I am trying to understand the proof of proposition 5.11 from these notes.
Given two measure spaces $(X, \Sigma_X, \mu)$ and $(Y, \Sigma_Y, \nu)$, define a measure $\lambda$ on $(X \times Y, \Sigma_X \otimes \Sigma_Y)$ by $\lambda(A \times B) = \mu(A)\nu(B)$.
Proposition 5.11. If a measurable rectangle $A \times B$ is a countable disjoint union of measurable rectangles $\{A_i \times B_i \mid i \in \mathbb{N}\}$, then $$\lambda(A \times B) = \sum^{\infty}_{i = 1}\lambda(A_i \times B_i).$$
In the proof, we arrive at an expression $$\chi_A(x)\chi_B(y) = \sum^{\infty}_{i = 1}\chi_{A_i}(x)\chi_{B_i}(y),$$ where $\chi_A$ is the characteristic function of $A$. The next step is where my understanding fails. We fix an $x \in X$ and integrate over $Y$, to get $$\chi_A(x) \nu(B) = \sum^{\infty}_{i = 1}\chi_{A_i}(x)\nu(B_i).$$ I feel like there is a lot of things going on in this one step, and I'm not able to follow it. Most crucially, the step is supposed to involve the monotone converge theorem, and I'm simply not able to see how that theorem plays any role. Can someone help elucidate this step for me?
Define for a positive integer $n$, $$f_n(y):=\sum_{i=1}^n\chi_{A_i}(x) \chi_{B_i}(y).$$ For each $y$, the sequence $(f_n(y))_{n\geqslant 1}$ is non-decreasing and non-negative. Since each function $f_n$ is $\Sigma_Y$-measurable, we have by the monotone convergence theorem $$\lim_{n\to +\infty }\int_Yf_n(y)d\nu(y)=\int_Y\lim_{n\to +\infty }f_n(y)d\nu(y) .$$