Proof. Let $A$ be a finite subset of $S$. Let $I$ be the greatest index (by $g$) of the elements in $A$, then $$\sum_{a\in A} f(a)\le\sum_{a\in\brace{g(0),\ldots,g(I)}} f(a)=\sum_{i=0}^I f(g(i))=\sum_{i=0}^I a_i\le\sum a_i$$ Thus $\sum_{s\in S} f(s)\le\sum a_i$.

Note that for each $J\in N$, $$\sum_{i=0}^J a_i=\sum_{a\in\brace{g(0),\ldots,g(J)}} f(a)\le\sum_{s\in S} f(s)$$ Hence $\sum a_i=\sup_J\sum_{i=0}^J a_i\le\sum_{s\in S} f(s)$. $\blacksquare$

Proof. Let $A$ be a finite subset of $S$, then for some $n\in N$, $$\sum_{a\in A} f(a)\le\sum_{i=0}^n\sum_{j=0}^n a_{i,j}\le\sum_{i=0}^\infty\sum_{j=0}^\infty a_{i,j}$$ Thus $$\sum_{s\in S} f(s)\le\sum_{i=0}^\infty\sum_{j=0}^\infty a_{i,j}$$
Note that $$ \sum_{i=0}^\infty\sum_{j=0}^\infty a_{i,j} =\lim_{I\to\infty}\sum_{i=0}^I\lim_{J\to\infty}\sum_{j=0}^J a_{i,j} =\lim_{I\to\infty}\lim_{J\to\infty}\sum_{i=0}^I\sum_{j=0}^J a_{i,j}$$ Since $$\sum_{i=0}^I\sum_{j=0}^J a_{i,j}\le\sum_{s\in S} f(s)$$ for all $I,J$, we have $$\sum_{i=0}^\infty\sum_{j=0}^\infty a_{i,j}\le\sum_{s\in S} f(s)$$
We have shown that $$\sum_{i=0}^\infty\sum_{j=0}^\infty a_{i,j}=\sum_{s\in S} f(s)$$ And by a symmetric argument, we also have $$\sum_{j=0}^\infty\sum_{i=0}^\infty a_{i,j}=\sum_{s\in S} f(s)$$ $\blacksquare$

Proof. There exists a bijection $h:N\to N\times N$ such that $b=a\circ h$. Then $a=b\circ h^{-1}$. Thus $$\sum_k b_k=\sum_{k\in N}b_k=\sum_i\sum_j a_{i,j}$$ $\blacksquare$

Proof. Note that $$\sum_i\sum_j\abs{a_{i,j}}=\sum_j\sum_i\abs{a_{i,j}}$$ Thus either is real implies the other is also real. Let $(b_k)$ be a reindex of $(a_{i,j})$, meaning for some bijection $h:N\to N\times N$, we have $b=a\circ h$. Then $\sum\abs{b_k}$ is also real, implying $\sum b_k$ exists and is real. Let $\varepsilon\gt0$. Then there exists $N$ such that $\abs{\sum_{k=0}^Nb_k-\sum b_k}\lt\frac{\varepsilon}{2}$ and $\abs{\sum_{k=0}^N\abs{b_k}-\sum\abs{b_k}}\lt\frac{\varepsilon}{2}$. And there exists $M$ such that $\{h(k):k\le N\}\subseteq\{(i,j):i,j\le M\}$. Let $m,n\ge M$, then $$ \abs{\sum_{i\le m,j\le n}a_{i,j}-\sum b_k} \le\abs{\sum_{i\le m,j\le n}a_{i,j}-\sum_{k=0}^Nb_k}+\abs{\sum_{k=0}^Nb_k-\sum b_k} \lt\abs{\sum_{k\in\{h^{-1}(i,j):i\le m,j\le n\}}b_k-\sum_{k=0}^Nb_k}+\frac{\varepsilon}{2} \le\p{\sum_{k\in\{h^{-1}(i,j):i\le m,j\le n\}}\abs{b_k}-\sum_{k=0}^N\abs{b_k}}+\frac{\varepsilon}{2} \le\p{\sum\abs{b_k}-\sum_{k=0}^N\abs{b_k}}+\frac{\varepsilon}{2} \lt\varepsilon $$ We have shown that $$\lim_{m\to\infty,n\to\infty}\sum_{i\le m,j\le n}a_{i,j}=\sum b_k$$
Since $\sum_i\sum_j\abs{a_{i,j}}$ is real, for each $i$, $\sum_j\abs{a_{i,j}}$ is real, thus $\sum_ja_{i,j}$ exists and is real. Note that $\abs{\sum_ja_{i,j}}\le\sum_j\abs{a_{i,j}}$. Thus $\sum_i\abs{\sum_ja_{i,j}}$ is real, implying $\sum_i\sum_ja_{i,j}$ exists and is real. Note that $$ \sum_i\sum_j a_{i,j} =\lim_{m\to\infty}\sum_{i=0}^m\lim_{n\to\infty}\sum_{j=0}^na_{i,j} =\lim_{m\to\infty}\lim_{n\to\infty}\sum_{i=0}^m\sum_{j=0}^na_{i,j} =\lim_{m\to\infty,n\to\infty}\sum_{i=0}^m\sum_{j=0}^na_{i,j} =\sum b_k $$ By a symmetric argument, $\sum_ia_{i,j}$ exists and is real for each $j$, $\sum_j\sum_ia_{i,j}$ exists and is real, and $$\sum_j\sum_i a_{i,j}=\sum b_k$$ $\blacksquare$

Proof. We will use induction to show that, given an elementary set $E=\bigcup_{i=1}^k B_i$, we can always partition $R^d$ into a grid (finite in every dimension, same below) of disjoint boxes, where $E$ is the union of a subset of the bounded boxes. The base case is just $R^d$ being a single box, and $\bigcup_{i=1}^0 B_i$ trivially being the union of no boxes. Now suppose we have a grid of disjoint boxes, where $\bigcup_{i=1}^m B_i$ is the union of a subset of the bounded boxes. Add all the endpoints of $B_{m+1}$ to the grid (different kinds of endpoints should be treated differently even at the same point), and we still have a grid of disjoint boxes, where $\bigcup_{i=1}^m B_i$ is the union of a subset of the bounded boxes. But for this new grid of disjoint boxes, $B_{m+1}$ is also the union of a subset of the bounded boxes, hence $\bigcup_{i=1}^{m+1} B_i$ is the union of a subset of the bounded boxes. Thus completing the induction. Since the grid is finite in every direction, the subset of boxes that forms $E$ must be finite. $\blacksquare$

Proof. Obviously, given an interval $I$, we have $$\abs{I}=\lim_{N\to\infty}\frac{1}{N}\abs{I\cap\left\{\frac{k}{N}:k\in Z\right\}}$$ Thus given a bounded box $B=\bigtimes I_k$, we have $$\abs{B}=\prod\abs{I_k}=\prod\lim_{N\to\infty}\frac{1}{N}\abs{I_k\cap\left\{\frac{k}{N}:k\in Z\right\}} =\lim_{N\to\infty}\prod\frac{1}{N}\abs{I_k\cap\left\{\frac{k}{N}:k\in Z\right\}} =\lim_{N\to\infty}\frac{1}{N^d}\abs{B\cap\left\{\frac{k}{N}:k\in Z\right\}^d}$$ Hence $$\sum\abs{B_i}=\sum\lim_{N\to\infty}\frac{1}{N^d}\abs{B_i\cap\left\{\frac{k}{N}:k\in Z\right\}^d} =\lim_{N\to\infty}\sum\frac{1}{N^d}\abs{B_i\cap\left\{\frac{k}{N}:k\in Z\right\}^d} =\lim_{N\to\infty}\frac{1}{N^d}\abs{E\cap\left\{\frac{k}{N}:k\in Z\right\}^d}$$ The same goes for the other partition. $\blacksquare$

Proof. Suppose $E_1=\bigcup_{i=1}^k B_i$ and $E_1=\bigcup_{i=1}^k' B'_i$ such that $B_1,\ldots,B_k$ are disjoint bounded boxes, and $B'_1,\ldots,B'_{k'}$ are also disjoint bounded boxes. Then $E_1\cup E_2=(\bigcup_{i=1}^k B_i)\cup(\bigcup_{i=1}^k' B'_i)$ where $B_1,\ldots,B_k,B'_1,\ldots,B'_{k'}$ are disjoint bounded boxes. Hence $m(E_1\cup E_2)=\sum\abs{B_i}+\sum\abs{B'_i}=m(E_1)+m(E_2)$. $\blacksquare$

Proof. Note that $E_2\setminus E_1$ is an elementary set, disjoint with $E_1$. Hence $m(E_2)=m(E_1\cup(E_2\setminus E_1))=m(E_1)+m(E_2\setminus E_1)\ge m(E_1)$. $\blacksquare$

Proof. $$m(E_1\cup E_2)=m(E_1\cup(E_2\setminus E_1))=m(E_1)+m(E_2\setminus E_1)\le m(E_1)+m(E_2)$$ $\blacksquare$

Proof. Let $S$ be a Jordan measurable set. Let $\varepsilon\gt0$. Then there exist elementary sets $E_1,E_2$ such that $E_1\subseteq S\subseteq E_2$ and $m(E_2)-m(E_1)\lt\frac{\varepsilon}{2}$. Suppose $E_2=\bigcup_{i=1}^kB_i$ where $B_1,\ldots,B_k$ are disjoint bounded boxes. We can enlarge each $B_i$ into an open bounded box $B'_i$ such that $\abs{B'_i}-\abs{B_i}\lt\frac{\varepsilon}{2k}$. Let $U=\bigcup_{i=1}^kB'_i$, then $U$ is an open elementary set and $S\subseteq U$, and we have $$m(U)\le\sum\abs{B'_i}\lt\sum\p{\abs{B_i}+\frac{\varepsilon}{2k}}=m(E_2)+\frac{\varepsilon}{2}\lt m(E_1)+\varepsilon$$ Thus $$m^*(U\setminus S)\le m^*(U\setminus E_1)\le m(U\setminus E_1)\lt\varepsilon$$ We have shown that $S$ is Lebesgue measurable.

Consider a compact elementary set $E$. Let $\varepsilon\gt0$. Let $B_1,B_2,\ldots$ be bounded boxes such that $E\subseteq\bigcup B_i$ and $\sum\abs{B_i}\lt m^*(E)+\frac{\varepsilon}{2}$. We can enlarge each $B_i$ into an open bounded box $B'_i$ such that $\abs{B'_i}-\abs{B_i}\lt\frac{\varepsilon}{2^{i+1}}$. Then $C=\brace{B'_i:i\in N^+}$ is an open cover of $E$, and thus has a finite subcover $C'$. Thus $$m(E)\le m\p{\bigcup C'}\le\sum_{B\in C'} m(B)\le\sum_{B\in C} m(B)=\sum\abs{B'_i}\le\sum\p{\abs{B_i}+\frac{\varepsilon}{2^{i+1}}}=\sum\abs{B_i}+\frac{\varepsilon}{2}\lt m^*(E)+\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $m(E)\le m^*(E)$. Clearly, $m^*(E)\le m(E)$, thus $m^*(E)=m(E)$.

Now consider any elementary set $E$. Let $\varepsilon\gt0$. Let $B_1,\ldots,B_k$ be disjoint bounded boxes such that $E=\bigcup_{i=1}^kB_i$. For each $B_i$, there exists a closed subbox $B'_i$ such that $\abs{B_i}-\abs{B'_i}\lt\frac{\varepsilon}{k}$. Note that $\bigcup_{i=1}^kB'_i$ is closed and bounded, and thus compact. Thus $$m^*(E)\ge m^*\p{\bigcup_{i=1}^kB'_i}=m\p{\bigcup_{i=1}^kB'_i}=\sum\abs{B'_i}\gt\sum\p{\abs{B_i}-\frac{\varepsilon}{k}}=m(E)-\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $m^*(E)\ge m(E)$. Clearly, $m^*(E)\le m(E)$, thus $m^*(E)=m(E)$.

Now we go back to $S$. Let $\varepsilon\gt0$. Then for some elementary subset $E$ of $S$, we have $m^*(S)\ge m^*(E)=m(E)\gt m(S)-\varepsilon$. Since $\varepsilon$ is arbitrary, we have $m^*(S)\ge m(S)$. Clearly, $m^*(S)\le m(S)$, thus $m^*(S)=m(S)$. $\blacksquare$

Proof. Suppose there exists $S_i$ with $m^*\p{S_i}=\infty$. Then both sides are clearly $\infty$.

Now suppose every $S_i$ has $m^*\p{S_i}\neq\infty$. Let $\varepsilon\gt0$. For each $S_i$, there exist bounded boxes $B_{i,1},B_{i,2},\ldots$ such that $S_i\subseteq\bigcup_j B_{i,j}$ and $\sum_j\abs{B_{i,j}}\lt m^*(S_i)+\frac{\varepsilon}{2^i}$. Note that we can reindex the double sequence $(B_{i,j})$ into a sequence $(B'_k)$. Then we have $$m^*\p{\bigcup S_i}\le m^*\p{\bigcup B_{i,j}}=m^*\p{\bigcup B'_k}\le\sum\abs{B'_k}=\sum_i\sum_j\abs{B_{i,j}}\le\sum\p{m^*(S_i)+\frac{\varepsilon}{2^i}}=\sum m^*(S_i)+\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $m^*\p{\bigcup S_i}\le\sum m^*(S_i)$. $\blacksquare$

Proof. Let $S$ be an open set. For $i\in N$, $$C_i=\brace{\bigtimes_j\left[\frac{k_j}{2^i},\frac{k_j+1}{2^i}\right]:k_1,\ldots,k_d\in Z}$$ is a countable collection of almost disjoint closed bounded boxes that covers $R^d$, and thus $S$. Then $\bigcup C_i$ is also a countable collection of almost disjoint closed bounded boxes that covers $S$. Now take a subset $D$ of $\bigcup C_i$ such that a box $B\in C_i$ is in $D$ if and only if $B\subseteq S$, and for all $j\lt i$, the box $B'\in C_j$ that contains $B$ is not contained by $S$. Then $D$ is an at most countable collection of almost disjoint closed bounded boxes, and $S=\bigcup D$. $\blacksquare$

Proof. Using the construction from the proposition above, but changing the definition of $C_i$ to $$C_i=\brace{\bigtimes_j\left[\frac{k_j}{2^i},\frac{k_j+1}{2^i}\right):k_1,\ldots,k_d\in Z}$$ then we have the desired partition. $\blacksquare$

Proof. Note that $m^*\p{\bigcup B_i}\le\sum m^*\p{B_i}=\sum\abs{B_i}$. Also note that given $\varepsilon\gt0$, we can find a subbox $B'_i$ for each $B_i$ such that $\sum\abs{B'_i}\gt\sum\abs{B_i}-\varepsilon$ and $B'_1,\ldots,B'_n$ are disjoint. Then $$m^*\p{\bigcup B_i}\ge m^*\p{\bigcup B'_i}=\sum\abs{B'_i}\gt\sum\abs{B_i}-\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $m^*\p{\bigcup B_i}\ge\sum\abs{B_i}$. $\blacksquare$

Proof. Note that $m^*\p{\bigcup B_i}\le\sum m^*\p{B_i}=\sum\abs{B_i}$. Also note that given any $n\in N^+$, $$m^*\p{\bigcup B_i}\ge m^*\p{\bigcup_{i=1}^n B_i}=\sum_{i=1}^n\abs{B_i}$$ Hence $m^*\p{\bigcup B_i}\ge\sum\abs{B_i}$. $\blacksquare$

Proof. For every open set $U$ containing $S$, we have $m^*(S)\le m^*(U)$, hence $m^*(S)\le\inf\brace{m^*(U):U\in\tau,S\subseteq U}$. The other direction is trivial when $m^*(S)=\infty$, so we assume $m^*(S)\neq\infty$. Let $\varepsilon\gt0$. We can find a sequence $(B_i)$ of bounded boxes covering $S$ such that $\sum\abs{B_i}\lt m^*(S)+\frac{\varepsilon}{2}$. Then we can enlarge each $B_i$ into an open bounded box $B'_i$ such that $\sum\abs{B'_i}\lt m^*(S)+\varepsilon$. Note that $\bigcup B'_i$ is open and contains $S$. Thus $$\inf\brace{m^*(U):U\in\tau,S\subseteq U}\le m^*(\bigcup B'_i)\le\sum m^*(B'_i)=\sum\abs{B'_i}\lt m^*(S)+\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $\inf\brace{m^*(U):U\in\tau,S\subseteq U}\le m^*(S)$. $\blacksquare$

Proof. Let $\varepsilon\gt0$. For each $i\in N^+$, there exists an open set $U_i$ containing $S_i$ such that $m^*(U_i\setminus S_i)\lt\frac{\varepsilon}{2^{i+1}}$. Then $$m^*\p{\p{\bigcup U_i}\setminus\p{\bigcup S_i}}\le m^*\p{\bigcup\p{U_i\setminus S_i}}\le\sum m^*(U_i\setminus S_i)\le\frac{\varepsilon}{2}\lt\varepsilon$$ Note that any union of a collection of open sets is also open, so $\bigcup U_i$ is open, which also contains $\bigcup S_i$. $\blacksquare$

Proof. Let $S$ be a compact set. Then $S$ is bounded, thus $m^*(S)\neq\infty$. Note that given $\varepsilon\gt0$, there exists an open set $U$ containing $S$ such that $m^*(U)\lt m^*(S)+\frac{\varepsilon}{2}$. Since $U\setminus S=U\cap S^c$ and $S$ is closed, $U\setminus S$ is open, and thus is the union of a sequence $(B_i)$ of almost disjoint closed bounded boxes. Let $n\in N$, then $E=\bigcup_{i=1}^n B_i$ is compact and disjoint with $S$.

Now suppose for contradiction that for every $r\gt0$ there exists $a\in S$ and $b\in E$ such that $d(a,b)\lt r$. Then there exist $(a_i)$ and $(b_i)$ such that $a_i\in S$, $b_i\in E$ and $d(a_i,b_i)\lt\frac{1}{i}$. Note that $(a_i)$ and $(b_i)$ are bounded, and thus each has a convergent subsequence. Clearly, these subsequences converge to the same point $p$. Note that $p$ is a limit point of both $S$ and $E$. Since $S$ and $E$ are both closed, $p$ is in both. But $E$ and $S$ are disjoint, a contradiction. We have shown that there exists $r\gt0$ such that for all $a\in S$ and $b\in E$, $d(a,b)\gt r$.

Since $E$ and $S$ are both bounded, so is $E\cup S$, thus $m^*(E\cup S)\neq\infty$. Let $\sigma\gt0$. Then there exists a sequence $(A_i)$ of bounded boxes covering $E\cup S$ such that $\sum\abs{A_i}\lt m^*(E\cup S)+\sigma$. Since $E$ and $S$ are separated by $r$, for each $i\in N^+$, we can find a finite collection $C_i$ of almost disjoint subboxes with $\bigcup C_i=A_i$, so that each subbox intersects at most one of $E$ and $S$. Then the collection $\bigcup_i C_i$ of all subboxes is countable. Let $C_S$ be the collection of subboxes that intersect $S$, and $C_E$ be the collection of subboxes that intersect $E$, then $C_S$ covers $S$ and $C_E$ covers $E$. Hence we have $$m^*(S)+m^*(E)\le\sum_{B\in C_S}\abs{B}+\sum_{B\in C_E}\abs{B}\le\sum_{B\in\bigcup_i C_i}\abs{B}\le\sum_i\sum_{B\in C_i}\abs{B}=\sum m^*(A_i)=\sum\abs{A_i}\lt m^*(E\cup S)+\sigma$$ Since $\sigma$ is arbitrary, we have $m^*(S)+m^*(E)\le m^*(E\cup S)$, and thus $m^*(S)+m^*(E)=m^*(E\cup S)$.

Now we have $$m^*(S)+m^*(E)=m^*(E\cup S)\le m^*(U)\lt m^*(S)+\frac{\varepsilon}{2}$$ and thus $m^*(\bigcup_{i=1}^n B_i)=m^*(E)\lt\frac{\varepsilon}{2}$. Since $n$ is arbitrary, we have $m^*(U\setminus S)=m^*(\bigcup B_i)\le\frac{\varepsilon}{2}\lt\varepsilon$. We have shown that $S$ is Lebesgue measurable.

Now let $S$ be a closed set. Let $S_i=S\cap\overline{B_{i}(\vb 0)}$ for all $i\in N^+$. Then $S=\cup S_i$ and each $S_i$ is compact, and thus Lebesgue measurable. Hence $S$ is Lebesgue measurable. $\blacksquare$

Proof. Let $S$ be a set with Lebesgue outer measure $0$. Let $\varepsilon\gt0$. Then there exists a sequence $(B_i)$ of bounded boxes covering $S$ such that $\sum\abs{B_i}\lt\frac{\varepsilon}{2}$. Then we can extend each $B_i$ into an open bounded box $B'_i$ with $\sum\abs{B'_i}\lt\varepsilon$. Note that $\bigcup B'_i$ is open, covering $S$, and $$m^*(\bigcup B'_i\setminus S)\le m^*(\bigcup B'_i)\le\sum m^*(B'_i)=\sum\abs{B'_i}\lt\varepsilon$$ $\blacksquare$

Proof. Let $S$ be a Lebesgue measurable set. Note that for every $i\in N^+$, there exists an open set $U_i$ containing $S$ such that $m^*(U_i\setminus S)\lt\frac{1}{i}$. Thus $$m^*\p{S^c\setminus\bigcup U_i^c}\le m^*(S^c\setminus U_i^c)=m^*(U_i\setminus S)\lt\frac{1}{i}$$ for every $i\in N^+$, and we have $m^*\p{S^c\setminus\bigcup U_i^c}=0$, implying $S^c\setminus\bigcup U_i^c$ is Lebesgue measurable. Note that $S^c=\p{S^c\setminus\bigcup U_i^c}\cup\p{\bigcup U_i^c}$. Since each $U_i$ is open, each $U_i^c$ is closed and thus Lebesgue measurable. Hence $S^c$ is Lebesgue measurable. $\blacksquare$

Proof. Since each $S_i$ is Lebesgue measurable, each $S_i^c$ is Lebesgue measurable, thus $\bigcup S_i^c$ is Lebesgue measurable. Then $\p{\bigcap S_i}^c=\bigcup S_i^c$ is Lebesgue measurable. Hence $\bigcap S_i$ is Lebesgue measurable. $\blacksquare$

Proof. Since $B$ is Lebesgue measurable, $B^c$ is Lebesgue measurable, thus $A\setminus B=A\cap B^c$ is also Lebesgue measurable. $\blacksquare$

Proof. Note that translations are clearly isometric. Thus $I'$ is an isometry that fixes $\vb0$.

Let $t\in R$ and $x\in R^d$. If $x=\vb0$, then $I'(tx)=I'(\vb0)=\vb0=tI'(x)$. Now suppose $x\neq\vb0$. Then $d(I'(x),\vb0)=d(I'(x),I'(\vb0))=d(x,\vb0)\neq0$, implying $I'(x)\neq\vb0$.

• If $t\ge1$, then $$\norm{I'(tx)}=d(I'(tx),I'(\vb0))=d(tx,\vb0)=d(x,\vb0)+d(tx,x)=d(I'(x),I'(\vb0))+d(I'(tx),I'(x))=\norm{I'(x)}+\norm{I'(tx)-I'(x)}$$ implying for some non-negative $t'$, $t'I'(x)=I'(tx)-I'(x)$, thus $$t\nnorm{x}=\norm{tx}=\norm{I'(tx)}=\norm{(t'+1)I'(x)}=(t'+1)\norm{I'(x)}=(t'+1)\nnorm{x}$$ Since $\nnorm{x}\neq0$, we have $I'(tx)=(t'+1)I'(x)=tI'(x)$.
• If $0\lt t\lt 1$, then $I'(tx)\neq\vb0$, and $$\norm{I'(x)}=d(I'(x),I'(\vb0))=d(x,\vb0)=d(tx,\vb0)+d(x,tx)=d(I'(tx),I'(\vb0))+d(I'(x),I'(tx))=\norm{I'(tx)}+\norm{I'(x)-I'(tx)}$$ implying for some non-negative $t'$, $t'I'(tx)=I'(x)-I'(tx)$, thus $$\nnorm{x}=\norm{I'(x)}=\norm{(t'+1)I'(tx)}=(t'+1)\norm{I'(tx)}=(t'+1)\norm{tx}=(t'+1)t\nnorm{x}$$ Since $\nnorm{x}\neq0$, we have $I'(tx)=\frac{I'(x)}{t'+1}=tI'(x)$.
• If $t\lt0$, then $$\norm{I'(tx)-I'(x)}=d(I'(tx),I'(x))=d(tx,x)=d(\vb0,x)+d(tx,\vb0)=d(I'(\vb0),I'(x))+d(I'(tx),I'(\vb0))=\norm{-I'(x)}+\norm{I'(tx)}$$ implying for some non-negative $t'$, $-t'I'(x)=I'(tx)$, thus $$-t\nnorm{x}=\norm{tx}=\norm{I'(tx)}=\norm{-t'I'(x)}=t'\norm{I'(x)}=t'\nnorm{x}$$ Since $\nnorm{x}\neq0$, we have $I'(tx)=-t'I'(x)=tI'(x)$.

In all cases, $I'(tx)=tI'(x)$.

Let $x,y\in R^d$.

• If $x=y$, then $x=\frac{x+y}{2}=y$, then $I'(x)=I'(\frac{x+y}{2})=I'(y)$, hence $I'(\frac{x+y}{2})=\frac{I'(x)+I'(y)}{2}$.
• If $x\neq y$, then $$\norm{I'(x)-I'(y)}=d(I'(x),I'(y))=d(x,y)=d\p{x,\frac{x+y}{2}}+d\p{\frac{x+y}{2},y}=d\p{I'(x),I'\p{\frac{x+y}{2}}}+d\p{I'\p{\frac{x+y}{2}},I'(y)}=\norm{I'(x)-I'\p{\frac{x+y}{2}}}+\norm{I'\p{\frac{x+y}{2}}-I'(y)}$$ implying for some non-negative $t'$, $t'(I'(x)-I'(\frac{x+y}{2}))=I'(\frac{x+y}{2})-I'(y)$, thus $$t'd\p{x,\frac{x+y}{2}}=t'd\p{I'(x),I'\p{\frac{x+y}{2}}}=t'\norm{I'(x)-I'\p{\frac{x+y}{2}}}=\norm{t'\p{I'(x)-I'\p{\frac{x+y}{2}}}}=\norm{I'\p{\frac{x+y}{2}}-I'(y)}=d\p{I'\p{\frac{x+y}{2}},I'(y)}=d\p{\frac{x+y}{2},y}=d\p{x,\frac{x+y}{2}}$$ hence $I'(x)-I'(\frac{x+y}{2})=I'(\frac{x+y}{2})-I'(y)$, and we have $I'(\frac{x+y}{2})=\frac{I'(x)+I'(y)}{2}$.

Therefore, $$I'(x+y)=2I'\p{\frac{x+y}{2}}=I'(x)+I'(y)$$
We have shown that $I'$ is linear. $\blacksquare$

Proof. Let $I$ be an isometry. Then $I'(x)=I(x)-I(\vb0)$ is linear. Note that $I(x)=I'(x)+I(\vb0)$, which is affine. $\blacksquare$

Proof. Let $I$ be a linear isometry. Suppose $I(x)=I(y)$, then $d(x,y)=d(I(x),I(y))=0$, implying $x=y$. We have shown that $I$ is injective. Let $i,j\in\{1,\ldots,n\}$ and $i\neq j$. Then $$\norm{I(\vb e_i)+I(\vb e_j)}^2=\norm{I(\vb e_i+\vb e_j)}^2=\norm{\vb e_i+\vb e_j}^2=\norm{\vb e_i}^2+\norm{\vb e_j}^2=\norm{I(\vb e_i)}^2+\norm{I(\vb e_j)}^2$$ Thus $I(\vb e_i)$ and $I(\vb e_j)$ are orthogonal. Hence $(I(\vb e_1),\ldots,I(\vb e_n))$ is orthonormal, and thus independent, and therefore a basis of $R^d$. We have shown that $I$ is surjective.

Now let $I$ be any isometry, then $I'(x)=I(x)-I(\vb0)$ is a linear isometry and is bijective, thus $I(x)=I'(x)+I(\vb0)$ is bijective. $\blacksquare$

Proof. Since $L$ and $L^{-1}$ are linear, they are differentiable, and thus continuous. If $L(U)$ is open, then $U=L^{-1}(L(U))$ is open. If $U$ is open, then $L(U)={L^{-1}}^{-1}(U)$ is open. $\blacksquare$

Proof. Let $x,y\in R^d$. Then $$\braket{A^TAx}{y}=\p{A^TAx}^Ty=(Ax)^T(Ay)=\braket{Ax}{Ay}=\frac{1}{4}\p{\norm{Ax+Ay}^2-\norm{Ax-Ay}^2}=\frac{1}{4}\p{\norm{x+y}^2-\norm{x-y}^2}=\braket{x}{y}$$ Thus $$\p{A^TA}_{ij}=\braket{A^TA\vb e_j}{\vb e_i}=\braket{\vb e_j}{\vb e_i}$$ implying $A^TA$ is the identity matrix $I_d$. Hence $\det(A)^2=\det(A^T)\det(A)=\det(A^TA)=\det(I_d)=1$. $\blacksquare$

Proof. Suppose $A$ is invertible. Then it can be decomposed into a product of elementary row operations. Let $E$ be an elementary row operation (as a matrix), and let $F$ be its corresponding linear map.

• Suppose $E$ swaps row $i$ with row $j$ where $i\neq j$. Given a bounded box $B=\bigtimes I_k$ where each $I_k$ is an interval, we have $F(B)=\bigtimes I'_k$ where $I'_i=I_j$, $I'_j=I_i$, and $I'_k=I_k$ otherwise. Note that $F(B)$ is a bounded box, and $\abs{(F(B))}=\abs{(B)}$.
  
  Now let $(B_k)$ be a sequence of bounded boxes that covers $S$. Then $\bigcup F(B_k)=F(\bigcup B_k)$ contains $F(S)$. Hence $(F(B_k))$ is a sequence of bounded boxes that covers $F(S)$, and $\sum\abs{F(B_k)}=\sum\abs{B_k}$. And we have shown that $m^*(F(S))\le m^*(S)$. Note that $E^{-1}$ also swaps row $i$ with row $j$, and $F^{-1}$ is its corresponding linear map. Thus $m^*(S)=m^*(F^{-1}(F(S)))\le m^*(F(S))$. Therefore $m^*(F(S))=m^*(S)=\abs{\det(E)}m^*(S)$.
  
  
• Suppose $E$ multiplies row $i$ with non-zero real $c$. Given a bounded box $B=\bigtimes I_k$ where each $I_k$ is an interval, we have $F(B)=\bigtimes I'_k$ where $I'_i=c(I_i)$ and $I'_k=I_k$ otherwise, where we treat $c$ as a function. Note that $F(B)$ is a bounded box and $\abs{(F(B))}=\abs{c}\abs{(B)}$.
  
  Now let $(B_k)$ be a sequence of bounded boxes that covers $S$. Then $\bigcup F(B_k)=F(\bigcup B_k)$ contains $F(S)$. Hence $(F(B_k))$ is a sequence of bounded boxes that covers $F(S)$, and $\sum\abs{F(B_k)}=\abs{c}\sum\abs{B_k}$. And we have shown that $m^*(F(S))\le\abs{c}m^*(S)$. Note that $E^{-1}$ multiplies row $i$ with $\frac{1}{c}$, and $F^{-1}$ is its corresponding linear map. Thus $\abs{c}m^*(S)=\abs{c}m^*(F^{-1}(F(S)))\le\abs{c}\abs{\frac{1}{c}}m^*(F(S))=m^*(F(S))$. Therefore $m^*(F(S))=\abs{c}m^*(S)=\abs{\det(E)}m^*(S)$.
  
  
• Suppose $E$ adds row $i$ to row $j$ where $i\neq j$. Suppose we have a bounded box $B=\bigtimes I_k$ where each $I_k$ is an interval. Let $l\in N^+$ and for $m\in\{1,\ldots,l\}$. Let $a,b$ (in that order) be the endpoints of $I_i$ and $c,d$ (in that order) be the endpoints of $I_j$. If $a\gt b$ or $c\gt d$, then both $B$ and $F(B)$ are empty, and trivially, $\abs{(F(B))}=0=\abs{(B)}$. Thus we suppose $a\le b$ and $c\le d$. Let $B_m=\bigtimes I'_k$ where $I'_i=[a+\frac{(m-1)(b-a)}{l},a+\frac{m(b-a)}{l}]$, $I'_j=[c+\frac{(m-1)(b-a)}{l},d+\frac{m(b-a)}{l}]$, and $I'_k=I_k$ otherwise. Then each $B_m$ is a bounded box, $\bigcup B_m$ clearly contains $E(B)$, and $\sum\abs{B_m}-\abs{B}=\frac{b-a}{l}\prod_{k\neq j}\abs{I_k}$. Since $l$ is arbitrary, for every $\varepsilon\gt0$, we can find a finite number of bounded boxes $B_1,\ldots,B_l$ that covers $E(B)$ and $\sum\abs{B_m}\lt\abs{B}+\varepsilon$.
  
  Now let $(B_k)$ be a sequence of bounded boxes that covers $S$. Let $\varepsilon\gt0$. We can use the "$\frac{\varepsilon}{2^{k+1}}$ trick" to find a sequence $(B'_m)$ of boxes that covers each $F(B_k)$, and thus $F(S)$, and $\sum\abs{B'_m}\lt\sum\abs{B_k}+\varepsilon$. And we have shown that $m^*(F(S))\le m^*(S)$.
  
  Now suppose $E$ adds non-zero $c$ times row $i$ to row $j$ where $i\neq j$. Let $E_1$ be the elementary row operation that multiplies row $i$ with $c$, let $E_2$ be the elementary row operation that adds row $i$ to row $j$, and let $E_3$ be the elementary row operation that multiplies row $i$ with $\frac{1}{c}$. Then $E=E_3E_2E_1$. Let $F_1,F_2,F_3$ be the corresponding linear maps of $E_1,E_2,E_3$ respectively. Then $$m^*(F(S))=m^*(F_3(F_2(F_1(S))))=\abs{\frac{1}{c}}m^*(F_2(F_1(S)))\le\abs{\frac{1}{c}}m^*(F_1(S))=\abs{\frac{1}{c}}\abs{c}m^*(S)=m^*(S)$$ Note that $E^{-1}$ adds $-c$ times row $i$ to row $j$, and $F^{-1}$ is its corresponding linear map. Thus $m^*(S)=m^*(F^{-1}(F(S)))\le m^*(F(S))$. Therefore $m^*(F(S))=m^*(S)=\abs{\det(E)}m^*(S)$.
  
  If $E$ adds $0$ times row $i$ to row $j$ where $i\neq j$, then $E$ is the identity matrix, and $F$ is an identity map. Hence $m^*(F(S))=m^*(S)=\abs{\det(E)}m^*(S)$.

In all cases, we have $m^*(F(S))=m^*(S)=\abs{\det(E)}m^*(S)$. Thus let $A=E_k\ldots E_1$ where each $E_i$ where $i\in\{1,\ldots,k\}$ is an elementary row operation, and let $F_i$ be the corresponding linear map of $E_i$, then $$m^*(L(S))=m^*(F_k(\ldots F_1(S)\ldots))=\prod\abs{\det(E_i)}m^*(S)=\abs{\prod\det(E_i)}m^*(S)=\abs{\det(A)}m^*(S)$$
Now suppose $A$ is not invertible. Then the echelon form $E$ of $A$ has at least one zero row $i$. Let $F$ be the corresponding linear map of $E$. Then every $x\in F(S)$ has $x_i=0$. Given $\varepsilon\gt0$, using the "$\frac{\varepsilon}{2^{k+1}}$ trick", we can easily find a sequence $(B_i)$ of bounded boxes covering $F(S)$ with $\sum\abs{B_i}\lt\varepsilon$, implying $m^*(F(S))=0$. Note that $\det(E)=0$, thus $m^*(F(S))=\abs{\det(E)}m^*(S)$. Thus let $A=E_k\ldots E_1E$ where each $E_i$ where $i\in\{1,\ldots,k\}$ is an elementary row operation and $E$ is the echelon form of $A$, and let $F_i$ be the corresponding linear map of $E_i$ and $F$ be the corresponding linear map of $E$, then $$m^*(L(S))=m^*(F_k(\ldots F_1(F(S))\ldots))=\prod\abs{\det(E_i)}m^*(F(S))=0=\abs{\det(E)}\abs{\prod\det(E_i)}m^*(S)=\abs{\det(A)}m^*(S)$$ $\blacksquare$

Proof. Suppose $A$ is invertible. Let $\varepsilon\gt0$, then there exists an open subset $U$ of $R^d$ such that $S\subseteq U$ and $m^*(U\setminus S)\lt\frac{\varepsilon}{\abs{\det(A)}}$. Thus $L(U)$ is open, $L(S)\subseteq L(U)$, and $$m^*(L(U)\setminus L(S))=m^*(L(U\setminus S))=\abs{\det(A)}m^*(U\setminus S)\lt\varepsilon$$ We have shown that $L(S)$ is Lebesgue measurable. Now suppose $A$ is not invertible. Then $m^*(L(S))=\abs{\det(A)}m^*(S)=0$, implying $L(S)$ is measurable. Thus $$m(L(S))=m^*(L(S))=\abs{\det(A)}m^*(S)=\abs{\det(A)}m(S)$$ $\blacksquare$

Proof. Note that $I'(x)=I(x)-I(\vb0)$ is a linear isometry. Let $A$ be the standard matrix representation of $I'$, then $I'(S)$ is Lebesgue measurable and $$m(I'(S))=\abs{\det(A)}m(S)=m(S)$$ Let $T(x)=x+I(\vb0)$, then $I(x)=T(I'(x))$. Since $I'(S)$ is Lebesgue measurable, $I(S)=T(I'(S))$ is clearly Lebesgue measurable. And we clearly have $$m(I(S))=m(T(I'(S)))=m(I'(S))=m(S)$$ $\blacksquare$

Proof. Suppose each $S_i$ is closed and bounded, and thus compact. Let $j\neq k$. By the "convergent subsequence argument" in a proof above, there exists some $r\gt0$ such that for all $a\in S_j$ and $b\in S_k$, $s(a,b)\gt r$. And by the following argument in to same proof, $$m(S_j\cup S_k)=m^*(S_j\cup S_k)=m^*(S_j)+m^*(S_k)=m(S_j)+m(S_k)$$ Let $n\in N^+$, then $m\p{\bigcup S_i}\ge m\p{\bigcup_{i=1}^nS_i}=\sum_{i=1}^nm(S_i)$, thus $m\p{\bigcup S_i}\ge\sum m(S_i)$, and hence $m\p{\bigcup S_i}=\sum m(S_i)$.

Now suppose each $S_i$ is bounded. Let $j\in N^+$. Then $S_j^c$ is Lebesgue measurable. Let $\varepsilon_j\gt0$. Then there exists an open set $U$ containing $S_j^c$ with $m^*(U\setminus S_j^c)\lt\varepsilon_j$. Note that $U^c$ is closed and contained by $S_j$, and thus compact. And we have $$m(S_j)=m^*(S_j)\le m^*(U^c)+m^*(S_j\setminus U^c)=m(U^c)+m^*(U\setminus S_j^c)\lt m(U^c)+\varepsilon_j$$ Thus given arbitrary $\varepsilon\gt0$, we can use the "$\frac{\varepsilon}{2^{i+1}}$ trick" to find a compact subset $E_i$ of $S_i$ for each $i$, such that $$\sum m(S_i)\lt\sum m(E_i)+\varepsilon=m\p{\bigcup E_i}+\varepsilon\le m\p{\bigcup S_i}+\varepsilon$$ hence we have $\sum m(S_i)\le m\p{\bigcup S_i}$, and therefore $m\p{\bigcup S_i}=\sum m(S_i)$.

Now suppose there are no additional constraints on $(S_i)$. We can partition each $S_i$ into at most countable partitions $(S_{i,j})$ such that each $S_{i,j}$ is Lebesgue measurable and bounded, by applying a grid on $S_i$. Note that we can reindex $(S_{i,j})$ into $(S'_k)$, and we have $$\sum m(S_i)=\sum_i\sum_jm(S_{i,j})=\sum m(S'_k)=m\p{\bigcup S'_k}=m\p{\bigcup_i\bigcup_jS_{i,j}}=m\p{\bigcup S_i}$$ $\blacksquare$

Proof. $$A\setminus B=A\cap B^c$$ $\blacksquare$

Proof. $$\mu(A)\le\mu(A)+\mu(B\setminus A)=m(B)$$ $\blacksquare$

Proof. Define $$T_n=S_n\setminus\p{\bigcup_{m=1}^{n-1}S_m}$$ Then $(T_n)$ is a disjoint sequence of measurable sets. Thus $$ \mu\p{\bigcup S_n} =\mu\p{\bigcup T_n} =\sum\mu(T_n) \le\sum\mu(S_n) $$ $\blacksquare$

Proof. $$\mu(A\cup B)=\mu(A\setminus B)+\mu(A\cap B)+\mu(B\setminus A)=(\mu(A\setminus B)+\mu(A\cap B))+(\mu(B\setminus A)+\mu(B\cap A))-\mu(A\cap B)=\mu(A)+\mu(B)-\mu(A\cap B)$$ $\blacksquare$

Proof. Let $E_i=\bigcup_{j\neq i}S_j\cap S_i$ for each $i$. Then each $E_i$ is contained by $S_i$, measurable, and $$\mu(E_i)=\mu\p{\bigcup_{j\neq i}\p{S_j\cap S_i}}\le\sum_{j\neq i}\mu(S_j\cap S_i)=0$$ Note that $\p{S_j\setminus E_j}\cap\p{S_k\setminus E_k}=\emptyset$ whenever $j\neq k$, and each $S_i\setminus E_i$ is measurable. Thus $$\mu\p{\bigcup S_i}\le\sum\mu\p{S_i}=\sum\mu\p{S_i\setminus E_i}=\mu\p{\bigcup\p{S_i\setminus E_i}}\le\mu\p{\bigcup S_i}$$ Therefore, $$\mu\p{\bigcup S_i}=\sum\mu(S_i)$$ $\blacksquare$

Proof. Since $\mu(S_i)$ is increasing, the limit $\lim_{i\to\infty}\mu(S_i)$ exists. Define $T_i=S_i\setminus\p{\bigcup_{j\lt i}S_j}$, then $$\lim_{i\to\infty}\mu(S_i)=\lim_{i\to\infty}\mu\p{\bigcup_{j\le i}T_j}=\lim_{i\to\infty}\sum_{j\le i}\mu(T_j)=\sum\mu(T_j)=\mu\p{\bigcup T_j}=\mu\p{\bigcup S_i}$$ $\blacksquare$

Proof. Trivially, $\mathcal P(X)$ is a $\sigma$-algebra of $X$ and $\tau\subseteq\mathcal P(X)$. Let $\Sigma$ denote the collection of $\sigma$-algebras $\sigma$ over $X$ such that $\tau\subseteq\sigma$, then $\Sigma$ is non-empty. Thus $\bigcap\Sigma$ is a $\sigma$-algebras over $X$. And clearly, $\tau\subseteq\bigcap\Sigma$. Given any $\sigma$-algebra $\sigma$ over $X$ such that $\tau\subseteq\sigma$, we have $\sigma\in\Sigma$, thus $\bigcap\Sigma\subseteq\sigma$. We have shown that $\bigcap\Sigma$ is minimal. $\blacksquare$

Proof. The Lebesgue $\sigma$-algebra contains all open sets of $R^d$, and is thus a superset of the Borel $\sigma$-algebra. $\blacksquare$

Proof. Suppose $f=\sum_{k=1}^nc_k\mathcal X_{S_k}$ where $n\in N$, $S_1,\ldots,S_n\subseteq U$ are measurable, and $c_1,\ldots,c_n\in R$ (for simple $f$) or $c_1,\ldots,c_n\in[0,\infty]$ (for unsigned simple $f$). We can partition $U$ into $2^{n}$ partitions depending on whether a point in $U$ is in $S_1,\ldots,S_n$. Let $A_1,\ldots,A_l$ denote this partition, where $l=2^n\gt0$. Note that each of $S_1,\ldots,S_n$ is the union of some of $A_1,\ldots,A_l$. We define $J_k$ to be the set of indexes such that $S_k=\bigcup_{i\in J_k}A_i$. And we define $g_k:\{1,\ldots,l\}\to\{0,1\}$ such that $g_k(i)=1$ if $i\in J_k$ and $g_k(i)=0$ otherwise. Let $x\in U$. Then $$f(x) =\sum_{k=1}^nc_k\mathcal X_{S_k}(x) =\sum_{k=1}^nc_k\sum_{i\in J_k}\mathcal X_{A_i}(x) =\sum_{k=1}^n\sum_{i\in J_k}c_k\mathcal X_{A_i}(x) =\sum_{k=1}^n\sum_{i=1}^l\p{c_kg_k(i)}\mathcal X_{A_i}(x) =\sum_{i=1}^l\sum_{k=1}^n\p{c_kg_k(i)}\mathcal X_{A_i}(x) =\sum_{i=1}^l\p{\sum_{k=1}^nc_kg_k(i)}\mathcal X_{A_i}(x)$$ Note that each $A_i$ is the intersection of finite measurable sets (including $U$) or their complements with respect to $U$, and is thus measurable. $\blacksquare$

Proof. We can partition $U$ into $2^{n+n'}$ partitions depending on whether a point in $U$ is in $S_1,\ldots,S_n,S'_1,\ldots,S'_{n'}$. Let $A_1,\ldots,A_l$ denote the non-empty partitions of this partition, and let $a_i\in A_i$. Since each $A_i$ is the intersection of finite measurable sets (including $U$) or their complements with respect to $U$, each $A_i$ is measurable. Note that each of $S_1,\ldots,S_n,S'_1,\ldots,S'_{n'}$ is the union of some of $A_1,\ldots,A_l$. We define $J_k$ to be the set of indexes such that $S_k=\bigcup_{i\in J_k}A_i$, and $J'_k$ to be the set of indexes such that $S'_k=\bigcup_{i\in J'_k}A_i$. Then $$\sum_{k=1}^nc_k\mu(S_k) =\sum_{k=1}^nc_km\p{\bigcup_{i\in J_k}A_i} =\sum_{k=1}^nc_k\sum_{i\in J_k}\mu(A_i) =\sum_{k=1}^n\sum_{i\in J_k}c_k\mu(A_i) =\sum_{k=1}^n\sum_{i=1}^lc_k\mu(A_i)\mathcal X_{S_k}(a_i) =\sum_{i=1}^l\sum_{k=1}^nc_k\mu(A_i)\mathcal X_{S_k}(a_i) =\sum_{i=1}^l\mu(A_i)\p{\sum_{k=1}^nc_k\mathcal X_{S_k}}(a_i)$$ Similarly, we have $$\sum_{k=1}^{n'}c'_k\mu(S'_k)=\sum_{i=1}^l\mu(A_i)\p{\sum_{k=1}^{n'}c'_k\mathcal X_{S'_k}}(a_i)$$ Since $\sum_{k=1}^nc_k\mathcal X_{S_k}=\sum_{k=1}^{n'}c'_k\mathcal X_{S'_k}$, we have $$\sum_{k=1}^nc_k\mu(S_k)=\sum_{k=1}^{n'}c'_k\mu(S'_k)$$ $\blacksquare$

Proof. Suppose $f$ is unsigned measurable. Then there exists a sequence $(f_n)$ of unsigned simple functions that converges to $f$. Thus $f$ is trivially unsigned. Define $(g_n)$ from $(f_n)$ such that for each $n$ and each $p\in U$, $g_n(p)=f_n(p)$ if $f_n(p)\in R$ and $g_n(p)=n$ if $f_n(p)=\infty$. Then each $g_n$ is a simple function and $(g_n)$ converges to $f$, implying $f$ is measurable.

Suppose $f$ is both unsigned and measurable. Then there exists a sequence $(f_n)$ of simple functions that converges to $f$. Define $(g_n)$ from $(f_n)$ such that for each $n$ and each $p\in U$, $g_n(p)=f_n(p)$ if $f_n(p)\ge0$ and $g_n(p)=0$ if $f_n(p)\lt0$. Then each $g_n$ is an unsigned simple function and $(g_n)$ converges to $f$, implying $f$ is unsigned measurable. $\blacksquare$

Proof. We will first show that, given a simple or unsigned simple function $g:U\to\overline R$ and $c\in\overline R$, $\{x\in U|g(x)\gt c\}$ is measurable. Assume $g$ is a simple function first. Note that for some $l\in N^+$, for some measurable partition $A_1,\ldots,A_l$ of $U$ and $c_1,\ldots,c_l\in R$, we have $g=\sum_{k=1}^lc_k\mathcal X_{A_k}$. Then $$\{x\in U|g(x)\gt c\}=\bigcup\{A_k:c_k\gt c\}$$ which is measurable. The case when $g$ is unsigned simple is proven similarly.

Let $I=(c,\infty]$ for some $c\in R$. Then we have $f^{-1}(I)=\{x\in U|f(x)\gt c\}$. Suppose $f$ is measurable, then there exists a sequence $(f_n)$ of simple functions on $U$, such that for all $x\in U$, we have $$f(x)=\lim_{n\to\infty}f_n(x)=\inf_{n\in N}\sup_{m\ge n}f_m(x)$$ Let $x\in U$. Suppose $x\in\{x\in U|f(x)\gt c\}$, then for some $k\in N^+$, $\inf_{n\in N}\sup_{m\ge n}f_m(x)=f(x)\gt c+\frac{1}{k}$, thus $\sup_{m\ge n}f_m(x)\gt c+\frac{1}{k}$ for all $n\in N$. Now suppose for some $k\in N^+$, for all $n\in N$, $\sup_{m\ge n}f_m(x)\gt c+\frac{1}{k}$. Then $f(x)=\inf_{n\in N}\sup_{m\ge n}f_m(x)\ge c+\frac{1}{k}\gt c$. Thus $x\in\{x\in U|f(x)\gt c\}$. Therefore, $$\{x\in U|f(x)\gt c\}=\bigcup_{k\in N^+}\bigcap_{n\in N}\{x\in U|\sup_{m\ge n}f_m(x)\gt c+\frac{1}{k}\} =\bigcup_{k\in N^+}\bigcap_{n\in N}\bigcup_{m\ge n}\{x\in U|f_m(x)\gt c+\frac{1}{k}\}$$ Note that each $\{x\in U|f_m(x)\gt c+\frac{1}{k}\}$ is measurable. Hence $f^{-1}(I)$ is measurable. The case when $f$ is unsigned measurable is proven similarly.

The following applies to both measurable and unsigned measurable $f$. Since $f^{-1}((-\infty,\infty])=\bigcup_{k\in N}f^{-1}((-k,\infty])$ and $f^{-1}((\infty,\infty])=\emptyset$ are measurable, given any $c\in\overline R$, $f^{-1}((c,\infty])$ is measurable, thus $f^{-1}([-\infty,c])=U\setminus f^{-1}((c,\infty])$ is measurable. Let $c\in R$, then $f^{-1}([c,\infty])=\bigcap_{k\in N^+}f^{-1}((c-\frac{1}{k},\infty])$ is measurable. Since $f^{-1}([-\infty,\infty])=U$ and $f^{-1}([\infty,\infty])=\bigcap_{k\in N}f^{-1}((k,\infty])$ are measurable, given any $c\in\overline R$, $f^{-1}([c,\infty])$ is measurable, thus $f^{-1}([-\infty,c))=U\setminus f^{-1}([c,\infty])$ is measurable. From here, we can see that given any interval $I$, $f^{-1}(I)$ is measurable. $\blacksquare$

Proof. If any of these conditions is satisfied, then $f^{-1}(I)$ is measurable for any interval $I$. Note that $U$ can be partitioned into $U_{-\infty}=f^{-1}(-\infty)$, $U_R=f^{-1}(R)$, and $U_\infty=f^{-1}(\infty)$, and they are all measurable.

Let $n\in N^+$. Then for $k\in\{1,\ldots,n^2\}$, $S^+_k=f^{-1}((\frac{k}{n},\infty))$ and $S^-_k=f^{-1}((-\infty,-\frac{k}{n}))$ are measurable subsets of $U_R$. Thus $$f_n=\sum_{k=1}^{n^2}\p{\frac{1}{n}\mathcal X_{S^+_k}+\frac{-1}{n}\mathcal X_{S^-_k}}$$ defined on $U_R$ is a simple function. Note that given $x\in U_R$, for all natural $n\gt\abs{f(x)}$, $\abs{f(x)-f_n(x)}\le\frac{1}{n}$. We have shown that $(f_n)$ converges pointwise to $f$ on $U_R$.

Note that let $g_n=n\mathcal X_{U_\infty}$ and $h_n=-n\mathcal X_{U_{-\infty}}$ for $n\in N^+$, then $(g_n)$ converges pointwise to $f$ on $U_\infty$ and $(h_n)$ converges pointwise to $f$ on $U_{-\infty}$. If we extend the domains of $(f_n),(g_n),(h_n)$ to $U$, then $(f_n+g_n+h_n)$ converges pointwise to $f$ on $U$, thus $f$ is measurable.

Suppose $f$ is signed. Note that for each $n\in N^+$, for some $l_n\in N^+$, for some measurable partition $A_{n,1},\ldots,A_{n,l_n}$ of $U$ and $c_{n,1},\ldots,c_{n,l_n}\in R$, we have $f_n+g_n+h_n=\sum_{k=1}^{l_n}c_{n,k}\mathcal X_{A_{n,k}}$. Let $f^*_n=\sum_{k=1}^{l_n}\sup\{c_{n,k},0\}\mathcal X_{A_{n,k}}$, then $f^*_n$ is an unsigned simple function on $U$. Let $x\in U_R$ and $\varepsilon\gt0$. Then there exists $m\in N^+$ such that for all $n\ge m$, $d((f_n+g_n+h_n)(x),f(x))\lt\varepsilon$. Note that for some $j_n\in\{1,\ldots,l_n\}$, $x\in A_{n,j_n}$. Thus $$d(f^*_n(x),f(x))=d(\sup\{c_{n,j_n},0\},f(x))\le d(c_{n,j_n},f(x))=d((f_n+g_n+h_n)(x),f(x))\lt\varepsilon$$ Hence $(f^*_n)$ converges pointwise to $f$ on $U_R$. A similar argument goes for $x\in U_\infty$, and $U_{-\infty}=\emptyset$. Therefore, $(f^*_n)$ converges pointwise to $f$ on $U$, implying $f$ is unsigned measurable. $\blacksquare$

Proof. There exist a sequence $(f_n)$ of simple functions that converges pointwise to $f$ and a sequence $(g_n)$ of simple functions that converges pointwise to $g$. Suppose $f+g$ is well-defined, then $(f_n+g_n)_n$ is a sequence of simple functions that converges pointwise to $f+g$, implying $f+g$ is measurable. If $c\in R$, then $(cf_n)_n$ is a sequence of simple functions that converges pointwise to $cf$, implying $cf$ is measurable. $\blacksquare$

Proof. Suppose $(f_n)$ are extended-real-valued measurable functions on measurable $U$. Let $c\in R$. Then $\{x\in U|f_n(x)\gt c\}$ is measurable for each $n$. Thus $\{x\in U|\sup_n f_n(x)\gt c\}=\bigcup_n\{x\in U|f_n(x)\gt c\}$ is measurable. Thus $\sup_nf_n$ is measurable. By a symmetric argument, $\inf_nf_n$ is also measurable.

Now suppose $(f_n)$ are extended-real-valued measurable functions on measurable $U$ that converge pointwise to an extended-real-valued function $f$. For each $n$, $\sup_{m\ge n}f_m$ is measurable, and $\inf_n\sup_{m\ge n}f_m$ is measurable. Thus $f=\inf_n\sup_{m\ge n}f_m$ is also measurable. $\blacksquare$

Proof. There is an implicit condition to this: for each $p\in U$, there do not exist $i,j$ such that $f_i(p)=\infty$ and $f_j(p)=-\infty$, to ensure that $\sum_{n\le m}f_n(p)$ is defined for any $m$. Then we can define $(g_m)$ such that $g_m(p)=\sum_{n\le m}f_n(p)$ for all $p\in U$. Then $(g_m)$ converges pointwise to $f$. Since each $g_m$ is measurable, $f$ is measurable. $\blacksquare$

Proof. Clearly, $$\sup\{I(g):g\in\mathcal S^*_U,g\le f\}\le\sup\{I(g):g\in\mathcal S_U,g\le f\}$$
Suppose $r=\sup\{I(g):g\in\mathcal S_U,g\le f\}$ is finite. Let $\varepsilon\gt0$, then for some $g\in\mathcal S_U$ with $g\le f$, $I(g)\in(r-\varepsilon,r]$. Note that for some $l\in N^+$, for some measurable partition $A_1,\ldots,A_l$ of $U$ and $c_1,\ldots,c_l\in[0,\infty]$ with at most one being $\infty$, we have $g=\sum_{k=1}^lc_k\mathcal X_{A_k}$. If $c_1,\ldots,c_l$ are finite, then $I(g)\in\{I(g):g\in\mathcal S^*_U,g\le f\}$. If some $c_j$ is $\infty$, then $\mu(A_j)=0$, because otherwise $r\ge I(g)=\infty$. Then $I(g)=I(\sum_{k\neq j}c_k\mathcal X_{A_k})\in\{I(g):g\in\mathcal S^*_U,g\le f\}$. Thus $\sup\{I(g):g\in\mathcal S^*_U,g\le f\}\ge I(g)\gt r-\varepsilon$. Since $\varepsilon$ is arbitrary, $\sup\{I(g):g\in\mathcal S^*_U,g\le f\}\ge\sup\{I(g):g\in\mathcal S_U,g\le f\}$.

Now suppose $\sup\{I(g):g\in\mathcal S_U,g\le f\}=\infty$. Let $s\in R$, then for some $g\in\mathcal S_U$ with $g\le f$, $I(g)\in(s,\infty]$. Again, for some $l\in N^+$, for some measurable partition $A_1,\ldots,A_l$ of $U$ and $c_1,\ldots,c_l\in[0,\infty]$ with at most one being $\infty$, we have $g=\sum_{k=1}^lc_k\mathcal X_{A_k}$. If $c_1,\ldots,c_l$ are finite or if some $c_j$ is $\infty$ with $\mu(A_j)=0$, then $\sup\{I(g):g\in\mathcal S^*_U,g\le f\}\ge I(g)\gt s$. If some $c_j$ is $\infty$ with $\mu(A_j)\gt0$, then for some $t\in[0,\infty)$, $t\mu(A_j)\gt s$. Note that $t\mathcal X_{A_j}\in\mathcal S^*_U$ and $t\mathcal X_{A_j}\le g\le f$. Thus $\sup\{I(g):g\in\mathcal S^*_U,g\le f\}\ge I(t\mathcal X_{A_j})=t\mu(A_j)\gt s$. Since $s$ is arbitrary, $\sup\{I(g):g\in\mathcal S^*_U,g\le f\}=\infty=\sup\{I(g):g\in\mathcal S_U,g\le f\}$. $\blacksquare$

Proof. Define $U^*_i=U_i\setminus\p{\bigcup_{j\lt i}U_j}$, then $(U^*_i)$ is a measurable partition of $\bigcup U_i$. Since $f$ is integrable on each $U_i$, it is measurable on each $U_i$, and thus each $U^*_i$. Suppose $(f_{j})_i$ are simple functions that converge pointwise to $f$ on $U^*_i$. Define $f^*_j=\sum_{i\le j}f_{j,i}$ such that each indicator function is extended to $\bigcup U^*_i$, then $(f^*_j)$ are simple functions that converge pointwise to $f$ on $\bigcup U^*_i$. Thus $f$ is measurable on $\bigcup U^*_i$. Since $f$ is also unsigned on $\bigcup U^*_i$, it is integrable on $\bigcup U^*_i$ (which is also $\bigcup U_i$).

If some $\int_{U_i}fd\mu$ is $\infty$, then both sides are $\infty$. Now suppose all $\int_{U_i}fd\mu$ are finite, then all $\int_{U^*_i}fd\mu$ are also finite. Let $J$ be any finite set of indexes. Suppose $\abs{J}\gt0$. Let $\varepsilon\gt0$, then for all $j\in J$, there exists an unsigned simple function $f_j$ on $U^*_j$ bounded above by $f$ such that $I_*(f,U^*_j)-I(f_j)\lt\frac{\varepsilon}{\abs{J}}$. Then $\sum_{j\in J}f_j$ is an unsigned simple function on $\bigcup U^*_i$ bounded above by $f$. Thus $$I_*\p{f,\bigcup U^*_i}\ge I\p{\sum_{j\in J}f_j}=\sum_{j\in J}I(f_j)\gt\sum_{j\in J}I_*(f,U^*_j)-\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $$\int_{\bigcup U^*_i}fd\mu=I_*\p{f,\bigcup U^*_i}\ge\sum_{j\in J}I_*(f,U^*_j)=\sum_{j\in J}\int_{U^*_j}fd\mu$$ Trivially, if $\abs{J}=0$, this still holds. Hence $$\int_{\bigcup U^*_i}fd\mu\ge\sup\brace{\sum_{j\in J}\int_{U^*_j}fd\mu:J\subseteq N,\abs{J}\in N}=\sum\int_{U^*_i}fd\mu$$
Suppose $\int_{\bigcup U^*_i}fd\mu$ is finite. Again, let $\varepsilon\gt0$. Then there exists an unsigned simple function $f^*$ on $\bigcup U^*_i$ bounded above by $f$ such that $I_*\p{f,\bigcup U^*_i}-I(f^*)\lt\varepsilon$. Note that $f^*|_{U^*_i}$ is an unsigned simple function on $U^*_i$ bounded above by $f$ for each $i$, and $$\sum I_*(f,U^*_i)\ge\sum I(f^*|_{U^*_i})=I(f^*)\gt I_*\p{f,\bigcup U^*_i}-\varepsilon$$ Since $\varepsilon$ is arbitrary, $$\sum\int_{U^*_i}fd\mu=\sum I_*(f,U^*_i)\ge I_*\p{f,\bigcup U^*_i}=\int_{\bigcup U^*_i}fd\mu$$ By a similar argument, if $\int_{\bigcup U^*_i}fd\mu=\infty$, this still holds. We have shown that $$\int_{\bigcup U_i}fd\mu=\int_{\bigcup U^*_i}fd\mu=\sum\int_{U^*_i}fd\mu\le\sum\int_{U_i}fd\mu$$ Note that if in addition, $(U_i)$ is pairwise disjoint, then $U^*_i=U_i$ for each $i$. Then we have $$\int_{\bigcup U_i}fd\mu=\sum\int_{U_i}fd\mu$$ This can be easily reduced to the finite case.

Now suppose $\mu(U_i\cap U_j)=0$ whenever $i\neq j$. Note that $U_i\cap\p{\bigcup_{j\lt i}U_j}$ and $U^*_i$ is a measurable partition of $U_i$ for each $i$. Thus $f$ is integrable on both $U_i\cap\p{\bigcup_{j\lt i}U_j}$ and $U^*_i$. Also note that $$\mu\p{U_i\cap\p{\bigcup_{j\lt i}U_j}}=\mu\p{\bigcup_{j\lt i}\p{U_i\cap U_j}}\le\sum_{j\lt i}\mu(U_i\cap U_j)=0$$ Therefore $\int_{U_i}fd\mu=\int_{U^*_i}fd\mu$ for each $i$, and we have $$\int_{\bigcup U_i}fd\mu=\int_{\bigcup U^*_i}fd\mu=\sum\int_{U^*_i}fd\mu=\sum\int_{U_i}fd\mu$$ $\blacksquare$

Proof. Suppose $(f_n)$ are simple functions that converge to $f$ on $U$, then $(\abs{f_n})$ are simple functions that converge to $\abs{f}$ on $U$. Thus $\abs{f}$ is measurable on $U$. Since $\abs{f}$ is also unsigned, it is integrable on $U$, and $$\int_U\abs{f}d\mu=I_*(\abs{f},U)$$ Suppose $I_*(f,U_+),I_*(-f,U_-),I_*(\abs{f},U)$ are finite. Let $\varepsilon\gt0$. Then there exists an unsigned simple function $g\le\abs{f}$ on $U$ such that $I(g)\gt I_*(\abs{f},U)-\varepsilon$. Note that $g|_{U_+}$ and $g|_{U_-}$ are unsigned simple functions such that $g|_{U_+}\le\abs{f}|_{U_+}=f|_{U_+}$, $g|_{U_-}\le\abs{f}|_{U_-}=(-f)|_{U_-}$, and $I(g)=I(g|_{U_+})+I(g|_{U_-})$. Thus $$I_*(f,U_+)+I_*(-f,U_-)\ge I(g|_{U_+})+I(g|_{U_-})=I(g)\gt I_*(\abs{f},U)-\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $$I_*(f,U_+)+I_*(-f,U_-)\ge I_*(\abs{f},U)$$ Let $\varepsilon\gt0$. Then there exists an unsigned simple function $g_+\le f$ on $U_+$ and an unsigned simple function $g_-\le -f$ on $U_-$ such that $I(g_+)+I(g_-)\gt I_*(f,U_+)+I_*(-f,U_-)-\varepsilon$. Let $g$ be the sum of $g_+$ and $g_-$ extended to $U$ with zeros, then $g$ is an unsigned simple function on $U$ such that $g\le\abs{f}$ and $I(g)=I(g_+)+I(g_-)$. Thus $$I_*(\abs{f},U)\ge I(g)=I(g_+)+I(g_-)\gt I_*(f,U_+)+I_*(-f,U_-)-\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $$I_*(\abs{f},U)\ge I_*(f,U_+)+I_*(-f,U_-)$$ Hence $$\int_U\abs{f}d\mu=I_*(\abs{f},U)=I_*(f,U_+)+I_*(-f,U_-)$$ The general case where $I_*(f,U_+),I_*(-f,U_-),I_*(\abs{f},U)$ may be infinite can be proven similarly. $\blacksquare$

Proof. By integrability, $f$ is measurable on $U$, thus $\abs{f}$ is integrable on $U$, and $$\int_U\abs{f}d\mu=I_*(f,U_+)+I_*(-f,U_-)\ge\abs{I_*(f,U_+)-I_*(-f,U_-)}=\abs{\int_Ufd\mu}$$ $\blacksquare$

Proof. Since $(f_n)$ is increasing, $\int_Uf_nd\mu$ is also increasing, implying $\lim_{n\to\infty}\int_Uf_nd\mu$ exists. Since $\int_Uf_nd\mu\le\int_Ufd\mu$ for each $n$, we have $$\lim_{n\to\infty}\int_Uf_nd\mu\le\int_Ufd\mu$$ Let $f^*$ be an unsigned simple function on $U$ with finite coefficients bounded by $f$, then there exists a measurable partition $A_1,\ldots,A_l$ of $U$ and finite unsigned coefficients $c_1,\ldots,c_l$ such that $f^*=\sum_{k=1}^lc_k\mathcal X_{A_k}$. Let $\varepsilon\in(0,1)$. Define $A_{k,n}=\{p\in A_k|f_n(p)\gt(1-\varepsilon)c_k\}$ for each $k\le l$ and each $n$, then each $A_{k,n}$ is measurable, $A_{k,n}\subseteq A_{k,n+1}$ for each $n$, and $\bigcup_nA_{k,n}=A_k$. Then $$ \lim_{n\to\infty}\int_Uf_nd\mu =\lim_{n\to\infty}\sum_k\int_{A_k}f_nd\mu =\sum_k\lim_{n\to\infty}\int_{A_k}f_nd\mu \ge\sum_k\lim_{n\to\infty}\int_{A_{k,n}}(1-\varepsilon)c_kd\mu =(1-\varepsilon)\sum_kc_k\lim_{n\to\infty}\mu(A_{k,n}) =(1-\varepsilon)\sum_kc_k\mu\p{\bigcup_nA_{k,n}} =(1-\varepsilon)\sum_kc_k\mu(A_k) =(1-\varepsilon)I(f^*) $$ Since $\varepsilon$ is arbitrary in $(0,1)$, we have $\lim_{n\to\infty}\int_Uf_nd\mu\ge I(f^*)$. Since $f^*$ is arbitrary in $\{g\in\mathcal S^*_U|g\le f\}$, we have $$ \lim_{n\to\infty}\int_Uf_nd\mu \ge\sup\{I(g):g\in\mathcal S^*_U,g\le f\} =\sup\{I(g):g\in\mathcal S_U,g\le f\} =\int_Ufd\mu $$ $\blacksquare$

Proof. Trivially, both $f$ and $g$ are measurable and thus integrable. Thus $f+g$ is measurable and integrable. Clearly, $f+g$ is also an unsigned simple function, and $$\int_U(f+g)d\mu=I(f+g)=I(f)+I(g)=\int_Ufd\mu+\int_Ugd\mu$$ $\blacksquare$

Proof. It is obvious that $f+g$ is integrable on $U$. Let $(f_n)$ be a sequence of increasing unsigned simple functions bounded above by $f$ that converge pointwise to $f$, and define $(g_n)$ similarly for $g$. Then $$ \int_U(f+g)d\mu =\lim_{n\to\infty}\int_U(f_n+g_n)d\mu =\lim_{n\to\infty}\p{\int_Uf_nd\mu+\int_Ug_nd\mu} =\lim_{n\to\infty}\int_Uf_nd\mu+\lim_{n\to\infty}\int_Ug_nd\mu =\int_Ufd\mu+\int_Ugd\mu $$ $\blacksquare$

Proof. Since each $f_n$ is unsigned on $U$, $\sum f_n$ is defined and unsigned on $U$. Since each $f_n$ is integrable on $U$, each $f_n$ is measurable on $U$. Then $\sum f_n$ is measurable and thus integrable on $U$. And we have $$ \sum\int_Uf_nd\mu =\lim_{m\to\infty}\sum_{n=1}^m\int_Uf_nd\mu =\lim_{m\to\infty}\int_U\sum_{n=1}^mf_nd\mu =\int_U\sum f_nd\mu $$ $\blacksquare$

Proof. Since $f$ is integrable on $U$ and $V$, $f$ is measurable on $U$ and $V$. Then $f$ is measurable on $U\cap V$, $U\setminus V$ and $V\setminus U$, which are a measurable partition of $U\cup V$. Thus $f$ is measurable on $U\cup V$. Hence both $f^+$ and $f^-$ are measurable on all the sets above.

Suppose both $I_*(f,(U\cap V)_+)$ and $I_*(-f,(U\cap V)_-)$ are $\infty$, then $$I_*(f,U_+)=\int_{U} f^+d\mu\ge\int_{U\cap V} f^+d\mu=I_*(f,(U\cap V)_+)=\infty$$ $$I_*(-f,U_-)=-\int_{U}f^-d\mu=\int_{U}-f^-d\mu\ge\int_{U\cap V}-f^-d\mu=I_*(-f,(U\cap V)_-)=\infty$$ implying $f$ is not integrable on $U$, a contradiction. Therefore, $f$ is integrable on $U\cap V$.

Now suppose both $I_*(f,(U\cup V)_+)$ and $I_*(-f,(U\cup V)_-)$ are $\infty$, then $$I_*(f,U_+)+I_*(f,V_+)=\int_{U} f^+d\mu+\int_{V} f^+d\mu\ge\int_{U\cup V} f^+d\mu=I_*(f,(U\cup V)_+)=\infty$$ $$I_*(-f,U_-)+I_*(-f,V_-)=\p{-\int_{U}f^-d\mu}+\p{-\int_{V}f^-d\mu}=\int_{U}-f^-d\mu+\int_{V}-f^-d\mu\ge\int_{U\cup V}-f^-d\mu=I_*(-f,(U\cup V)_-)=\infty$$ Thus $I_*(f,U_+)=\infty$ or $I_*(f,V_+)=\infty$, and $I_*(-f,U_-)=\infty$ or $I_*(-f,V_-)=\infty$. Then $f$ is not integrable on $U$, or $f$ is not integrable on $V$, or $\int_{U} fd\mu+\int_{V} fd\mu$ is not defined, a contradiction. Hence $f$ is integrable on $U\cup V$.

Note that $$\int_{U\cup V} f^+d\mu=\int_{U\cap V} f^+d\mu+\int_{U\setminus V} f^+d\mu+\int_{V\setminus U} f^+d\mu$$ $$\int_{U\cup V} f^+d\mu+\int_{U\cap V} f^+d\mu=\int_{U} f^+d\mu+\int_{V} f^+d\mu$$ And similarly, $$\int_{U\cup V} f^-d\mu+\int_{U\cap V} f^-d\mu=\int_{U} f^-d\mu+\int_{V} f^-d\mu$$ Since $f$ is integrable on both $U$ and $V$, and $\int_{U} fd\mu+\int_{V} fd\mu$ is defined, if $\int_{U} f^+d\mu=\infty$ or $\int_{V} f^+d\mu=\infty$, then $\int_{U} f^-d\mu\neq-\infty$ and $\int_{V} f^-d\mu\neq-\infty$. Then $$\int_{U} f^+d\mu+\int_{V} f^+d\mu+\int_{U} f^-d\mu+\int_{V} f^-d\mu$$ is defined. If both $\int_{U} f^+d\mu$ and $\int_{V} f^+d\mu$ are finite, then this still holds. Similarly, $$\int_{U\cup V} f^+d\mu+\int_{U\cap V} f^+d\mu+\int_{U\cup V} f^-d\mu+\int_{U\cap V} f^-d\mu$$ is also defined. Therefore, $$ \int_{U\cup V} fd\mu+\int_{U\cap V} fd\mu =\int_{U\cup V} f^+d\mu+\int_{U\cap V} f^+d\mu+\int_{U\cup V} f^-d\mu+\int_{U\cap V} f^-d\mu =\int_{U} f^+d\mu+\int_{V} f^+d\mu+\int_{U} f^-d\mu+\int_{V} f^-d\mu =\int_{U} fd\mu+\int_{V} fd\mu $$ $\blacksquare$

Proof. Since $f\le g$ on $U$, we have $f^+\le g^+$ and $f^-\le g^-$ on $U$, thus $-f^-\ge-g^-$ on $U$. Note that $$-\int_Ug^-d\mu=\int_U-g^-d\mu\le\int_U-f^-d\mu=-\int_Uf^-d\mu$$ Hence $$\int_Ufd\mu=\int_Uf^+d\mu+\int_Uf^-d\mu\le\int_Ug^+d\mu+\int_Ug^-d\mu=\int_Ugd\mu$$ $\blacksquare$

Proof. Since $f$ and $g$ are integrable on $U$,

• $f$ and $g$ are measurable on $U$,
• not both $I_*(f,U_+)$ and $I_*(-f,U_-)$ are $\infty$, and
• not both $I_*(g,U_+)$ and $I_*(-g,U_-)$ are $\infty$.

Now let $U_{c+}=\{x\in U|cf\ge0\}$ and $U_{c-}=\{x\in U|cf\le0\}$. Suppose $c\in(0,\infty)$. Then $cf$ is clearly measurable on $U$. It is also clear that $I_*(cf,U_{c+})=cI_*(f,U_+)$ and $I_*(-(cf),U_{c-})=cI_*(-f,U_-)$, and not both are $\infty$. Therefore $\int_Ucfd\mu=c\int_Ufd\mu$. Suppose $c=0$, this trivially holds. Suppose $c\in(-\infty,0)$, we have $$\int_Ucfd\mu=\int_U-\abs{c}fd\mu=-\int_U\abs{c}fd\mu=-\abs{c}\int_Ufd\mu=c\int_Ufd\mu$$
Clearly, if $f+g$ is defined, then $f+g$ is measurable on $U$. Then $(f+g)^+$ and $(f+g)^-$ are measurable and thus integrable on $U$. Note that, when restricted on $U$, $$(f+g)^++(f+g)^-=f+g=f^++f^-+g^++g^-$$ Let $x\in U$.

• If $(f+g)(x)=\infty$, then $f^-(x),g^-(x),(f+g)^-(x)$ are finite, thus $$(f+g)^+(x)-f^-(x)-g^-(x)=(f+g)^+(x)+(f+g)^-(x)-(f^-(x)+g^-(x)+(f+g)^-(x))=f^+(x)+f^-(x)+g^+(x)+g^-(x)-(f^-(x)+g^-(x)+(f+g)^-(x))=f^+(x)+g^+(x)-(f+g)^-(x)$$
• If $(f+g)(x)=-\infty$, then $f^+(x),g^+(x),(f+g)^+(x)$ are finite, thus $$(f+g)^-(x)-f^+(x)-g^+(x)=(f+g)^+(x)+(f+g)^-(x)-(f^+(x)+g^+(x)+(f+g)^+(x))=f^+(x)+f^-(x)+g^+(x)+g^-(x)-(f^+(x)+g^+(x)+(f+g)^+(x))=f^-(x)+g^-(x)-(f+g)^+(x)$$
• If $(f+g)(x)=R$, then all terms are finite, and we have $$(f+g)^+(x)+(-f^-)(x)+(-g^-)(x)=f^+(x)+g^+(x)+(-(f+g)^-)(x)$$

Therefore, when restricted on $U$, $$(f+g)^++(-f^-)+(-g^-)=f^++g^++(-(f+g)^-)$$ Note that each term is unsigned on $U$, thus $$\int_U(f+g)^+d\mu+\int_U-f^-d\mu+\int_U-g^-d\mu=\int_U\p{(f+g)^++(-f^-)+(-g^-)}d\mu=\int_U\p{f^++g^++(-(f+g)^-)}d\mu=\int_Uf^+d\mu+\int_Ug^+d\mu+\int_U-(f+g)^-d\mu$$ Note that $$f^++g^+\ge(f+g)^+$$ $$f^-+g^-\le(f+g)^-$$ Suppose $\int_Ufd\mu+\int_Ugd\mu$ is defined.

• If $\int_Ufd\mu=\infty$ or $\int_Ugd\mu=\infty$, then $\int_Uf^+d\mu=\infty$ or $\int_Ug^+d\mu=\infty$, and $\int_U-f^-d\mu$ and $\int_U-g^-d\mu$ are finite, implying $\int_U(f+g)^+d\mu=\infty$. Since $0\le\int_U-(f+g)^-d\mu\le\int_U-f^-d\mu+\int_U-g^-d\mu$, $\int_U-(f+g)^-d\mu$ is finite, then $f+g$ is integrable on $U$, and both $\int_U(f+g)d\mu$ and $\int_Ufd\mu+\int_Ugd\mu$ are $\infty$.
• If $\int_Ufd\mu=-\infty$ or $\int_Ugd\mu=-\infty$, then $\int_U-f^-d\mu=\infty$ or $\int_U-g^-d\mu=\infty$, and $\int_Uf^+d\mu$ and $\int_Ug^+d\mu$ are finite, implying $\int_U-(f+g)^-d\mu=\infty$. Since $0\le\int_U(f+g)^+d\mu\le\int_Uf^+d\mu+\int_Ug^+d\mu$, $\int_U(f+g)^+d\mu$ is finite, then $f+g$ is integrable on $U$, and both $\int_U(f+g)d\mu$ and $\int_Ufd\mu+\int_Ugd\mu$ are $-\infty$.
• If $\int_Ufd\mu$ and $\int_Ugd\mu$ are finite, then all terms are finite, thus $f+g$ is integrable on $U$, and we have $$\int_U(f+g)d\mu=\int_U(f+g)^+d\mu-\int_U-(f+g)^-d\mu=\int_Uf^+d\mu+\int_Ug^+d\mu-\int_U-f^-d\mu-\int_U-g^-d\mu=\int_Ufd\mu+\int_Ugd\mu$$

Now suppose $f-g$ and $\int_Ufd\mu-\int_Ugd\mu$ are defined, then we have $$\int_U(f-g)d\mu=\int_U(f+(-g))d\mu=\int_Ufd\mu+\int_U-gd\mu=\int_Ufd\mu-\int_Ugd\mu$$ $\blacksquare$

Proof. Note that $(\abs{f_n})$ are extended-real-valued functions defined and unsigned on measurable $U$, and each $\abs{f_n}$ is integrable on $U$. Thus $\sum\abs{f_n}$ is integrable on $U$ and $$\int_U\sum\abs{f_n}d\mu=\sum\int_U\abs{f_n}d\mu$$ Hence either is real implies the other is also real. Since $\sum\int_U\abs{f_n}d\mu$ is real, each $\int_U\abs{f_n}d\mu$ is real. Note that $$\sum\abs{\int_Uf_nd\mu}\le\sum\int_U\abs{f_n}d\mu$$ implying $\sum\abs{\int_Uf_nd\mu}$ is real and each $\int_Uf_nd\mu$ is real. Thus $\sum\int_Uf_nd\mu$ converges to a real number.

Since $\sum\abs{f_n}$ converges pointwise to a real-valued function on $U$, $\sum f_n$ also converges pointwise to a real-valued function on $U$. Since each $f_n$ is integrable, it is measurable. Thus $\sum f_n$ is measurable on $U$. Then $\abs{\sum f_n}$ is integrable on $U$. Let $p\in U$, then $\abs{\sum f_n(p)}\le\sum\abs{f_n(p)}$. Thus $$\int_U\abs{\sum f_n}d\mu\le\int_U\sum\abs{f_n}d\mu$$ implying $\int_U\abs{\sum f_n}d\mu$ is real. Hence $\sum f_n$ is integrable on $U$.

Let $f_n^+$ denote the function $\sup(f_n,0)$ and let $f_n^-$ denote the function $\inf(f_n,0)$. Then $f_n^+$ and $-f_n^-$ are clearly integrable. Note that $0\le f_n^+\le\abs{f_n}$ and $0\le f_n^-\le\abs{f_n}$. Thus $$ \int_U\sum f_nd\mu =\int_U\sum(f_n^+-(-f_n^-))d\mu =\int_U\sum f_n^+d\mu-\int_U\sum -f_n^-d\mu =\sum\int_U f_n^+d\mu-\sum\int_U -f_n^-d\mu =\sum\int_U(f_n^+-(-f_n^-))d\mu =\sum\int_U f_nd\mu $$ $\blacksquare$

Proof. Since $\sigma(\tau)$ is a $d$-system of $X$, $d(\tau)\subseteq\sigma(\tau)$.

Define $$\mathcal A=\{U\in d(\tau)|\forall S\in\tau, U\cap S\in d(\tau)\}$$ Trivially, $X\in\mathcal A$. If $A,B\in\mathcal A$ and $B\subseteq A$, then $A\setminus B\in d(\tau)$ and given $S\in\tau$, $B\cap S\subseteq A\cap S$ and $(A\setminus B)\cap S=(A\cap S)\setminus(B\cap S)\in d(\tau)$, implying $A\setminus B\in\mathcal A$. If $(U_n)$ is an increasing sequence of $\mathcal A$, then $\bigcup U_n\in d(\tau)$ and given $S\in\tau$, for each $n$, $U_n\cap S\in d(\tau)$, and $(U_n\cap S)_n$ is increasing, thus $(\bigcup U_n)\cap S=\bigcup_n(U_n\cap S)\in d(\tau)$, implying $\bigcup U_n\in\mathcal A$. We have shown that $\mathcal A$ is a $d$-system on $X$. Since $\tau$ is a $\pi$-system, $\tau\subseteq\mathcal A$, implying $d(\tau)=\mathcal A$.

Define $$\mathcal B=\{U\in d(\tau)|\forall S\in d(\tau), U\cap S\in d(\tau)\}$$ Trivially, $X\in\mathcal B$. If $A,B\in\mathcal B$ and $B\subseteq A$, then $A\setminus B\in d(\tau)$ and given $S\in d(\tau)$, $B\cap S\subseteq A\cap S$ and $(A\setminus B)\cap S=(A\cap S)\setminus(B\cap S)\in d(\tau)$, implying $A\setminus B\in\mathcal B$. If $(U_n)$ is an increasing sequence of $\mathcal B$, then $\bigcup U_n\in d(\tau)$ and given $S\in d(\tau)$, for each $n$, $U_n\cap S\in d(\tau)$, and $(U_n\cap S)_n$ is increasing, thus $(\bigcup U_n)\cap S=\bigcup_n(U_n\cap S)\in d(\tau)$, implying $\bigcup U_n\in\mathcal B$. We have shown that $\mathcal B$ is a $d$-system on $X$. Since $d(\tau)=\mathcal A$, $\tau\subseteq\mathcal B$, implying $d(\tau)=\mathcal B$.

Trivially, $\emptyset=X\setminus X\in d(\tau)$, and if $S\in d(\tau)$, then $S^c=X\setminus S\in d(\tau)$. Let $(S_n)$ be any sequence of $d(\tau)$. Since $d(\tau)=\mathcal B$, given $A,B\in d(\tau)$, $A\cap B\in d(\tau)$. Thus $\bigcup_{k=1}^nS_k=(\bigcap_{k=1}^n(S_k)^c)^c\in d(\tau)$. Then $(\bigcup_{k=1}^nS_k)_n$ is an increasing sequence of $d(\tau)$, and we have $\bigcup S_n=\bigcup_n\bigcup_{k=1}^nS_k\in d(\tau)$. We have shown that $d(\tau)$ is a $\sigma$-algebra over $X$, thus $\sigma(\tau)\subseteq d(\tau)$. $\blacksquare$

Proof. Let $\mathcal F=\{U\in\sigma|\mu(U)=\nu(U)\}$. Trivially, $X\in\mathcal F$. If $A,B\in\mathcal F$ and $B\subseteq A$, then $\mu(A\setminus B)=\mu(A)-\mu(B)=\nu(A)-\nu(B)=\nu(A\setminus B)$, implying $A\setminus B\in\mathcal F$. Let $(S_n)$ be an increasing sequence of $\mathcal F$, then $\mu(\bigcup S_n)=\lim_{n\to\infty}\mu(S_n)=\lim_{n\to\infty}\nu(S_n)=\nu(\bigcup S_n)$, implying $\bigcup S_n\in\mathcal F$. We have shown that $\mathcal F$ is a $d$-system of $X$ that contains $\tau$, thus $\sigma=\sigma(\tau)=d(\tau)\subseteq\mathcal F$, implying $\mu=\nu$. $\blacksquare$

Proof. Let $(S_n)$ be the sequence of $\tau$ as described. For each $n$, we can define measures $\mu_n$ and $\nu_n$ on $\sigma$ by $\mu_n(U)=\mu(U\cap S_n)$ and $\nu_n(U)=\nu(U\cap S_n)$. Note that, since $\tau$ is a $\pi$-system, given $U\in\tau$, $U\cap S_n\in\tau$, thus $\mu_n(U)=\nu_n(U)$. Also note that $\mu_n(X)=\mu(S_n)=\nu(S_n)=\nu_n(X)\lt\infty$. Thus $\mu_n=\nu_n$. Let $U\in\sigma$, then $$\mu(U)=\mu\p{\p{\bigcup S_n}\cap U}=\mu\p{\bigcup(S_n\cap U)}=\lim_{n\to\infty}\mu(S_n\cap U)=\lim_{n\to\infty}\mu_n(U)=\lim_{n\to\infty}\nu_n(U)=\nu(U)$$ $\blacksquare$

Proof. Let $x\in X$. Define $$\mathcal F_x=\{S\subseteq X\times Y|S_x\in\tau\}$$ Then trivially, $X\times Y\in\mathcal F_x$. Let $S\in\mathcal F_x$, then $S_x\in\tau$, thus $(S^c)_x=(S_x)^c\in\tau$, implying $S^c\in\mathcal F_x$. Let $(S_n)$ be a sequence of $\mathcal F_x$, then for each $n$, $(S_n)_x\in\tau$, thus $(\bigcup S_n)_x=\bigcup_n((S_n)_x)\in\tau$, implying $\bigcup S_n\in\mathcal F_x$. We have shown that $\mathcal F_x$ is a $\sigma$-algebra on $X\times Y$. Note that $\{S\times T:S\in\sigma,T\in\tau\}\subseteq\mathcal F_x$, thus $\sigma\otimes\tau\subseteq\mathcal F_x$. Hence given $S\in\sigma\otimes\tau$, for all $x\in X$, $S_x\in\tau$. Similarly, given $S\in\sigma\otimes\tau$, for all $y\in Y$, $S^y\in\sigma$.

Let $f:X\times Y\to\overline R$ be $(\sigma\otimes\tau)$-measurable, then for all $c\in R$, $\{(x,y)\in X\times Y|f(x,y)\gt c\}\in\sigma\otimes\tau$. Let $c\in R$ and $x\in X$, then $\{y\in Y|f_x(y)\gt c\}=\{(x,y)\in X\times Y|f(x,y)\gt c\}_x\in\tau$. Thus for all $x\in X$, $f_x$ is $\tau$-measurable. Similarly, for all $y\in Y$, $f^y$ is $\sigma$-measurable. $\blacksquare$

Proof. Suppose $\nu$ is finite. Define $\mathcal F$ to be the collection of sets $U$ in $\sigma\otimes\tau$ such that $\nu[U]$ is $\sigma$-measurable. Trivially, $X\times Y\in\mathcal F$. Let $A,B\in\mathcal F$ with $B\subseteq A$. Then $$\nu[A\setminus B](x)=\nu((A\setminus B)_x)=\nu((A_x)\setminus(B_x))=\nu(A_x)-\nu(B_x)=\nu[A](x)-\nu[B](x)$$ Since $\nu[A]$ and $\nu[B]$ are $\sigma$-measurable functions, $\nu[A\setminus B]$ is $\sigma$-measurable, implying $A\setminus B\in\mathcal F$. Let $(S_n)$ be an increasing sequence of $\mathcal F$. Then $$\nu\left[\bigcup S_n\right](x)=\nu\p{\p{\bigcup S_n}_x}=\nu\p{\bigcup_n((S_n)_x)}=\lim_{n\to\infty}\nu((S_n)_x)=\lim_{n\to\infty}\nu[S_n](x)$$ Since each $\nu[S_n]$ is $\sigma$-measurable, $\nu[\bigcup S_n]$ is $\sigma$-measurable, implying $\bigcup S_n\in\mathcal F$. We have shown that $\mathcal F$ is a $d$-system on $X\times Y$. Note that $\{S\times T:S\in\sigma,T\in\tau\}$ is a $\pi$-system on $X\times Y$, thus $$\sigma\otimes\tau=\sigma\p{\{S\times T:S\in\sigma,T\in\tau\}}=d\p{\{S\times T:S\in\sigma,T\in\tau\}}\subseteq\mathcal F$$ Hence given $S\in\sigma\otimes\tau$, $\nu[S]$ is $\sigma$-measurable.

Now let $\nu$ be $\sigma$-finite. Then $Y$ is the union of a sequence $(V_n)$ of disjoint $\tau$-measurable sets with finite $\nu$-measures. For each $n$, we can define a finite measure $\nu_n$ on $\tau$ by $\nu_n(U)=\nu(U\cap V_n)$. Then given $S\in\sigma\otimes\tau$, $\nu_n[S]$ is $\sigma$-measurable. Let $S\in\sigma\otimes\tau$, then $$\nu[S](x)=\nu(S_x)=\nu\p{\bigcup_n(S_x\cap V_n)}=\sum_n\nu(S_x\cap V_n)=\sum_n\nu_n(S_x)=\sum_n\nu_n[S](x)$$ Since each $\nu_n[S]$ is $\sigma$-measurable, $\nu[S]$ is $\sigma$-measurable.

Similarly, if $S\in\sigma\otimes\tau$, then $\mu[S]$ is $\tau$-measurable. $\blacksquare$

Proof. Let $S\in\sigma\otimes\tau$, then $\nu[S]$ is $\sigma$-measurable and $\mu[S]$ is $\tau$-measurable. Since they are also unsigned, they are integrable, thus $\int_X\nu[S]d\mu$ and $\int_Y\mu[S]d\nu$ are well-defined and unsigned. Then we can define functions $f:\sigma\otimes\tau\to[0,\infty]$ and $g:\sigma\otimes\tau\to[0,\infty]$ by $f(S)=\int_X\nu[S]d\mu$ and $g(S)=\int_Y\mu[S]d\nu$. Trivially, $\int_X\nu[\emptyset]d\mu=0$. Let $(S_n)$ be a disjoint sequence of $\sigma\otimes\tau$, then given $x\in X$, $((S_n)_x)_n$ is a disjoint sequence of $\tau$, thus $$ f\p{\bigcup S_n} =\int_X\nu\p{\p{\bigcup S_n}_x}d\mu =\int_X\nu\p{\bigcup_n((S_n)_x)}d\mu =\int_X\sum_n\nu[S_n]d\mu =\sum_n\int_X\nu[S_n]d\mu =\sum_nf(S_n) $$ We have shown that $f$ is a measure on $\sigma\otimes\tau$. Similarly, $g$ is a measure on $\sigma\otimes\tau$. Note that given $A\in\sigma$ and $B\in\tau$, $$f(A\times B)=\int_X\nu((A\times B)_x)d\mu=\int_X\nu(B)\mathcal X_Ad\mu=I(\nu(B)\mathcal X_A)=\mu(A)\nu(B)$$ Similarly, $g(A\times B)=\mu(A)\nu(B)$. We have shown the existence of the desired measure on $\sigma\otimes\tau$.

Suppose we have measures $\alpha,\beta$ of $\sigma\otimes\tau$ with the desired property. Note that $\{S\times T:S\in\sigma,T\in\tau\}$ is a $\pi$-system on $X\times Y$, $\sigma\otimes\tau=\sigma\p{\{S\times T:S\in\sigma,T\in\tau\}}$, and $\alpha$ and $\beta$ agree on $\{S\times T:S\in\sigma,T\in\tau\}$. Since $(X,\sigma,\mu)$ and $(Y,\tau,\nu)$ are $\sigma$-finite, there exist an increasing sequence $(U_n)$ of $\sigma$-measurable sets with finite $\mu$-measure such that $\bigcup U_n=X$, and an increasing sequence $(V_n)$ of $\tau$-measurable sets with finite $\nu$-measure such that $\bigcup V_n=Y$. Then $(U_n\times V_n)_n$ is an increasing sequence of $\sigma\otimes\tau$ with finite measures with respect to $\alpha$ and $\beta$, such that $\bigcup_n(U_n\times V_n)=X\times Y$. By a lemma above, we have $\alpha=\beta$. We have shown that the desired measure on $\sigma\otimes\tau$ uniquely exists. Let $\mu\times\nu$ denote this measure, then $f=\mu\times\nu=g$, thus given $S\in\sigma\otimes\tau$, we have $$\int_X\nu[S]d\mu=f(S)=(\mu\otimes\nu)(S)=g(S)=\int_Y\mu[S]d\nu$$ $\blacksquare$

Proof. Let $S\in\sigma\otimes\tau$, then for all $x\in X$, $$\int_Y(\mathcal X_S)_xd\nu=\nu[S](x)$$ thus $\int_Y(\mathcal X_S)_xd\nu$ is integrable with respect to $(X,\sigma,\mu)$, and $$\int_X\int_Y(\mathcal X_S)_xd\nu d\mu=\int_X\nu[S](x)d\mu=(\mu\otimes\nu)(S)=\int_{X\times Y}\mathcal X_Sd(\mu\otimes\nu)$$
Now suppose $f$ is a simple function with non-negative coefficients, then $f=\sum_{k=1}^nc_k\mathcal X_{S_k}$ for some $n\in N$, $c_1,\ldots,c_n\in[0,\infty)$, and $S_1,\ldots,S_n\in\sigma\otimes\tau$. For all $x\in X$, $$ \int_Yf_xd\nu =\int_Y\p{\sum_{k=1}^nc_k\mathcal X_{S_k}}_xd\nu =\int_Y\sum_{k=1}^nc_k(\mathcal X_{S_k})_xd\nu =\sum_{k=1}^nc_k\int_Y(\mathcal X_{S_k})_xd\nu =\sum_{k=1}^nc_k\nu[S_k](x) $$ Thus $\int_Yf_xd\nu$ is integrable with respect to $(X,\sigma,\mu)$, and $$ \int_X\int_Yf_xd\nu d\mu =\int_X\sum_{k=1}^nc_k\nu[S_k]d\mu =\sum_{k=1}^nc_k\int_X\nu[S_k]d\mu =\sum_{k=1}^nc_k\int_{X\times Y}\mathcal X_{S_k}d(\mu\otimes\nu) =\int_{X\times Y}\sum_{k=1}^nc_k\mathcal X_{S_k}d(\mu\otimes\nu) =\int_{X\times Y}fd(\mu\otimes\nu) $$
Now suppose $f$ is unsigned measurable, then there exists an increasing sequence $(f_n)$ of simple functions with non-negative coefficients that converges pointwise to $f$. For all $x\in X$, $((f_n)_x)_n$ is an increasing sequence of unsigned $\tau$-measurable functions. Thus $$ \int_Yf_xd\nu =\int_Y\p{\lim_{n\to\infty}f_n}_xd\nu =\int_Y\lim_{n\to\infty}(f_n)_xd\nu =\lim_{n\to\infty}\int_Y(f_n)_xd\nu $$ implying $\int_Yf_xd\nu$ is integrable with respect to $(X,\sigma,\mu)$. Note that $(\int_Y(f_n)_xd\nu)_n$ is an increasing sequence of unsigned $\sigma$-measurable functions. Thus $$ \int_X\int_Yf_xd\nu d\mu =\lim_{n\to\infty}\int_X\int_Y(f_n)_xd\nu d\mu =\lim_{n\to\infty}\int_{X\times Y}f_nd(\mu\otimes\nu) =\int_{X\times Y}fd(\mu\otimes\nu) $$
The other equality can be proven by a symmetric argument. $\blacksquare$

Proof. Since $f$ is measurable, $\abs{f}$ is unsigned measurable and thus integrable. And we have $$\int_X\int_Y\abs{f}_xd\nu d\mu=\int_{X\times Y}\abs{f}d(\mu\otimes\nu)\lt\infty$$ Since $f$ is measurable and $\int_{X\times Y}\abs{f}d(\mu\otimes\nu)\lt\infty$, $f$ is integrable. Since $f$ is measurable, $f^+$ and $-f^-$ are unsigned measurable, then $\int_Y(f^+)_xd\nu$ and $\int_Y(-f^-)_xd\nu$, as functions of $x$, are unsigned measurable. Let $X^*$ denote the set of $x\in X$ such that $f_x$ is not integrable. Since $(f_x)^+=(f^+)_x$ and $-(f_x)^-=(-f^-)_x$, we have $$X^*=\left\{x\in X\bigg|\int_Y(f^+)_xd\nu=\int_Y(-f^-)_xd\nu=\infty\right\}$$ thus $X^*$ is measurable. Since $\int_X\int_Y\abs{f}_xd\nu d\mu\lt\infty$, $X^*$ has zero measure. Define $g:X\times Y\to\overline R$ by $g(x,y)=0$ if $x\in X^*$ and $g(x,y)=f(x,y)$ otherwise, then $$ \int_{X\times Y}fd(\mu\otimes\nu) =\int_{X\times Y}f^+d(\mu\otimes\nu)-\int_{X\times Y}-f^-d(\mu\otimes\nu) =\int_X\int_Y(f^+)_xd\nu d\mu-\int_X\int_Y(-f^-)_xd\nu d\mu =\int_X\int_Y(g^+)_xd\nu d\mu-\int_X\int_Y(-g^-)_xd\nu d\mu =\int_X\int_Y((g^+)_x-(-g^-)_x)d\nu d\mu =\int_X\int_Yg_xd\nu d\mu =\int_X\int^*_Yf_xd\nu d\mu $$ The other equality can be proven by a symmetric argument. $\blacksquare$

Proof. Let $A$ be an open subset of $U$, as a topological subspace of $R^d$, then $A$ is open with respect to the standard topology of $U$, as a metric subspace of $R^d$. Let $\varepsilon\gt0$. Since $U$ is measurable, there exists an open set $B$ of $R^d$ containing $U$ such that $m^*(B\setminus U)\lt\varepsilon$. Let $p\in A$, then there exists $r\gt0$ such that $\{x\in U|d(x,p)\lt r\}\subseteq A$ and $\{x\in R^d|d(x,p)\lt r\}\subseteq B$. Note that let $x\in R^d$ with $d(x,p)\lt r$, if $x\notin A$, then $x\notin\{x\in U|d(x,p)\lt r\}$, thus $x\notin U$, and hence $x\in B\setminus U$. This implies that for all $x\in R^d$ with $d(x,p)\lt r$, either $x\in A$ or $x\in B\setminus U$, thus $$\{x\in R^d|d(x,p)\lt r\}\subseteq A\cup(B\setminus U)$$ And $p$ is thus an interior point of $A\cup(B\setminus U)$. Since $p$ is arbitrarily chosen from $A$, $A$ is a subset of the interior of $A\cup(B\setminus U)$, which is open in $R^d$. Denote the interior of $A\cup(B\setminus U)$ by $C$, then $$m^*(C\setminus A)\le m^*((A\cup(B\setminus U))\setminus A)=m^*(B\setminus U)\lt\varepsilon$$ Therefore $A$ is measurable.

Since $f$ is continuous and real-valued, for all $c\in R$, $f^{-1}((c,\infty])=f^{-1}((c,\infty))$, which is an open subset of $U$, as a topological subspace of $R^d$, and is thus measurable. Hence $f$ is measurable. $\blacksquare$

Proof. Define $g:U\to[0,\infty]$ such that $$g(p)=\inf_{\delta\gt0}\sup_{x,y\in U\cap B_\delta(p)}d(f(x),f(y))$$ Clearly, given $p\in U$, $g(p)\ge0$, and $f$ is continuous at $p$ if and only if $g(p)=0$. Thus the set of discontinuity of $f$ on $B$ is $G=\{p\in B|g(p)\gt0\}$.

Suppose $f$ is bounded on $B$ and $m^*(G)\gt0$. Let $G_k=\{x\in B|g(x)\ge\frac{1}{k}\}$ for $k\in N^+$, then $G=\bigcup_{k\in N^+}G_k$. Note that $$\sum_{k\in N^+} m^*\p{G_k}\ge m^*\p{\bigcup_{k\in N^+}G_k}=m^*(G)\gt0$$ Thus for some $k\in N^+$, $m^*\p{G_k}\gt0$. Suppose $P(j_1,\ldots,j_d)$ is a $d$-partition on $B$. Let $P_1,\ldots,P_l$ denote $P(j_1,\ldots,j_d)$ reindexed into a tuple. Let $P^*$ denote the union of boundaries of $P_1,\ldots,P_l$, then clearly $m^*(P^*)=0$. Thus $$m^*\p{G_k}\le m^*\p{G_k\setminus P^*}+m^*(G_k\cap P^*)\le m^*\p{G_k\setminus P^*}+m^*(P^*)=m^*\p{G_k\setminus P^*}\le m^*\p{G_k}$$ implying $m^*\p{G_k\setminus P^*}=m^*\p{G_k}$. Let $P^*_j$ denote the interior of $P_j$. Let $J=\{j\in\{1,\ldots,l\}|P^*_j\cap(G_k\setminus P^*)\neq\emptyset\}$. Then $\{P^*_j:j\in J\}$ covers $G_k\setminus P^*$. Thus $$U(P,f)-L(P,f) \ge\sum_{j\in J}\p{\sup_{x\in P_j}f(x)-\inf_{x\in P_j}f(x)}m(P_j) \ge\sum_{j\in J}\frac{1}{k}m(P_j) =\frac{1}{k}\sum_{j\in J}m^*(P^*_j) \ge\frac{1}{k}m^*\p{\bigcup_{j\in J}P^*_j} \ge\frac{1}{k}m^*\p{G_k\setminus P^*} =\frac{1}{k}m^*\p{G_k}$$ Note that $\varepsilon=\frac{1}{k}m^*\p{G_k}\gt0$ is independent on the $d$-partition $P$. Suppose for contradiction that $\inf_PU(P,f)=\sup_PL(P,f)$. Then for some $d$-partitions $P_1,P_2$, $U(P_1,f)-L(P_2,f)\lt\varepsilon$. Then their common refinement $P^*$ has $U(P^*,f)-L(P^*,f)\le U(P_1,f)-L(P_2,f)\lt\varepsilon$, a contradiction. Hence $f$ is not Riemann integrable on $U$. Taking the contraposition, if $f$ is Riemann integrable on $U$, then either $f$ is not bounded on $B$ or $m^*(G)=0$. But $f$ is bounded on $B$ by definition of Riemann integrability. Thus $m^*(G)=0$, and hence $m(G)=0$.

Now suppose $f$ is bounded on $B$ and $m(G)=0$. Note that given $r\gt0$, $G_r=\{p\in B|g(p)\ge r\}$ contains its limit points, and is thus closed. Since $G_r$ is also bounded, it is compact. Let $\varepsilon\gt0$ and $$\varepsilon^*=\inf\brace{\frac{\varepsilon}{2\p{\sup_{x\in B}f(x)-\inf_{x\in B}f(x)}+1},\frac{\varepsilon}{2m(B)+1}}$$ Since $m^*(G_{\varepsilon^*})\le m^*(G)=m(G)=0$, there exists a sequence of bounded open boxes $(B_n)$ covering $G_{\varepsilon^*}$ such that $\sum\abs{B_n}\lt\varepsilon^*$. Since $G_{\varepsilon^*}$ is compact, there exists a finite subcover $B^*_1,\ldots,B^*_k$ of $G_{\varepsilon^*}$. Let $S_j=\overline{B^*_j}\cap B$ for $j\in\{1,\ldots,k\}$. Then $S_1,\ldots,S_k$ are closed subboxes of $B$ that cover $G_{\varepsilon^*}$. Note that $$m\p{\bigcup S_j}\le\sum m(S_j)\le\sum m(B^*_j)\le\sum\abs{B_n}\lt\varepsilon^*$$ Now define a $d$-partition $P$ by the endpoints of the intervals that form each of $S_j$. Then there are two kinds of partitions in $P$: those that are subboxes of some $S_j$, denoted $P_1$, and the others, denoted $P_2$. Note that

• $\bigcup P_1=\bigcup S_j$,
• $\p{\bigcup P_2}\cap\p{\bigcup B^*_j}=\emptyset$ and thus $\bigcup P_2\subseteq\{p\in B|g(p)\lt\varepsilon^*\}$, and
• $\bigcup P_2$ is closed and bounded and thus compact.

Let $p\in\bigcup P_2$, then there exists $\delta_p\gt0$ such that $\sup_{x,y\in U\cap B_{\delta_p}(p)}d(f(x),f(y))\lt\varepsilon^*$. Note that $\{B_{\delta_p/2}(p):p\in\bigcup P_2\}$ is an open cover of $\bigcup P_2$ and thus has a finite subcover $\{B_{\delta_{p_i}/2}(p_i):i\in\{1,\ldots,l\}\}$ for some $p_1,\ldots,p_l\in\bigcup P_2$. Let $\delta=\frac{1}{2}\inf\{\delta_{p_i}:i\in\{1,\ldots,l\}\}$. Note that for all $x,y\in\bigcup P_2$ with $d(x,y)\lt\delta$, $x\in B_{\delta_{p_i}/2}(p_i)$ for some $i\in\{1,\ldots,l\}$. Then $d(y,p_i)\le d(x,p_i)+d(x,y)\lt\delta_{p_i}$. Thus $x,y\in U\cap B_{\delta_{p_i}}(p_i)$, implying $d(f(x),f(y))\lt\varepsilon^*$. Now let $P^*$ be a refinement of $P$ such that each partition $A$ in $P^*$ has $d(x,y)\lt\delta$ for all $x,y\in A$. Let $P^*_1$ denote partitions of $P^*$ in $\bigcup P_1$, and $P^*_2$ denote partitions of $P^*$ in $\bigcup P_2$, then we have $$\inf_PU(P,f)-\sup_PL(P,f) \le U(P^*,f)-L(P^*,f) =\sum_{A\in P^*_1}\p{\sup_{x\in A}f(x)-\inf_{x\in A}f(x)}m(A)+\sum_{A\in P^*_2}\p{\sup_{x\in A}f(x)-\inf_{x\in A}f(x)}m(A) $$ $$ \le\sum_{A\in P^*_1}m(A)\p{\sup_{x\in B}f(x)-\inf_{x\in B}f(x)}+\sum_{A\in P^*_2}m(A)\varepsilon^* =m\p{\bigcup P_1}\p{\sup_{x\in B}f(x)-\inf_{x\in B}f(x)}+m\p{\bigcup P_2}\varepsilon^* \lt\varepsilon^*\p{\sup_{x\in B}f(x)-\inf_{x\in B}f(x)}+m(B)\varepsilon^* \lt\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon$$ Since $\varepsilon$ is arbitrary, $f$ is integrable on $B$. $\blacksquare$

Proof. Suppose $f$ is Riemann integrable on $B$. Then the set $D$ of discontinuity of $f$ on $B$ has measure zero, thus the set of continuity of $f$ on $B$, which is $C=B\setminus D$, is measurable. Note that $f|_C$ is a continuous function on a measurable set, and is thus measurable. Then both $C_+$ and $C_-$ are measurable. Since $D$ has measure zero, $f|_D$ is measurable, and both $D_+$ and $D_-$ has measure zero. Hence both $B_+=C_+\cup D_+$ and $B_-=C_-\cup D_-$ are measurable, and $f|_B$ is measurable. Since $B$ is bounded and $f$ is bounded on $B$, both $I_*(f,B_+)$ and $I_*(-f,B_-)$ are finite, hence $f$ is Lebesgue integrable on $B$.

Let $\varepsilon\gt0$. Since $f$ is Riemann integrable on $B=\bigtimes_{i=1}^d[a_i,b_i]$, there exists a $d$-partition $P$ of $B$, reindexed into $P_1,\ldots,P_k$, such that $U(P,f)-L(P,f)\lt\varepsilon$. For each partition $P_j=\bigtimes_{i=1}^d[a_{i,j},b_{i,j}]$, let $P^*_j=\bigtimes_{i=1}^d(a_{i,j},b_{i,j}]$. Define $B^*=\bigtimes_{i=1}^d(a_i,b_i]$, then $P^*_1,\ldots,P^*_k$ is a measurable partition of $B^*$. If we further define $P_0=P^*_0=B\setminus B^*$, then $P^*_0,\ldots,P^*_k$ is a measurable partition of $B$. Now define ${P^*_\gt}_j=P^*_j\cap\p{B\setminus B_-}$ and ${P^*_\le}_j=P^*_j\cap B_-$ for $j\in\{0,\ldots,k\}$, then they are both measurable. Note that ${P^*_\gt}_0,\ldots,{P^*_\gt}_k$ is a measurable partition of $B\setminus B_-$, ${P^*_\le}_0,\ldots,{P^*_\le}_k$ is a measurable partition of $B_-$, and $\{{P^*_\gt}_j,{P^*_\le}_j\}$ is a partition of $P^*_j$ for $j\in\{0,\ldots,k\}$. Define $$L_\gt=\sum_{j=0}^k\sup\{0,\inf_{x\in P_j}f(x)\}\mathcal X_{{P^*_\gt}_j}$$ on $B_+$, then $L_\gt$ is an unsigned simple function on $B_+$ with $L_\gt\le f|_{B_+}$. Thus $$I_*(f,B_+)\ge I(L_\gt)$$ Define $$L_\le=\sum_{j=0}^k\inf\{0,\inf_{x\in P_j}f(x)\}\mathcal X_{{P^*_\le}_j}$$ on $B_-$, then $-L_\le$ is an unsigned simple function on $B_-$. Suppose $g$ is an unsigned simple function on $B_-$ such that $g\le(-f)|_{B_-}$, then for some $l\in N^+$, $c_1,\ldots,c_l\in[0,\infty]$ and measurable partition $A_1,\ldots,A_l$ of $B_-$, $g=\sum_{j=1}^lc_j\mathcal X_{A_j}$. Let $S_{i,j}={P^*_\le}_i\cap A_j$ for $i\in\{0,\ldots,k\}$ and $j\in\{1,\ldots,l\}$, then the double tuple $(S_{i,j})$ is a measurable partition of $B_-$. Also, given $i$, $S_{i,j}$ is a measurable partition of ${P^*_\le}_i$, and given $j$, $S_{i,j}$ is a measurable partition of $A_j$. Thus $$-L_\le =\sum_{i=0}^k\p{-\inf\{0,\inf_{x\in P_i}f(x)\}}\mathcal X_{{P^*_\le}_i} =\sum_{i=0}^k\p{-\inf\{0,\inf_{x\in P_i}f(x)\}}\sum_{j=1}^l\mathcal X_{S_{i,j}} =\sum_{i=0}^k\sum_{j=1}^l\p{-\inf\{0,\inf_{x\in P_i}f(x)\}}\mathcal X_{S_{i,j}}$$ and $$g =\sum_{j=1}^lc_j\mathcal X_{A_j} =\sum_{j=1}^lc_j\sum_{i=0}^k\mathcal X_{S_{i,j}} =\sum_{j=1}^l\sum_{i=0}^kc_j\mathcal X_{S_{i,j}} =\sum_{i=0}^k\sum_{j=1}^lc_j\mathcal X_{S_{i,j}}$$ Note that for all $i\in\{0,\ldots,k\}$ and $j\in\{1,\ldots,l\}$, either $S_{i,j}=\emptyset$, implying $m(S_{i,j})=0$, or $S_{i,j}\neq\emptyset$, in which case let $p\in S_{i,j}$, then $$-\inf\{0,\inf_{x\in P_{i}}f(x)\}\ge-\inf_{x\in P_{i}}f(x)\ge(-f)(p)\ge g(p)=c_j$$ Hence $$I(-L_\le) =\sum_{i=0}^k\sum_{j=1}^l\p{-\inf\{0,\inf_{x\in P_i}f(x)\}}m(S_{i,j}) \ge\sum_{i=0}^k\sum_{j=1}^lc_jm(S_{i,j}) =I(g)$$ And we thus have $$I(-L_\le)\ge I_*(-f,B_-)$$ Note that let $J_-=\{j\in\{1,\ldots,k\}|\inf_{x\in P_j}f(x)\le0\}$, then if $j\in\{1,\ldots,k\}\setminus J_-$, we have $\inf_{x\in P_j}f(x)\gt0$, implying ${P^*_\le}_j=\emptyset$. Therefore $$\int_Bf(\vb x)dm =I_*(f,B_+)-I_*(-f,B_-) \ge I(L_\gt)-I(-L_\le) $$ $$ =\p{\sum_{j=0}^k\sup\{0,\inf_{x\in P_j}f(x)\}m({P^*_\gt}_j)}-\p{\sum_{j=0}^k\p{-\inf\{0,\inf_{x\in P_j}f(x)\}}m({P^*_\le}_j)} =\p{\sum_{j=1}^k\sup\{0,\inf_{x\in P_j}f(x)\}m({P^*_\gt}_j)}+\p{\sum_{j\in J_-}\inf_{x\in P_j}f(x)m({P^*_\le}_j)} $$ $$ \ge\p{\sum_{j=1}^k\inf_{x\in P_j}f(x)m({P^*_\gt}_j)}+\p{\sum_{j=1}^k\inf_{x\in P_j}f(x)m({P^*_\le}_j)} =\sum_{j=1}^k\inf_{x\in P_j}f(x)\p{m({P^*_\gt}_j)+m({P^*_\le}_j)} =\sum_{j=1}^k\inf_{x\in P_j}f(x)m(P^*_j) =\sum_{j=1}^k\inf_{x\in P_j}f(x)m(P_j) =L(P,f)$$ By a symmetric argument, $\int_Bf(\vb x)dm\le U(P,f)$. Let $\int_{[\vb a,\vb b]}f(\vb x)d\vb x$ denote the Riemann integral of $f$ on $B$. Note that $$L(P,f)\le \int_{[\vb a,\vb b]}f(\vb x)d\vb x\le U(P,f)$$ therefore, $$\abs{\int_Bf(\vb x)dm-\int_{[\vb a,\vb b]}f(\vb x)d\vb x}\le U(P,f)-L(P,f)\lt\varepsilon$$ Since $\varepsilon$ is arbitrary, we have $$\int_Bf(\vb x)dm=\int_{[\vb a,\vb b]}f(\vb x)d\vb x$$ $\blacksquare$

Proof. Since $\abs{I_j}\gt0$ for each $j\in\{1,\ldots,d\}$, we have $a_j\lt b_j$, and the interior of $B$ is non-empty. Let $(c_1,\ldots,c_d)$ be a point in the interior of $B$, denoted $B^*$. For all $n\in N^+$ and $j\in\{1,\ldots,d\}$, define

• $a_{n,j}=a_j+\frac{1}{n}(c_j-a_j)$ if $a_j\in R$;
• $a_{n,j}=c_j-n$ if $a_j=-\infty$;
• $b_{n,j}=b_j-\frac{1}{n}(b_j-c_j)$ if $b_j\in R$;
• $b_{n,j}=c_j+n$ if $b_j=\infty$;
• $B_n=\bigtimes_{j=1}^d(a_{n,j},b_{n,j})$.
• $B^*_n=B_n\setminus\bigcup_{m=1}^{n-1}B_m$.

Then $(B^*_n)$ is a measurable partition of $B^*$. And we have $$\int_B f=\int_{B^*} f=\sum\int_{B^*_n} f=\sup_k\sum_{n=1}^k\int_{B^*_n} f=\sup_k\int_{\bigcup_{n=1}^kB^*_n} f=\sup_k\int_{B_k} f$$ From here, it is clear that the limit on the right-hand-side converges to $\int_B f$. $\blacksquare$

Proof. Since $f$ is continuous, it is measurable on $(a,b)$ and integrable on any $[r,s]$ where $a\lt r\le s\lt b$. Since $f$ is also unsigned, it is integrable on $(a,b)$. Note that given $a\lt r\le s\lt b$, $$F(s)-F(r)=\int_r^sf(x)dx=\int_{[r,s]}f\ge0$$ thus $F$ is increasing, which implies that $$\lim_{r\to a^+}F(r)=\inf_rF(r)\neq\infty$$ $$\lim_{s\to b^-}F(r)=\sup_rF(r)\neq-\infty$$ Then $$ \int_{(a,b)}f =\lim_{r\to a^+,s\to b^-}\int_{[r,s]}f =\lim_{r\to a^+,s\to b^-}\int_r^sf(x)dx =\lim_{r\to a^+,s\to b^-}(F(s)-F(r)) =\lim_{r\to a^+,s\to b^-}F(s)-\lim_{r\to a^+,s\to b^-}F(r) =\lim_{s\to b^-}F(s)-\lim_{r\to a^+}F(r)$$ $\blacksquare$

Proof. Let $\varepsilon\in(0,1)$ and let $\delta=\frac{\varepsilon}{m^*(U)+m^*(V)+1}$, then $\delta\in(0,1)$. Suppose $(B_i)$ is a sequence of bounded boxes covering $U$ with $\sum\abs{B_i}\lt m^*(U)+\delta$, and $(B'_j)$ is a sequence of bounded boxes covering $V$ with $\sum\abs{B'_j}\lt m^*(V)+\delta$. Note that for each $i$ and $j$, $B^*_{i,j}=B_i\times B'_j$ is a bounded box with $\abs{B^*_{i,j}}=\abs{B_i}\abs{B'_j}$. Also note that $(B^*_{i,j})$ covers $U\times V$. Thus $$ m^*(U\times V) \le\sum_i\sum_j\abs{B^*_{i,j}} =\sum_i\sum_j\abs{B_i}\abs{B'_j} =\sum_i\abs{B_i}\sum_j\abs{B'_j} \le\sum_i\abs{B_i}(m^*(V)+\delta) \lt(m^*(U)+\delta)(m^*(V)+\delta) $$ $$ =m^*(U)m^*(V)+(m^*(U)+m^*(V))\delta+\delta^2 \lt m^*(U)m^*(V)+(m^*(U)+m^*(V)+1)\delta =m^*(U)m^*(V)+\varepsilon $$ Since $\varepsilon$ is arbitrary, we have $m^*(U\times V)\le m^*(U)m^*(V)$. $\blacksquare$

Proof. Since $U\times V$ is open, it is measurable. Note that $U$ can be partitioned into a sequence $(B_i)$ of bounded boxes, and $V$ can be partitioned into a sequence $(B'_j)$ of bounded boxes. Note that for each $i$ and $j$, $B^*_{i,j}=B_i\times B'_j$ is a bounded box with $\abs{B^*_{i,j}}=\abs{B_i}\abs{B'_j}$. And $(B^*_{i,j})$ is clearly a partition of $U\times V$. Thus $$m(U\times V) =\sum_i\sum_j\abs{B^*_{i,j}} =\sum_i\sum_j\abs{B_i}\abs{B'_j} =\sum_i\abs{B_i}\sum_j\abs{B'_j} =m(U)m(V)$$ $\blacksquare$

Proof. Suppose first that $U$ and $V$ are bounded, then both $m(U)$ and $m(V)$ are finite. Let $\varepsilon\in(0,1)$ and let $\delta=\frac{\varepsilon}{m(U)+m(V)+2}$, then $\delta\in(0,\frac{1}{2})$. Since $U$ and $V$ are measurable, there exists an open set $S_U$ containing $U$ with $m^*(S_U\setminus U)\lt\delta$, and an open set $S_V$ containing $V$ with $m^*(S_V\setminus V)\lt\delta$. Note that $S_U\times S_V$ is open, contains $U\times V$, and $$(S_U\times S_V)\setminus(U\times V)=((S_U\setminus U)\times S_V)\cup(S_U\times(S_V\setminus V))$$ Thus $$ m^*((S_U\times S_V)\setminus(U\times V)) \le m^*((S_U\setminus U)\times S_V)+m^*(S_U\times(S_V\setminus V)) \le m(S_U\setminus U)m(S_V)+m(S_U)m(S_V\setminus V) \le (m(S_U)+m(S_V))\delta \lt (m(U)+m(V))\delta+2\delta^2 \lt (m(U)+m(V)+1)\delta \lt\varepsilon $$ Since $\varepsilon$ is arbitrary, $U\times V$ is measurable.

Again, let $\varepsilon\in(0,1)$ and define $\delta, S_U, S_V$ as above, then $$ m(U\times V) =m(S_U\times S_V)-m((S_U\times S_V)\setminus(U\times V)) \gt m(S_U\times S_V)-\varepsilon =m(S_U)m(S_V)-\varepsilon \ge m(U)m(V)-\varepsilon $$ Since $\varepsilon$ is arbitrary, $$m(U\times V)\ge m(U)m(V)$$ Since we also have $$m(U\times V)=m^*(U\times V)\le m^*(U)m^*(V)=m(U)m(V)$$ We conclude that $$m(U\times V)=m(U)m(V)$$
Now for general $U$ and $V$, define $$B_i=B_i(\vb0_d)\setminus\p{\bigcup_{k\lt i}B_k(\vb0_d)}$$ $$U_i=U\times B_i$$ $$B'_j=B_j(\vb0_{d'})\setminus\p{\bigcup_{k\lt j}B_k(\vb0_{d'})}$$ $$V_j=V\times B'_j$$ Then $(U_i\times V_j)$ is a measurable partition of $U\times V$. Thus $U\times V$ is measurable and $$ m(U\times V) =\sum_i\sum_jm(U_i\times V_j) =\sum_i\sum_jm(U_i)m(V_j) =\sum_im(U_i)\sum_jm(V_j) =m(U)m(V) $$ $\blacksquare$

Proof. Suppose $U$ is bounded and $f$ is bounded by $1$. Since $f$ is integrable on $U$, $f$ is measurable on $U$. Let $U_c=\{x\in U|f(x)\gt c\}$ for $c\in R$, and define $${f_*}_k=\sum_{j=1}^{k-1}\frac{1}{k}\mathcal X_{U_{\frac{j}{k}}}$$ $${f^*}_k=\sum_{j=0}^{k-1}\frac{1}{k}\mathcal X_{U_{\frac{j}{k}}}$$ on $U$ for each $k\in N^+$. Note that

• if $x\in U$, $j\lt k$, and $f(x)\in(\frac{j}{k},\frac{j+1}{k}]$, then ${f_*}_k(x)=\frac{j}{k}$ and ${f^*}_k(x)=\frac{j+1}{k}$;
• let ${P_*}_{k,j}=\{x\in U|f(x)\in(\frac{j}{k},\frac{j+1}{k}]\}\times[0,\frac{j}{k}]$ for $j\lt k$, then these ${P_*}_{k,j}$, together with $\{x\in U|f(x)=0\}\times\{0\}$, is a measurable partition of $U\times {f_*}_k$;
• let ${P^*}_{k,j}=\{x\in U|f(x)\in(\frac{j}{k},\frac{j+1}{k}]\}\times[0,\frac{j+1}{k}]$ for $j\lt k$, then these ${P^*}_{k,j}$, together with $\{x\in U|f(x)=0\}\times\{0\}$, is a measurable partition of $U\times {f^*}_k$.

Let $U_+$ denote $\{x\in U|f(x)\gt 0\}$, then $U_+\times {f_*}_k=\bigcup_{j\lt k}{P_*}_{k,j}$ and $U_+\times {f^*}_k=\bigcup_{j\lt k}{P^*}_{k,j}$. Let $\varepsilon\gt0$. Let $n\in N^+$ such that $n\gt\frac{2m(U)}{\varepsilon}$. Since each ${P^*}_{n,j}$ is measurable, we can find an open superset $S_{n,j}$ for each ${P^*}_{n,j}$ such that $m^*(S_{n,j}\setminus{P^*}_{n,j})\lt\frac{\varepsilon}{2n}$. Then $S_n=\bigcup_{j\lt n}S_{n,j}$ is open, $U_+\times {f_*}_n\subseteq U_+\times f\subseteq U_+\times {f^*}_n\subseteq S_n$, and $$ m^*(S_n\setminus U_+\times f) \le m(S_n\setminus U_+\times {f^*}_n)+m(U_+\times {f^*}_n\setminus U_+\times {f_*}_n) \le m\p{\bigcup_{j\lt n}\p{S_{n,j}\setminus{P^*}_{n,j}}}+m\p{\bigcup_{j\lt n}\p{{P^*}_{n,j}\setminus{P_*}_{n,j}}} $$ $$ \le \sum_{j\lt n}m(S_{n,j}\setminus{P^*}_{n,j})+\sum_{j\lt n}m({P^*}_{n,j}\setminus{P_*}_{n,j}) \lt \sum_{j\lt n}\frac{\varepsilon}{2n}+\sum_{j\lt n}m\p{\brace{x\in U|f(x)\in\left(\frac{j}{n},\frac{j+1}{n}\right]}}\frac{1}{n} \le\frac{\varepsilon}{2}+\frac{m(U)}{n} \lt\varepsilon $$ Since $\varepsilon$ is arbitrary, $U_+\times f$ is measurable. Since $\{x\in U|f(x)=0\}\times f$ has measure zero, $U\times f$ is measurable.

Note that, if we let $U_j=\{x\in U|f(x)\in(\frac{j}{k},\frac{j+1}{k}]\}$ for $j\lt k$, then $${f_*}_k=\sum_{j\lt k}\frac{j}{k}\mathcal X_{U_j}$$ $${f^*}_k=\sum_{j\lt k}\frac{j+1}{k}\mathcal X_{U_j}$$ Let $U^0$ denote $\{x\in U|f(x)=0\}$, then $$m(U\times {f_*}_k)=m(U^0\times\{0\})+\sum_{j\lt k}m\p{U_j\times\left[0,\frac{j}{k}\right]}=\sum_{j\lt k}m(U_j)\frac{j}{k}=I({f_*}_k)$$ $$m(U\times {f^*}_k)=m(U^0\times\{0\})+\sum_{j\lt k}m\p{U_j\times\left[0,\frac{j+1}{k}\right]}=\sum_{j\lt k}m(U_j)\frac{j+1}{k}=I({f^*}_k)$$ Note that $$I({f_*}_k)\le \int_Uf\le I({f^*}_k)$$ $$m(U\times {f_*}_k)\le m(U\times f)\le m(U\times {f^*}_k)$$ Since $$I({f^*}_k)-I({f_*}_k)=\sum_{j\lt k}\frac{1}{k}m(U_j)\le\frac{m(U)}{k}$$ and $k$ is arbitrary, we have $$I({f^*}_k)=I({f_*}_k)$$ Therefore $$\int_Uf=m(U\times f)$$
Now for bounded $U$ and general $f$, let $f_n=\inf(\sup(f-n,0),1)$ for each $n\in N$. Then

• $f=\sum f_n$,
• each $f_n$ is unsigned, bounded by $1$ and integrable on $U$, and
• for each $n$, the isometry $I_n:R^{d+1}\to R^{d+1}$ such that $I_n(\vb x,y)=(\vb x,y+n)$ maps $(U\times f_n)\cap(R^d\times(0,1])$ to $(U\times f)\cap(R^d\times(n,n+1])$.

Since $R^d\times\{0\}$ has measure zero, $$ \int_U f =\int_U\sum f_n =\sum\int_Uf_n =\sum m(U\times f_n) =\sum m((U\times f_n)\cap(R^d\times(0,1])) =\sum m((U\times f)\cap(R^d\times(n,n+1])) =m(U\times f) $$
Now for general $U$ and general $f$, let $U_i=U\cap\p{B_i(\vb0_d)\setminus\bigcup_{j\lt i}B_j(\vb0_d)}$ for each $i\in N$. Then

• each $U_i$ is bounded,
• $(U_i)$ is a measurable partition of $U$, and
• $(U_i\times f)$ is a measurable partition of $U\times f$.

Therefore, $$ \int_U f =\sum\int_{U_i}f =\sum m(U_i\times f) =m(U\times f) $$ $\blacksquare$

Proof. $\blacksquare$