Folland pre-meausre for product measure is a pre-measure 
Let $(X, \mathcal{X}, \mu)$ and $(Y, N, v)$ be measure spaces. We have already discussed, product $\sigma$ algebra $\mathcal{N} \otimes \mathcal{N}$ on $X \times Y$; we now construct a measure on $\mathcal{M} \otimes \mathcal{N}$ thaf in an obvious sense, the product of $\mu$ and $\nu$.
To begin with, we define a (measurable) rectangle to be a set of the form $A x$, where $A \in \mathcal{M}$ and $B \in \mathcal{N}$. Clearly
$(A \times B) \cap(E \times F)=(A \cap E) \times(B \cap F), \quad(A \times B)^c=\left(X \times B^{c}\right) \cup\left(A^{c} \times B\right)$ Therefore, by Proposition 1.7, the collection $\mathcal{A}$ of finite disjoint unions of rectangls is an algebra, and of course the $\sigma$-algebra it generates is $\mathcal{M} \otimes \mathcal{N}$.
Suppose $A \times B$ is a rectangle that is a (finite or countable) disjoint union y rectangles $A_j \times B_j$. Then for $x \in X$ and $y \in Y$.
$$
\chi_A(x) \chi B(y)=\chi_{A \times B}(x, y)=\sum \chi_A, \times B_1(x, y)=\sum \chi_A,(x) \chi_B,(y) .
$$
If we integrate with respect to $x$ and use Theorem $2.15$, we obtain
$$
\begin{aligned}
\mu(A) \chi_B(y)=\int \chi_A(x) \chi_B(y) d \mu(x) &=\sum \int \chi_{A_1}(x) \chi_{B_j}(y) d \mu(x) \\
&=\sum \mu\left(A_j\right) \chi_{B_j}(y) .
\end{aligned}
$$
In the same way, integration in $y$ then yields
$$
\mu(A) \nu(B)=\sum \mu\left(A_j\right) \nu\left(B_j\right) .
$$
It follows that if $E \in A$ is the disjoint union of rectangles $A_1 \times B_1, \ldots, A_n \times B_n$, and we set
$$
\pi(E)=\sum_1^n \mu\left(A_j\right) \nu\left(B_j\right)
$$
(with the usual convention that $0 \cdot \infty=0$ ), then $\pi$ is well defined on $\mathcal{A}$ (since any two representations of $E$ as a finite disjoint union of rectangles have a common refinement), and $\pi$ is a premeasure on $\mathcal{A}$. According to Theorem $1.14$, therefore, $\pi$ generates an outer measure on $X \times Y$ whose restriction to $\mathcal{M}^{\prime} \otimes \mathcal{N}$ is a measure that extends $\pi$. Wo call this measure the product of $\mu$ and $\nu$ and denote it by $\mu \times \nu$. Moreover, if $\mu$ and $\nu$ are $\sigma$-finite - say, $X=\bigcup_1^{\infty} A_j$ and $Y=\bigcup_1^{\infty} B_k$ with $\mu\left(A_j\right)<$ $\infty$ and $\nu\left(B_k\right)<\infty$ - then $X \times Y=\bigcup_{j, k} A_j \times B_k$, and $\mu \times \nu\left(A_j \times B_k\right)<\infty$, so $\mu \times \nu$ is also $\sigma$-finite. In this case, by Theorem $1.14, \mu \times \nu$ is the unique measure on $\mathcal{N} \otimes \mathcal{N}$ such that $\mu \times \nu(A \times B)=\mu(A) \nu(B)$ for all rectangles $A \times B$.

Why is it that $\pi$ is a pre-measure? It seems we have only defined it for finite disjoint unions of rectangles. To show $\pi$ is a pre-measure, I need to show that for $\bigcup_{i=1}^{n_j}A_i^j\times B_i^j \in \mathcal{A}$, disjoint, then
$$\pi(\bigcup_{j=1}^\infty \bigcup_{i=1}^{n_j}A_i^j\times B_i^j) = \sum_{j=1}^\infty\pi(\bigcup_{i=1}^{n_j}A_i^j\times B_i^j)$$
but I am not sure how to show this holds, as we have not defined it for sets which are not finite unions of rectangles.
 A: Since $E_j\in \mathcal{A}$, $E_j$ is a finite disjoint union of rectangles;
$$E_j=\bigcup_{i=1}^{n_j} A_i^j \times B_i^j$$
by the definition of $\pi$;
$$\pi(E_j)=\sum_{i=1}^{n_j}\mu(A_i^j) v(B_i^j)$$
Since $$\bigcup_{j=1}^{\infty}E_j\in \mathcal{A}$$
$\bigcup_{j=1}^{\infty}E_j$ is a finite disjoint union of rectangles;
$$\bigcup_{j=1}^{\infty}E_j=\bigcup_{j=1}^{\infty} \bigcup_{i=1}^{n_j}A_i^j\times B_i^j=\bigcup_{k=1}^NA_k\times B_k$$
so
$$\pi\left (\bigcup_{j=1}^{\infty}E_j \right )=\sum_{k=1}^N \mu(A_k) v(B_k)$$
Similar to the derivation in the text using $\bigcup_{j=1}^{\infty} \bigcup_{i=1}^{n_j}A_i^j\times B_i^j=\bigcup_{k=1}^NA_k\times B_k$
$$\begin{aligned}\sum_{k=1}^N \chi_{A_k}(x) \chi_{B_k}(y)&=\sum_{k=1}^N \chi_{A_k\times B_k} (x,y)\\&= \sum_{j=1}^{\infty}\sum_{i=1}^{n_j} \chi_{A_i^j \times B_i^j}(x,y)\\ &= \sum_{j=1}^{\infty}\sum_{i=1}^{n_j} \chi_{A_i^j}(x)\chi_{B_i^j}(y)\end{aligned}$$
then integrating both sides of the above equation first wrt $x$ and then wrt $y$ we get;
$$\sum_{k=1}^N \mu(A_k) v(B_k)=\sum_{j=1}^{\infty} \sum_{i=1}^{n_j}\mu(A_i^j) v(B_i^j)=\sum_{j=1}^{\infty} \pi(E_j)$$
therefore
$$\pi\left (\bigcup_{j=1}^{\infty}E_j \right )=\sum_{k=1}^N \mu(A_k) v(B_k)=\sum_{j=1}^{\infty} \pi(E_j)$$
It follows that $\pi$ is a premeasure.
