Let $(X, \Sigma, \mu)$ be a measure space and let $F\subset L^1(X,\Sigma,\mu)$ be a Banach sublattice of $L^1$ with the following properties:
(1) If $f\in F$, $f$ real-valued, then $f\land 1\in F$
(2) For $g\in L^1(X,\Sigma,\mu)$, $g$ real-valued, there is $h\in F$ with $g\leq h$.
Then the following holds:
$\Sigma_F := \{A\subset X~\vert~ A = \bigcup_{n\in\mathbb N}A_n\text{ or }A^c = \bigcup_{n\in\mathbb N}A_n\text{ for }A_n\text{ with } 1_{A_n}\in F\}$ is a $\sigma$-algebra and $F = L^1(X,\Sigma_F,\mu)$.
I already showed that $\Sigma_F$ is a $\sigma-$algebra and it is clear that $L^1(X,\Sigma_F,\mu)\subset F$ since $F$ contains charactersitic functions of $\Sigma_F$-measurable sets with finite measure, is a subspace and $L^1-$closed. So it remains to show that $f\in F$ is $\Sigma_F$-measurable. We can assume without loss of generality that $f\geq 0$. Now it suffices to show $\{f>c\}\in \Sigma_F$ for $c>0$. I don't really know how to do that and would appreciate a hint.