Precursor to Fubini's Theorem - Question on Proof

Question

I'm reading about Fubini's theorem. The book I'm reading considers the following theorem:

Suppose $(X, \mathcal{M}, \mu), (Y, \mathcal{N}, \nu)$ are $\sigma$-finite measure spaces. If $E \in \mathcal{M} \times \mathcal{N}$, then the functions $x \mapsto \nu(E_x)$ and $y \mapsto \mu(E^y)$ are measureable on $X$ and $Y$ respectively and $$\mu \times \nu (E) = \int \nu(E_x)\,d\mu(x) = \int \mu(E^y)\,d\nu(y)$$ where $E_x = \{y \in Y:(x,y) \in E\}$ and $E^y = \{x \in X:(x,y) \in E\}$

They prove this by letting $\mathcal{C}$ be the set of all $E\in \mathcal{M} \times \mathcal{N}$ that satisfy the conclusions of the theorem and prove this is a monotone class. By a theorem proved earlier, this means $\mathcal{C}$ is a sigma algebra. They conclude the proof here.

My question: I'm confused why they are trying to prove $\mathcal{C}$ is a sigma algebra - how does the proposition follow from knowing this?

Answer 1

1. The conclusion of the theorem holds for all $A \times B \in \cal{M} \times \cal{N}$.
2. The set of all $E \in \cal{M} \times \cal{N}$ for which the theorem holds is a $\sigma$-algebra $\cal{C} \subset \cal{M} \times \cal{N}$.
3. Since $\cal{C}$ contains all sets of the form $A \times B$, it contains the $\sigma$-algebra generated by such sets.
4. That $\sigma$-algebra is exactly $\cal{M} \times \cal{N}$, hence $ \cal{M} \times \cal{N} \subset \cal{C}$.