Proof of Null Sets and Properties Holding Almost Everywhere

Every member of $\mathcal{F}$ is a subset of $X$, as $\mathcal{F}\subseteq\mathcal{P}(X)$ by σ-Algebra and Measurable Space (i).

Proof of (i). If $\mu(B)=0$, then $B\subseteq X$, $B\subseteq B$ and $B\in\mathcal{F}$, so $B$ is $\mu$-null by Null Set of a Measure (i), with $B$ itself as the witnessing set. Conversely, let $B$ be $\mu$-null; by Null Set of a Measure (i) there is $C\in\mathcal{F}$ with $B\subseteq C$ and $\mu(C)=0$. By Basic Properties of a Measure (iii), $\mu(B)\le\mu(C)=0$, and $\mu(B)\in[0,\infty]$, so $\mu(B)=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)).

Proof of (ii). By Null Set of a Measure (i) there is $C\in\mathcal{F}$ with $M\subseteq C$ and $\mu(C)=0$. Then $N\subseteq C$, and $N\subseteq X$ by hypothesis, so $N$ is $\mu$-null by Null Set of a Measure (i).

Proof of (iii). By Null Set of a Measure (i) there are $B,C\in\mathcal{F}$ with $N\subseteq B$, $M\subseteq C$ and $\mu(B)=\mu(C)=0$. Then $B\cup C\in\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vii), $N\cup M\subseteq B\cup C$ and $N\cup M\subseteq X$. By Basic Properties of a Measure (vii), $\mu(B\cup C)\le\mu(B)+\mu(C)=0+0$, and $0+0=0$ in $[0,\infty]$, as $0$ is a neutral element of the sum on $[0,\infty]$ by Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (i). Hence $\mu(B\cup C)=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)), and $N\cup M$ is $\mu$-null by Null Set of a Measure (i), with $B\cup C$ as the witnessing set.

Proof of (iv). Put $B=\bigcup_{m\in\mathbb{N}}B_{m}$; then $B\in\mathcal{F}$ by σ-Algebra and Measurable Space (i). By Basic Properties of a Measure (vi), $\mu(B)\le\sum_{m\in\mathbb{N}}\mu(B_{m})$, the sum being that of Sum of a Sequence in [0,∞] (ii). Every term of the sequence $m\mapsto\mu(B_{m})$ is $0$, so its sum is $0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (iv)). Hence $\mu(B)\le0$, and $\mu(B)=0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (i)).

Proof of (v). Let $(B_{m})$ and $(N_{m})$ be as stated, and put $B=\bigcup_{m\in\mathbb{N}}B_{m}$; then $B\in\mathcal{F}$ and $\mu(B)=0$ by (iv). The union $N=\bigcup_{m\in\mathbb{N}}N_{m}$ is a set by The Union and Product of a Family of Sets Indexed by a Set, and the Intersection of a Family with an Inhabited Index Class, Are Sets (i). If $u\in N$, then by Indexed Families of Sets and Their Union, Intersection and Product (ii) there is $m\in\mathbb{N}$ with $u\in N_{m}$; then $u\in X$, as $N_{m}\in\mathcal{P}(X)$, and $u\in B_{m}$, as $N_{m}\subseteq B_{m}$, so $u\in B$ by Indexed Families of Sets and Their Union, Intersection and Product (ii). Thus $N\subseteq X$ and $N\subseteq B$, with $B\in\mathcal{F}$ and $\mu(B)=0$, so $N$ is $\mu$-null by Null Set of a Measure (i). No choice is made here: the witnessing sets are the given terms $B_{m}$, and the single witness for $N$ is their union.

Proof of (vi). By A Property Holding Almost Everywhere (i), $X\setminus Q$ is $\mu$-null. Let $u\in X\setminus R$, so $u\in X$ and $u\notin R$. If $u\in Q$, then $u\in X\cap Q\subseteq R$, which is false; so $u\notin Q$ and $u\in X\setminus Q$. Thus $X\setminus R\subseteq X\setminus Q\subseteq X$, and $X\setminus R$ is a set, being a subclass of the set $X$ (Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper (i)). By (ii), $X\setminus R$ is $\mu$-null, that is, $R$ holds $\mu$-a.e. by A Property Holding Almost Everywhere (i).

Proof of (vii). By A Property Holding Almost Everywhere (i), $X\setminus Q$ and $X\setminus R$ are $\mu$-null, and they are subsets of $X$; so $(X\setminus Q)\cup(X\setminus R)$ is $\mu$-null by (iii). If $u\in X\setminus(Q\cap R)$, then $u\in X$ and $u\notin Q\cap R$, so $u\notin Q$ or $u\notin R$, that is, $u\in X\setminus Q$ or $u\in X\setminus R$. Hence $X\setminus(Q\cap R)\subseteq(X\setminus Q)\cup(X\setminus R)\subseteq X$, and $X\setminus(Q\cap R)$ is a set, being a subclass of the set $X$ (Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper (i)); so $X\setminus(Q\cap R)$ is $\mu$-null by (ii), and $Q\cap R$ holds $\mu$-a.e. by A Property Holding Almost Everywhere (i).