Proof of The Strict Level Sets of a Sequence of Nonnegative Measurable Maps above a Simple Function Form a Sequence of Measurable Sets

As $(X,\mathcal{F})$ is a measurable space, $\mathcal{F}$ is a $\sigma$-algebra on $X$ (σ-Algebra and Measurable Space (ii)). By Simple Functions, Nonnegative Simple Functions, and the Simple Integral (ii) and Simple Functions, Nonnegative Simple Functions, and the Simple Integral (i), $t:X\to\mathbb{R}$ is measurable as a map into $\mathbb{R}$ (Measurable Map, and the Set of Measurable Maps (iii)) and its image $t[X]$ is finite; so $t[X]$ is countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (iii). By Nonnegative Measurable Maps and Their Lebesgue Integral (i), each $f_{m}$ is a map $X\to[0,\infty]$ that is measurable as a map into $[0,\infty]$.

Each set is measurable. Fix $m\in\mathbb{N}$ and let $E=\{x\in X:t(x)<f_{m}(x)\}$, a subset of $X$. Let $d\in t[X]$, so $d\in\mathbb{R}$. Then $\{d\}\in\mathcal{B}(\mathbb{R})$ by Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (iii), applied to the metric space of Borel Sigma-Algebra of a Metric Space (ii), so the preimage $t^{-1}(\{d\})$ lies in $\mathcal{F}$ by Measurable Map, and the Set of Measurable Maps (i); and $\{x\in X:f_{m}(x)>d\}\in\mathcal{F}$ by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vi). Hence

$$A_{d}=t^{-1}(\{d\})\cap\{x\in X:f_{m}(x)>d\}\in\mathcal{F}$$

by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix). By Maps and Relations Given by Formulas (i) there is a map $A:t[X]\to\mathcal{F}$ with $A(d)=A_{d}$ for every $d\in t[X]$, and $\bigcup_{d\in t[X]}A_{d}\in\mathcal{F}$ by Unions and Intersections over a Countable Index Set in a σ-Algebra (i). Now $E=\bigcup_{d\in t[X]}A_{d}$. If $x\in E$, then $d=t(x)$ lies in $t[X]$ (The Image and the Preimage of a Class under a Class (i)), $x\in t^{-1}(\{d\})$ (The Image and the Preimage of a Class under a Class (ii)) and $f_{m}(x)>d$, as $>$ is the converse of $<$ (The Extended Real Line, Its Order, and [0,∞] (iii)); so $x\in A_{d}$. Conversely, if $x\in A_{d}$ for some $d\in t[X]$, then $t(x)=d$ and $d<f_{m}(x)$, so $x\in E$. Thus $E\in\mathcal{F}$.

The sequence. By the previous paragraph, $\{x\in X:t(x)<f_{m}(x)\}\in\mathcal{F}$ for every $m\in\mathbb{N}$. So Maps and Relations Given by Formulas (i), applied to the sets $\mathbb{N}$ and $\mathcal{F}$ and this expression in $m$, gives exactly one map $\mathbb{N}\to\mathcal{F}$, $m\mapsto E_{m}$, with $E_{m}=\{x\in X:t(x)<f_{m}(x)\}$ for every $m\in\mathbb{N}$, that is, by Sequences (i), exactly one such sequence $(E_{m})$ in $\mathcal{F}$.