_n)<\infty$ for all $n$. We apply result from 1. and get $f (x) \in F$ for almost all $x\in X_n$. This completes the proof.
