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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.

user202723
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1 Answers1

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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.

martini
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  • That is what I meant with standard machinery, but my doubt comes from the definition of $\mu[f]$ in his book. This is maybe the most important definition so I edit it to my question. – user202723 Oct 07 '15 at 09:25
  • Ok, I see. But what is $P(X)$ in the above? And if $X$ is a typo for $\Omega$, how to add the images the $A$s under $f$, as they are sets ... – martini Oct 07 '15 at 10:16
  • I got confused by a paper which gave that definition of $\mu[f]$ and the functions where defined on the powersets. But in Grimmett he stated that the function $f$ is indeed defined on $\Omega$ so I think he uses indeed the normal way of defining the average and this means that your proof is correct. – user202723 Oct 07 '15 at 12:26