This question comes reading Analysis III of Amann and Escher. We set
$$[f\le\alpha]:=\{x\in X: f(x)\le\alpha\}$$
Then it is stated that if $f_1,f_2\in\mathcal L_0(X,\lambda_n,\overline{\Bbb R}^+)$, that is, each $f_k$ is Lebesgue measurable for some $X\subset\Bbb R^n$ Lebesgue measurable, then the set
$$[0\le f_1\le f_2]:=\{x\in X: 0\le f_1(x)\le f_2(x)\}\tag1$$
is $\lambda_n$-measurable. The book says that this is implied by the fact that
$$f\in\mathcal L_0(X,\lambda_m,\overline{\Bbb R})\iff \forall\alpha\in\overline{\Bbb R}:[f\le\alpha]\in\mathcal L(m)\tag2$$
where $\mathcal L(m)$ is the Lebesgue $\sigma$-algebra in $\Bbb R^m$. However I dont see how $(2)$ imply $(1)$. can someone show me the relation? Or, if this is not clear, can someone show me other way to see why $(1)$ is measurable?
It is the intersection of the sets $$ [f_1 \ge 0],\ [g \ge 0]$$ Where $g$ is the measurable function $g=f_2-f_1$.