Proof of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions

Claim 1. $X_0\in\mathcal{F}|_{X_0}$. If $A\in\mathcal{F}|_{X_0}$ then $X_0\setminus A=X_0\cap(X\setminus A)$ belongs to $\mathcal{F}$ and is contained in $X_0$; if $A_1,A_2,\dots$ lie in $\mathcal{F}|_{X_0}$ then their union lies in $\mathcal{F}$ and is contained in $X_0$. Hence $\mathcal{F}|_{X_0}$ is a $\sigma$-algebra on $X_0$. Next, $\mu|_{X_0}(\emptyset)=0$, and a disjoint sequence in $\mathcal{F}|_{X_0}$ is a disjoint sequence in $\mathcal{F}$ whose union lies in $\mathcal{F}|_{X_0}$, so countable additivity is inherited from $\mu$.

For measurability, fix a real $c$. Then $\{\tilde f>c\}=\{f>c\}$ if $c\ge0$, while $\{\tilde f>c\}=\{f>c\}\cup(X\setminus X_0)$ if $c<0$; in both cases $\{f>c\}\subseteq X_0$. If $f$ is measurable with respect to $\mathcal{F}|_{X_0}$, every $\{f>c\}$ lies in $\mathcal{F}$, so every $\{\tilde f>c\}$ lies in $\mathcal{F}$ and $\tilde f$ is measurable in the sense of Lebesgue Integral of a Nonnegative Measurable Function. Conversely, if $\tilde f$ is measurable, then for $c\ge0$ the set $\{f>c\}=\{\tilde f>c\}$ lies in $\mathcal{F}$ and is contained in $X_0$, hence lies in $\mathcal{F}|_{X_0}$, and for $c<0$ the set $\{f>c\}$ is $X_0$ itself.

For the integrals, let $s$ be a simple function on $X$ with $0\le s\le\tilde f$; any term of its representation carrying the value $0$ contributes $0$ to the integral under the conventions of Measure, Measure Space, and Probability Measure, so we may take its remaining coefficients $c_p$ positive, on disjoint sets $A_p\in\mathcal{F}$. Since $\tilde f$ vanishes off $X_0$ and $s\le\tilde f$, each $A_p$ with $c_p>0$ is contained in $X_0$; hence $s$ vanishes off $X_0$, its restriction to $X_0$ is a simple function on $(X_0,\mathcal{F}|_{X_0})$ below $f$, and $\int_X s\,d\mu=\sum_p c_p\,\mu(A_p)$ equals the integral of that restriction with respect to $\mu|_{X_0}$. Conversely, every simple function on $X_0$ below $f$ extends by zero to a simple function on $X$ below $\tilde f$ with the same integral. Since the integral of a nonnegative measurable function is the least upper bound of the integrals of simple functions below it, the two integrals agree.

Claim 2. $\varphi(X)=Z$; for $A\in\mathcal{F}$, $Z\setminus\varphi(A)=\varphi(X\setminus A)$ because $\varphi$ is a bijection; and unions commute with $\varphi$. Hence $\varphi(\mathcal{F})$ is a $\sigma$-algebra on $Z$. By injectivity $\varphi^{-1}(\varphi(A))=A\in\mathcal{F}$, so $\varphi$ is measurable; and the preimage of $A\in\mathcal{F}$ under the inverse map is $\varphi(A)\in\varphi(\mathcal{F})$, so the inverse is measurable. Finally, by claim 1 of Image Measures, Measures with Densities, and Change of Variables, $\mu_\varphi(\varphi(A))=\mu\bigl(\varphi^{-1}(\varphi(A))\bigr)=\mu(A)$.

Claim 3. The only subsets of $Z$ are $\emptyset$ and $Z$, and the $\sigma$-algebra and measure axioms are immediate, as is measurability of every $g$. A simple function on $Z$ below $g$ takes a single value $c\le g(z)$ on $Z$ and has integral $c\,\delta(Z)=c$; the least upper bound over such $c$ is $g(z)$.

Claim 4(a). $X_\sqcup\cap X_i=X_i\in\mathcal{F}_i$, so $X_\sqcup\in\mathcal{F}_\sqcup$. For $A\in\mathcal{F}_\sqcup$, $(X_\sqcup\setminus A)\cap X_i=X_i\setminus(A\cap X_i)\in\mathcal{F}_i$; and countable unions intersect cellwise, so $\mathcal{F}_\sqcup$ is a $\sigma$-algebra. If $A\in\mathcal{F}_i$ then $A\cap X_i=A\in\mathcal{F}_i$ and $A\cap X_j=\emptyset\in\mathcal{F}_j$ for $j\neq i$ by disjointness, so $A\in\mathcal{F}_\sqcup$ and $\mu_\sqcup(A)=\mu_i(A)$.

For countable additivity, let $A_1,A_2,\dots$ be disjoint members of $\mathcal{F}_\sqcup$ with union $A$. Countable additivity of each $\mu_i$ gives $\mu_\sqcup(A)=\sum_{i\in I}\sum_{n}\mu_i(A_n\cap X_i)$, and $\sum_n\mu_\sqcup(A_n)=\sum_{n}\sum_{i\in I}\mu_i(A_n\cap X_i)$. Both iterated sums equal the least upper bound $S$ of the sums over finite subsets of the double family $\{\mu_i(A_n\cap X_i)\}$, in the finite-partial-sum sense of the statement's conventions. Indeed, a finite subset of the double family touches finitely many outer indices and, within each, finitely many inner indices, so its sum is at most either iterated sum; hence $S$ is at most either iterated sum. Conversely, for a finite set $J$ of outer indices, $\sum_{j\in J}(\text{inner sum at }j)$ is the least upper bound, over choices of finite sets of inner indices for each $j\in J$, of sums over finite subsets of the double family (finitely many nondecreasing least upper bounds add), so every finite partial sum of either iterated sum is at most $S$, and therefore each iterated sum is at most $S$. Hence $\mu_\sqcup(A)=\sum_n\mu_\sqcup(A_n)$.

The restriction of the disjoint union to $X_i$ is $(X_i,\mathcal{F}_i,\mu_i)$: $\{A\in\mathcal{F}_\sqcup:A\subseteq X_i\}=\mathcal{F}_i$, one inclusion having been shown above, the other holding because $A\subseteq X_i$ gives $A=A\cap X_i\in\mathcal{F}_i$; and the measures agree as shown. The final assertion holds since each $X_i\in\mathcal{F}_\sqcup$ has $\mu_\sqcup(X_i)=\mu_i(X_i)<\infty$ and there are countably many cells.

Claim 4(b). If $W$ is measurable then $W^{-1}(A)\in\mathcal{G}$ for $A\in\mathcal{F}_i\subseteq\mathcal{F}_\sqcup$. Conversely, the class of $A\subseteq X_\sqcup$ with $W^{-1}(A)\in\mathcal{G}$ is a $\sigma$-algebra (preimages commute with complements and countable unions); if it contains every $\mathcal{F}_i$, it contains every $A\in\mathcal{F}_\sqcup$, since $A$ is the countable union of the sets $A\cap X_i\in\mathcal{F}_i$.

Claim 4(c). For real $c$, $\{f>c\}\cap X_i=\{f|_{X_i}>c\}$, which gives the equivalence of the measurability statements. For the integral, let $(i_p)_{p\ge1}$ be a sequence whose set of values is $I$ (repeating an index if $I$ is finite) and let $E_m=X_{i_1}\cup\dots\cup X_{i_m}$. For $E\in\mathcal{F}_\sqcup$ the function $f\,\mathbf{1}_{E}$ is measurable, since $\{f\,\mathbf{1}_{E}>c\}=\{f>c\}\cap E$ for $c\ge0$ and $=X_\sqcup$ for $c<0$; the functions $f\,\mathbf{1}_{E_m}$, with the indicator $\mathbf{1}_{E_m}$, are nondecreasing in $m$ with pointwise limit $f$, so by the Monotone Convergence Theorem their integrals converge to $\int_{X_\sqcup}f\,d\mu_\sqcup$. Writing $\mathbf{1}_{E_m}$ as the finite sum of the indicators of the distinct cells among $X_{i_1},\dots,X_{i_m}$ and using the additivity of the integral, $\int f\,\mathbf{1}_{E_m}\,d\mu_\sqcup$ is the sum of the corresponding terms $\int f\,\mathbf{1}_{X_i}\,d\mu_\sqcup$. By claim 1 and claim 4(a), $\int_{X_\sqcup}f\,\mathbf{1}_{X_i}\,d\mu_\sqcup=\int_{X_i}f|_{X_i}\,d\mu_i$, since $f\,\mathbf{1}_{X_i}$ is the zero extension of $f|_{X_i}$ and the restriction of the disjoint union to $X_i$ is $(X_i,\mathcal{F}_i,\mu_i)$. The limit of these nondecreasing finite partial sums is the least upper bound of the finite partial sums of $\{\int_{X_i}f|_{X_i}\,d\mu_i\}$, that is, $\sum_{i\in I}\int_{X_i}f|_{X_i}\,d\mu_i$ in the sense of the statement's conventions.