I am studying the book Probability on Graphs by Grimett. Grimett tells us that $\mu_1 \leq_{st} \mu_2$ if and only if $\mu_1(f)\leq\mu_2(f)$ for all increasing functions $f:\Omega\to \mathbb{R}$. I want to proof this. I tried to use the standard machinery but I think this is not the right way to do it.
To prove:
$\mu_1 \leq_{st} \mu_2$ if and only if $\mu_1(f)\leq\mu_2(f)$
For completion I will also give the definitions Grimmett uses:
Let $E$ be a non-empty finite set, $\omega =\{0,1\}^E$. The sample space $\omega$ is partially ordered by $\omega_1\leq \omega_2 \iff \omega_1(e)\leq \omega_2(e) \forall e \in E$.
A non-empty set $A \subset \Omega$ is called increasing if for $\omega \in A$ we have that $\omega \leq \omega' \implies \omega'\in A$.
We write $\mu_1\leq_{st}\mu_2$ if $\mu_1(A)\leq\mu_2(A) \forall$ increasing events $A$.
Any help is appreciated.
This looks like the usual "from measurable sets to integrable functions" way. Let's start with $\implies$. If $f = \chi_A$ for a measurable $A \subseteq \{0,1\}^E$, as $f$ is increasing, $A$ is increasing, so we have $$ \int_{\Omega} f\, d\mu_1 = \mu_1(A) \le \mu_2(A) = \int_\Omega f \, d\mu_2 $$ Now suppose, $f \colon \Omega \to \mathbf R$ is finitely valued, say $x_1 < \ldots < x_n$ are the range of $f$. Write $$ f = x_1\chi_{\{f \ge x_1\}}+\sum_{i=2}^n (x_i - x_{i-1})\chi_{\{f \ge x_i\}} $$ As $f$ is increasing, the sets $\{f \ge x_i\}$ are increasing, hence \begin{align*} \int_\Omega f \, d\mu_1& = \int_\Omega x_1 f_{\{f \ge x_1\}} \, d\mu_1 + \sum_{i=2}^n (x_i - x_{i-1})\int_\Omega \chi_{\{f \ge x_i\}} \, d\mu_1\\ &\le \int_\Omega x_1 f_{\{f \ge x_1\}} \, d\mu_2 + \sum_{i=2}^n (x_i - x_{i-1})\int_\Omega \chi_{\{f \ge x_i\}} \, d\mu_2\\ &= \int_\Omega f\, d\mu_2 \end{align*} As $\{0,1\}^E$ is finite, we are done.
The other direction follows from the observation that $f = \chi_A$ is increasing iff $A$ is.