Approximation by simple functions on a product $\sigma$-algebra

Question

Let

• $(\Omega_i,\mathcal A_i)$ be a measureable space
• $\mathcal M_i\subseteq2^{\Omega_i}$ be a $\pi$-system with $\Omega_i\in\mathcal M_i$ and $\sigma(\mathcal M_i)=\mathcal A_i$
• $\mathcal E(M_1\times\mathcal M_2)$ be the $\mathbb R$-vector space of functions $H:\Omega_1\times\Omega_2\to\mathbb R$ with $$H=\sum_{i=1}^k1_{A_i}x_i$$ for some $k\in\mathbb N$, pairwise disjoint $A_1,\ldots,A_k\in\mathcal M_1\times\mathcal M_2$ and $x_1,\ldots,x_k\in\mathbb R$

Note that $\mathcal M:=\mathcal M_1\times\mathcal M_2$ is a $\pi$-system with $\Omega_1\times\Omega_2\in\mathcal M$. Let $$\mathcal H:=\left\{H:\Omega_1\times\Omega_2\to\mathbb R\mid\exists(H_n)_{n\in\mathbb N}\subseteq\mathcal E(\mathcal M_1\times\mathcal M_2):H_n\xrightarrow{n\to\infty}H\right\}\;.$$ Note that $\mathcal H$ is a $\mathbb R$-vector space with $$1_A\in\mathcal H\;\;\;\text{for all }A\in\mathcal M\;.$$

I would like to show that $\mathcal H$ contains any $\mathcal A_1\otimes\mathcal A_2$-measurable function $\Omega_1\times\Omega\to\mathbb R$.

This would follow from the functional monotone class theorem, if we are able to show the following: If $H:\Omega_1\times\Omega_2\to\mathbb R$ and $(H_n)_{n\in\mathbb N}\subseteq\mathcal H$ with $$0\le H_n\le H_{n+1}\;\;\;\text{for all }n\in\mathbb N$$ and $H_n\xrightarrow{n\to\infty}H$, then $H\in\mathcal H$.

How can we show that?

Remark: I guess this can be reduced to the following problem: If $(\Omega,\mathcal A)$ isa measureable space and $\mathcal R\subseteq 2^\Omega$ with $\sigma(\mathcal R)=\mathcal A$, then we can approximate a $\mathcal A$-measurable function $X:\Omega\to\overline{\mathbb R}$ by functions of the form $\sum_{i=1}^kx_i1_{R_i}$ with $k\in\mathbb N$, $x_1,\ldots,x_k\in\mathbb R$ and $R_1,\ldots,R_k\in\mathcal R$.

Answer 1

$\mathcal{H}$ does not contain every measurable function. Take $(\Omega_i,\mathcal{A}_i)=([0,1],\mathcal{B})$ to be the unit interval with the Borel $\sigma$-algebra and $\mathcal{M}_i=\mathcal{B}$ for $i=1,2$.

Let $H$ be the indicator function of the diagonal, certainly a measurable function. Let $H_n$ be a linear combination of measurable rectangles such that $0\leq H_n\leq H$. That $0\leq H_n$ is without loss of generality since for every linear combination $G$ of measurable rectangles, $G\vee 0$ is still a linear combination of measurable rectangles.

Since every measurable rectangle that is a subset of the diagonal must be a singleton of the form $\{(x,x)\}$, each $H_n$ has only finitely many nonzero values. Taking the union of these values over all $H_n$, we get a countable set of such values. But since the diagonal is uncountable, we cannot get convergence at every point.