Can we compare simple functions without looking at their values at specific points?

Question

Recently I realized that measurable functions can always be written as countable sum of indicator function of measurable sets. This is because we can write for $n\in\mathbb Z$, $A_n=\bigcup_{k=-\infty}^\infty \left[ (2k+1)2^{n}, (2k+2)2^{n} \right[$, i.e. $A_n$ is the set of real numbers that have a $1$ as it $n$'th bit in their binary decomposition. Then any positive measurable function $f$ can be written as $\sum_{n=-\infty}^\infty 2^{n} \mathbf 1_{f^{-1}(A_n)}$. So in a sense, the set of measurable functions is the set of countably simple functions. I am now trying to get rid of the dependence on the underlying space for some other reasons and therefore I have the following question :

Let $(X, \mathcal X)$ be a measure space, let $f$ and $g$ be two positive countably simple functions, that is $f=\sum_{i=1}^\infty a_i \mathbf 1_{A_i}$ and $g=\sum_{i=1}^\infty b_i \mathbf 1_{B_i}$. I want to know if there is an appealing formulation for the inequality $f\leq g$ involving no point $x\in X$, i.e. just saying things about $\{ a_i\}$, $\{b_i\}$, $\{ A_i\}$ and $\{B_i\}$. We are allowed to use elements of $\mathcal X$, unions, complements and inclusion, the main thing that we cannot use is $x\in X$ or $x\in A_i$ and such. We can also assume, if needed, that $a_i$ and $b_i$ are positive.

Answer 1

I claim that $f\leq g$ is equivalent to

For any $\mathcal I, \mathcal J\subseteq \mathbb N$, denote $U_{\mathcal I} = \bigcap_{i\in\mathcal I} A_i \cap \bigcap_{i\notin \mathcal I} A_i^c$ and $V_{\mathcal J}=\bigcap_{i\in\mathcal J} B_i \cap \bigcap_{i\notin \mathcal J} B_i^c$. Whenever $U_{\mathcal I}\cap V_{\mathcal J}\neq \emptyset$, $\sum_{i\in\mathcal I} a_i\leq \sum_{i\in\mathcal J} b_i$.

In order to prove this, observe that, whenever the condition is satisfied, $f$ is constant on $U_{\mathcal I}$ and $g$ is constant on $V_{\mathcal J}$, their respective values are $\sum_{i\in\mathcal I} a_i$ and $\sum_{i\in\mathcal J} b_i$ therefore $f\leq g$ on $U_{\mathcal I}\cap V_{\mathcal J}$, this also holds if $U_{\mathcal I}\cap V_{\mathcal J}$ is empty. Since this has to be true for all choices of $\mathcal I$ and $\mathcal J$ and that $\bigcup_{\mathcal I\subseteq \mathbb N} U_{\mathcal I} = \bigcup_{\mathcal J\subseteq \mathbb N} V_{\mathcal J} = X$, we get that $f\leq g$ on $\bigcup_{\mathcal I,\mathcal J\subset\mathbb N} U_{\mathcal I}\cap V_{\mathcal J}= X$.

If the condition is not satisfied, then we have $U_{\mathcal I}\cap V_{\mathcal J}\neq \emptyset$ such that $\sum_{i\in\mathcal J} b_i < \sum_{i\in\mathcal I} a_i$, those are the respective values of $g$ and $f$ on $U_{\mathcal I}\cap V_{\mathcal J}$ and therefore $g < f$ on $U_{\mathcal I}\cap V_{\mathcal J}$.