Proof of Arithmetic of Measurable Real- and [0,∞]-Valued Maps

For a map $u$ from $X$ to $\mathbb{R}$ or to $[0,\infty]$ and $a\in\mathbb{R}$, write $\{u>a\}=\{x\in X:u(x)>a\}$, and similarly $\{u=\infty\}=\{x\in X:u(x)=\infty\}$.

Continuous maps on the plane. With $\mathbb{R}^{2}$ and its metric $d_{E}$ as in Euclidean Space and Real Matrices: Standing Notation, the identity map $\iota$ of $\mathbb{R}^{2}$ is continuous on $\mathbb{R}^{2}$ from $(\mathbb{R}^{2},d_{E})$ to itself by Continuity of Identity Maps, Compositions and Reciprocals (i), with $\mathbb{R}^{2}$ for both its $X$ and its $A$. By Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (i) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v), applied with $S=\mathbb{R}^{2}$, its component maps $\pi_{1},\pi_{2}:\mathbb{R}^{2}\to\mathbb{R}$, $\pi_{k}(z)=z_{k}$, of Component Maps and Entry Maps (i) are continuous on $\mathbb{R}^{2}$. Hence, for $c\in\mathbb{R}$, the maps $z\mapsto z_{1}+z_{2}$, $z\mapsto z_{1}z_{2}$ and $z\mapsto c\,z_{1}$ from $\mathbb{R}^{2}$ to $\mathbb{R}$ are continuous for $d_{E}$ and $d_{\mathbb{R}}$, by Continuity of Sums and Products of Real-Valued Functions on a Metric Space (v) with $A=\mathbb{R}^{2}$.

Arithmetic. Let $f,g\in\mathcal{L}$ and $c\in\mathbb{R}$. By Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (i) and Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (ii), $[2]=[1]\cup\{2\}=\{1,2\}$ with $2\notin[1]$, where $2=1+1$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion (vii); so $\{(1,f),(2,g)\}$ is a map from $[2]$ to $\mathcal{L}$ with value $f$ at $1$ and $g$ at $2$. Applying Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (iii) to this map, with $n=2$ and the three continuous maps of the previous paragraph, shows that $f+g$, $fg$ and $cf$ are measurable, since $(f(x),g(x))$ has components $f(x)$ and $g(x)$. In particular $-f=(-1)f$ is measurable, and $f-g=f+(-1)g$ is measurable.

Let $a\in\mathbb{R}$. By The Maximum and Minimum of Two Elements of a Total Order (vi), applied to the total order of $\mathbb{R}$, $a<\max\{f(x),g(x)\}$ holds if and only if $a<f(x)$ or $a<g(x)$; so $\{\max(f,g)>a\}=\{f>a\}\cup\{g>a\}$. Both sets on the right belong to $\mathcal{F}$ by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v), so their union does by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii), and $\max(f,g)$ is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v). Likewise, by the second part of The Maximum and Minimum of Two Elements of a Total Order (vi), $\min\{f(x),g(x)\}<a$ holds if and only if $f(x)<a$ or $g(x)<a$; so $\{x\in X:\min\{f(x),g(x)\}<a\}=\{x\in X:f(x)<a\}\cup\{x\in X:g(x)<a\}$. Both sets on the right belong to $\mathcal{F}$ by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v), so their union does by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii), and $\min(f,g)$ is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v). By The Maximum and Minimum of Two Elements of an Ordered Field: Sum, Difference and Absolute Value (iii), $|f(x)|=\max\{f(x),-f(x)\}$, so $|f|=\max(f,-f)$ is measurable. The constant map $0$ on $X$ is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii), so $f^{+}=\max(f,0)$ and $f^{-}=\max(-f,0)$ are measurable. This proves (i).

Finite sums. Let $m\in\mathbb{N}$ and let $k\mapsto f_{k}$ be a map from $[m]$ to $\mathcal{L}$. With $\mathbb{R}^{m}$ and its metric $d_{E}$ as in Euclidean Space and Real Matrices: Standing Notation, the elements of $\mathbb{R}^{m}$ are the maps $[m]\to\mathbb{R}$, with $z_{k}=z(k)$, by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (ii), so $\mathbb{R}^{m}=\mathbb{R}^{[m]}$; and $[m]$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vii). By The Summation Map on Real-Valued Maps over a Finite Set, and the Dot Product, Euclidean Norm and Euclidean Distance as Maps (i), with $B=[m]$, there is a map $\varphi:\mathbb{R}^{m}\to\mathbb{R}$ with $\varphi(z)=\sum_{k\in[m]}z_{k}$ for every $z\in\mathbb{R}^{m}$.

The map $\varphi$ is continuous for $d_{E}$ and $d_{\mathbb{R}}$. Let $e\in\mathbb{R}^{m}$ be the constant map on $[m]$ with value $1$. For $z\in\mathbb{R}^{m}$, by The Dot Product on $\mathbb{R}^n$ (i), $z\cdot e=\sum_{k=1}^{m}z_{k}\cdot1=\sum_{k=1}^{m}z_{k}$, and this is $\sum_{k\in[m]}z_{k}=\varphi(z)$ by Natural Numbers as Indices: ℕ ⊆ ℕ₀, 1 ∈ ℕ₀, and the Two Readings of an Operation from 1 to n (v). The identity map of $\mathbb{R}^{m}$ is continuous on $\mathbb{R}^{m}$ by Continuity of Identity Maps, Compositions and Reciprocals (i), with $\mathbb{R}^{m}$ for both its $X$ and its $A$, and the constant map from $\mathbb{R}^{m}$ to $\mathbb{R}^{m}$ with value $e$ is continuous on $\mathbb{R}^{m}$ by Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v), applied with $(\mathbb{R}^{m},d_{E})$ for its metric space and $S=\mathbb{R}^{m}$. By the same two clauses, the map $z\mapsto z\cdot e$ from $\mathbb{R}^{m}$ to $\mathbb{R}$, the dot product of these two maps, is continuous on $\mathbb{R}^{m}$. It has the same values as $\varphi$, so it is $\varphi$, and $\varphi$ is continuous.

Let $F:X\to\mathbb{R}^{m}$ be the map with $F(x)_{k}=f_{k}(x)$ for all $x\in X$ and $k\in[m]$, as in Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (iii) with $n=m$. By that clause, applied to $k\mapsto f_{k}$ and $\varphi$, the map $x\mapsto\varphi(F(x))$ on $X$ is measurable. For $x\in X$, $F(x)$ is the map $[m]\to\mathbb{R}$, $k\mapsto f_{k}(x)$, so $\varphi(F(x))=\sum_{k\in[m]}f_{k}(x)$, which is the finite sum $\sum_{k=1}^{m}f_{k}(x)$ by Natural Numbers as Indices: ℕ ⊆ ℕ₀, 1 ∈ ℕ₀, and the Two Readings of an Operation from 1 to n (v). Hence the map $x\mapsto\sum_{k=1}^{m}f_{k}(x)$ has the same values as $x\mapsto\varphi(F(x))$, is equal to it, and is measurable. This proves (ii).

Maps into $[0,\infty]$. $\mathbb{Q}$ is countable by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (ix), and so is its subset $Q_{+}=\{q\in\mathbb{Q}:q>0\}$, by Countable Sets: a Bijection from ℕ onto ℕ×ℕ, Finite Sets, Subsets, Surjective Images, Images, Products, and the Integers and Rationals (iv). Hence, for $Q=\mathbb{Q}$ or $Q=Q_{+}$ and every map $q\mapsto B_{q}$ from $Q$ to $\mathcal{F}$, $\bigcup_{q\in Q}B_{q}\in\mathcal{F}$ by Unions and Intersections over a Countable Index Set in a σ-Algebra (i).

Let $f,g:X\to[0,\infty]$ be measurable. For $b\in\mathbb{R}$ the sets $\{f>b\}$ and $\{g>b\}$ belong to $\mathcal{F}$ by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vi), and $\{f=\infty\}=f^{-1}(\{\infty\})$ (Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii)) and likewise $\{g=\infty\}=g^{-1}(\{\infty\})$ belong to $\mathcal{F}$ by Borel Subsets of [0,∞]: Rays, the Point ∞, and Traces on ℝ (ii) and Measurable Map, and the Set of Measurable Maps (i). We use that an element of $[0,\infty]$ other than $\infty$ is a real number $t$ with $0\le t$ (The Extended Real Line, Its Order, and [0,∞] (ii)), that every element of $[0,\infty]$ is $\ge0$ (Order and Suprema in the Extended Real Line (viii)), that $b<\infty$ for $b\in\mathbb{R}$ (The Extended Real Line, Its Order, and [0,∞] (iii), as $\infty\notin\mathbb{R}$ by The Extended Real Line, Its Order, and [0,∞] (i)), that $\le$ is a total order on $[-\infty,\infty]$ which on $\mathbb{R}$ is the order of $\mathbb{R}$ (Order and Suprema in the Extended Real Line (i)), and that between two real numbers $b<b'$ there is $q\in\mathbb{Q}$ with $b<q<b'$ (The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (x)). Let $a\in\mathbb{R}$, $x\in X$, $s=f(x)$ and $t=g(x)$.

Sum. We show that $s+t>a$ if and only if $s=\infty$, or $t=\infty$, or there is $q\in\mathbb{Q}$ with $s>q$ and $t>a-q$. If $s=\infty$ or $t=\infty$, then $s+t=\infty>a$ by Addition and Multiplication on [0,∞] (i). If $s>q$ and $t>a-q$ with $s,t\neq\infty$, then $s,t$ are real and $s+t$ is their real sum, so $s+t>q+(a-q)=a$. Conversely, let $s+t>a$ with $s,t\neq\infty$; then $s,t$ are real and $a-t<s$, so there is $q\in\mathbb{Q}$ with $a-t<q<s$, that is, $s>q$ and $t>a-q$. Hence

$$\{f+g>a\}=\{f=\infty\}\cup\{g=\infty\}\cup\bigcup_{q\in\mathbb{Q}}\bigl(\{f>q\}\cap\{g>a-q\}\bigr).$$

Each $\{f>q\}\cap\{g>a-q\}$ belongs to $\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix). Here $a-q$, the value at $q$ of the map $q\mapsto a-q$ from $\mathbb{Q}$ to $\mathbb{R}$ given by this expression, is a term of type $\mathbb{R}$ under the variable $q\in\mathbb{Q}$ (Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (iii)). So $q\mapsto\{f>q\}\cap\{g>a-q\}$, its values lying in $\mathcal{F}$, is a map from $\mathbb{Q}$ to $\mathcal{F}$ (Preimages, Unions and Intersections over an Index Set, Suprema and Infima, and Superlevel Sets as Typed Terms (vi), Preimages, Unions and Intersections over an Index Set, Suprema and Infima, and Superlevel Sets as Typed Terms (iii), Preimages, Unions and Intersections over an Index Set, Suprema and Infima, and Superlevel Sets as Typed Terms (v)); its union over $q\in\mathbb{Q}$ belongs to $\mathcal{F}$ by the first paragraph of this part, applied with $Q=\mathbb{Q}$, and $\{f+g>a\}\in\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii). So $f+g$ is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vi).

Product. If $a<0$, then $s\cdot t\ge0>a$ for every $x$, so $\{f\cdot g>a\}=X\in\mathcal{F}$. Let $a\ge0$. We show that $s\cdot t>a$ if and only if $s=\infty$ and $t>0$, or $t=\infty$ and $s>0$, or there is $q\in Q_{+}$ with $s>q$ and $t>a/q$. By Addition and Multiplication on [0,∞] (ii), $s\cdot t=\infty$ if one of $s,t$ is $\infty$ and neither is $0$, and $s\cdot t=0$ if one of them is $0$. If $s=\infty$ and $t>0$, or $t=\infty$ and $s>0$, then $s\cdot t=\infty>a$. Let $q\in Q_{+}$ with $s>q$ and $t>a/q$; then $s>0$ and $t>0$, as $a/q\ge0$. If $s$ or $t$ is $\infty$, then $s\cdot t=\infty>a$; otherwise $s\cdot t$ is the real product and $st>qt>q(a/q)=a$. Conversely, let $s\cdot t>a$, so $s\cdot t>0$ and hence $s\neq0$ and $t\neq0$, that is, $s>0$ and $t>0$. If $s=\infty$ or $t=\infty$ we are in one of the first two cases. Otherwise $s,t$ are real with $st>a$, so $0\le a/t<s$, and there is $q\in\mathbb{Q}$ with $a/t<q<s$; then $q\in Q_{+}$, $s>q$, and $qt>a$, that is, $t>a/q$. Hence

$$\{f\cdot g>a\}=\bigl(\{f=\infty\}\cap\{g>0\}\bigr)\cup\bigl(\{g=\infty\}\cap\{f>0\}\bigr)\cup\bigcup_{q\in Q_{+}}\bigl(\{f>q\}\cap\{g>a/q\}\bigr).$$

The sets $\{f=\infty\}\cap\{g>0\}$, $\{g=\infty\}\cap\{f>0\}$ and, for $q\in Q_{+}$, $\{f>q\}\cap\{g>a/q\}$ belong to $\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix); as $a/q$, the value at $q$ of the map $q\mapsto a/q$ from $Q_{+}$ to $\mathbb{R}$ given by this expression ($q\neq0$ for $q\in Q_{+}$), is a term of type $\mathbb{R}$ under the variable $q\in Q_{+}$ (Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (iii)), $q\mapsto\{f>q\}\cap\{g>a/q\}$, its values lying in $\mathcal{F}$, is a map from $Q_{+}$ to $\mathcal{F}$ (Preimages, Unions and Intersections over an Index Set, Suprema and Infima, and Superlevel Sets as Typed Terms (vi), Preimages, Unions and Intersections over an Index Set, Suprema and Infima, and Superlevel Sets as Typed Terms (iii), Preimages, Unions and Intersections over an Index Set, Suprema and Infima, and Superlevel Sets as Typed Terms (v)), and its union over $q\in Q_{+}$ belongs to $\mathcal{F}$ by the first paragraph of this part, applied with $Q=Q_{+}$, and $\{f\cdot g>a\}\in\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii). So $f\cdot g$ is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vi).

Multiples. For $c\in[0,\infty]$, $c=\infty$ included, there is a constant map $x\mapsto c$ from $X$ to $[0,\infty]$ by Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (i), with $X$ for its $S$ and $[0,\infty]$ for its $Y$; it is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii), and $x\mapsto c\cdot f(x)$ is its product with $f$, so it is measurable by the previous paragraph. This proves (iii).