Additivity of the integral over finitely many Jordan pieces that fill a Jordan set up to content zero

Fix one nondegenerate rectangle $Q\supseteq A$; since each $A_i\subseteq A$, the same $Q$ bounds every one of the $N+1$ sets. Write $\tilde{f}$ for the zero extension of $f$ from $A$ to $Q$ and ${\tilde{f}}_i$ for the zero extension of $f{\mathrm{\mid }}_{A_i}$ from $A_i$ to $Q$. By hypothesis and [F1], with [L1] licensing the common choice of $Q$, all $N+1$ of these functions are integrable over $Q$, with ${\int }_Q\tilde{f}={\int }_{A}f$ and ${\int }_Q{\tilde{f}}_i={\int }_{A_i}f$. If $m=1$, the boundary of $Q=[u,v]$ is the two-point set $\{u,v\}$, and each point has content zero because for every $\epsilon >0$ it lies in a closed interval of length below $\epsilon$; if $m>1$, the boundary of $Q$ is the finite union of its coordinate faces, each a continuous graph over a compact nondegenerate rectangle, so [L5] makes every face content zero. Thus $\mathrm{\partial }Q$ has content zero by [F2] in every dimension, and therefore $Q$ is Jordan measurable by [L4].

Put $g:={\sum }_{i=1}^N{\tilde{f}}_i-\tilde{f}$ on $Q$. By [L2] it is integrable over $Q$, being a finite linear combination of the integrable functions of step 1.1. It is bounded as well: if $A=\mathrm{\varnothing }$ then every zero extension and hence $g$ is identically zero, while if $A\mathrm{\ne }\mathrm{\varnothing }$ the boundedness of $f$ supplies a real $M\ge 0$ with $\mathrm{\mid }f(x)\mathrm{\mid }\le M$ on $A$, and then $\mathrm{\mid }g\mathrm{\mid }\le (N+1)M$ on $Q$.

Let $S:=({\bigcup }_{i\mathrm{\ne }j}(A_i\cap A_j))\cup (A\setminus {\bigcup }_{i=1}^N{A}_i)$ and let $x\in Q\setminus S$. If $x\mathrm{\notin }A$ then $\tilde{f}(x)=0$ and every ${\tilde{f}}_i(x)=0$, because $A_i\subseteq A$, so $g(x)=0$. If $x\in A$ then $x$ lies in some $A_i$, since otherwise it would lie in the residual set, and in exactly one, since otherwise it would lie in one of the pairwise intersections; hence ${\sum }_i{\tilde{f}}_i(x)=f(x)=\tilde{f}(x)$ and again $g(x)=0$. So $\{x\in Q:g(x)\mathrm{\ne }0\}\subseteq S$.

The set $S$ is the union of the $N(N-1)$ pairwise intersections and the residual set, each of content zero by hypothesis. Given $\epsilon >0$, cover each of those finitely many sets by finitely many closed cubes of total volume at most $\epsilon \mathrm{/}(N(N-1)+1)$ and take all of those cubes together: this is a finite cover of $S$ by closed cubes of total volume at most $\epsilon$, so $S$ has content zero by [F2], and so does its subset $\{x\in Q:g(x)\mathrm{\ne }0\}$.

By step 1.1 the set $Q$ is bounded and Jordan measurable and $g$ is bounded on it, and by step 4.1 the set where $g$ differs from the zero function has content zero; so [L3] applies with the zero function and gives ${\int }_{Q}g=0$.

Expanding ${\int }_{Q}g$ by [L2] and using step 1.1, $0={\sum }_{i=1}^N{\int }_Q{\tilde{f}}_i-{\int }_Q\tilde{f}={\sum }_{i=1}^N{\int }_{A_i}f-{\int }_{A}f$, which is the asserted identity. For $N=1$ there is no pairwise intersection and $S$ is the residual set alone; the hypothesis $N\ge 1$ excludes the empty index set, for which the right-hand side would be $0$ while the left need not be.