I know that the Hahn-Banach Theorem, implies the following statement for Banach-space valued functions:
Let $(X, \mathcal F, \mu)$ be a $\sigma$-finite complete measure space, and $(E, |\cdot|_E)$ a (separable) Banach space. Assume $f:X \to E$ is $\mu$-integrable such that $$ \int_A f \mathrm d \mu = 0 \quad \forall A \in \mathcal F. $$ Then it follows that $f=0$ $\mu$-a.e.
If we now additionally assume that there exists an ordering on $E$, such that:
$$ x \le y \implies c \ge 0 \implies c \cdot x \le c \cdot y $$ $$ c \le d \implies x \ge 0 \implies c \cdot x \le d \cdot x $$ for $c,d \in \mathbb{R}$ and $x,y \in E$
Does this imply the following analogous statement:
$$ \int_A f \mathrm d \mu \ge 0 \quad \forall A \in \mathcal F. \implies f \ge 0 \quad \mu\text{-a.e.} $$
I am a beginner in this area, so I lack the intuition (although I feel like it should probably be true). Any help is appreciated!