Proof of Nonnegative Measurable Maps and Null Sets: Constant Maps, the Pointwise-Sum Operation on 𝓜⁺, and Complements of Measure-Zero Sets

Maps with a constant value. Let $\mathcal{F}$, $X$, $c$ and $g$ be as in (i). Then $(X,\mathcal{F})$ is a measurable space, and so is $([0,\infty],\mathcal{B}([0,\infty]))$, as recorded in Measurable Map, and the Set of Measurable Maps. Any two maps $X\to[0,\infty]$ with value $c$ at every point are equal (Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (iii)), so $g$ is the map $X\to[0,\infty]$ with value $c$ at every point, and by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii), applied with $(Y,\mathcal{G})=([0,\infty],\mathcal{B}([0,\infty]))$ and $y_{0}=c$, it is $\mathcal{F}/\mathcal{B}([0,\infty])$-measurable, that is, measurable as a map into $[0,\infty]$. Hence $g\in\mathcal{M}^{+}(X,\mathcal{F})$ by Nonnegative Measurable Maps and Their Lebesgue Integral (i).

The pointwise-sum operation. Let $(X,\mathcal{F},\mu)$ be as in (ii), and write $\mathcal{M}^{+}=\mathcal{M}^{+}(X,\mathcal{F})$. Throughout, $\mathcal{F}$ is a $\sigma$-algebra on the set $X$ (Measure, Measure Space, and Probability Measure (ii)) and $\mathcal{M}^{+}$ is a set (Nonnegative Measurable Maps and Their Lebesgue Integral (i)). For $f,g\in\mathcal{M}^{+}$ we have $f+g\in\mathcal{M}^{+}$ by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (i). So Maps and Relations Given by Formulas (ii), applied with each of its three sets equal to $\mathcal{M}^{+}$ and the expression $t(f,g)=f+g$ of Sums, Multiples and Products of Functions into [0,∞] (i), gives a binary operation $\oplus$ on $\mathcal{M}^{+}$ (Binary Operations on a Set (i)) with

$$f\oplus g=f+g\qquad\text{for all }f,g\in\mathcal{M}^{+}$$

(Binary Operations on a Set (ii)). Next, $0\in[0,\infty]$ (The Extended Real Line, Its Order, and [0,∞] (ii)), so Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (i) gives a map $z:X\to[0,\infty]$ with $z(x)=0$ for every $x\in X$, and $z\in\mathcal{M}^{+}$ by (i), applied with $c=0$.

Now let $f,g,h\in\mathcal{M}^{+}$. All maps below are maps $X\to[0,\infty]$, their values are given by Sums, Multiples and Products of Functions into [0,∞] (i), and two of them with the same value at every $x\in X$ are equal by Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (iii). By Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i), the sum on $[0,\infty]$ is associative and commutative with neutral element $0$, so for every $x\in X$

$$(f+g)(x)=f(x)+g(x)=g(x)+f(x)=(g+f)(x),$$ $$\bigl((f+g)+h\bigr)(x)=\bigl(f(x)+g(x)\bigr)+h(x)=f(x)+\bigl(g(x)+h(x)\bigr)=\bigl(f+(g+h)\bigr)(x),$$ $$(z+f)(x)=0+f(x)=f(x).$$

Hence $f\oplus g=f+g=g+f=g\oplus f$. As $f+g$ and $g+h$ lie in $\mathcal{M}^{+}$, also $(f\oplus g)\oplus h=(f+g)+h=f+(g+h)=f\oplus(g\oplus h)$. Finally $z\oplus f=z+f=f$, and $f\oplus z=z\oplus f=f$ by the commutativity just shown. So $\oplus$ is associative and commutative and $z$ is a neutral element of $\oplus$, by Associative and Commutative Binary Operations, and Neutral Elements (i), Associative and Commutative Binary Operations, and Neutral Elements (ii) and Associative and Commutative Binary Operations, and Neutral Elements (iii).

Complements of measure-zero sets. Let $(X,\mathcal{F},\mu)$ and $Z$ be as in (iii). If $u\in X\setminus(X\setminus Z)$, then $u\in X$ and $u\notin X\setminus Z$, so $u\in Z$, by the definition of the difference; thus $X\setminus(X\setminus Z)\subseteq Z$. Since $Z\in\mathcal{F}$ and $\mu(Z)=0$, the set $X\setminus(X\setminus Z)$ is $\mu$-null, with $Z$ as the set $B$ there. By A Property Holding Almost Everywhere (i), $X\setminus Z$ holds $\mu$-almost everywhere.