Approximate a non-negative function which is measurable in product $\sigma$-algebra

Question

$ \DeclareMathOperator*{\supp}{supp} \newcommand{\bR}{\mathbb{R}} \newcommand{\bT}{\mathbb{T}} \newcommand{\bN}{\mathbb{N}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bE}{\mathbb{E}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bD}{\mathbb{D}} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\cA}{\mathcal{A}} \newcommand{\cB}{\mathcal{B}} \newcommand{\cD}{\mathcal{D}} \newcommand{\cE}{\mathcal{E}} \newcommand{\cF}{\mathcal{F}} \newcommand{\cG}{\mathcal{G}} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\sD}{\mathscr{D}} \newcommand{\sE}{\mathscr{E}} \newcommand{\sG}{\mathscr{G}} \newcommand{\sH}{\mathscr{H}} \newcommand{\sK}{\mathscr{K}} \newcommand{\sP}{\mathscr{P}} %%%%%%%%%%%%%%%%%%%%%%%%%%%% $ Let $(E, \cE)$ and $(\Omega, \cG)$ be measurable spaces. We endow $E \times \Omega$ with the product $\sigma$-algebra $\cE \otimes \cG$. We endow $\bR_+$ with its Borel $\sigma$-algebra $\cB (\bR_+)$.

• Let $\sH^1_+$ be the collection of all measurable functions $f : E \times \Omega \to \bR_+$.

• Let $\sH^2_+$ be the subset of $\sH^1_+$ containing those $f$ with the property: there exists a sequence $(f^n)$ of functions $f^n : E \times \Omega \to \bR_+$ such that $f^n \uparrow f$ everywhere and that $f^n$ is of the form $$ f^n (x, \omega) = \sum_{k=1}^{n_f} r^n_k 1_{A^n_k} (x) 1_{B^n_k} (\omega), $$ for some $n_f \in \bN^*, r^n_k \in \bR_+, A^n_k \in \cE, B^n_k \in \cG$.

Is $\sH^1_+ = \sH^2_+$?

Thank you so much for your elaboration!

Answer 1

Take $(E,\mathcal E)=(\Omega,\mathcal G)= \mathbb R$ with the Borel $\sigma$-algebra, and let $C$ be a subset of the diagonal $\Delta\subset \mathbb R^2$. If for a linear combination $g$ of rectangles with positive coefficients $r_k$ one has $g:=\sum_{k=1}^{n } r_k {\bf1}_{A_k\times B_k} \le {\bf1}_C $, necessarily $A_k=B_k$ is a singleton, and $g$ has finite support. It follows that the only $C\subset \Delta$ for which there is a sequence $f^n\uparrow {\bf1}_C$ are the countable subsets; in particular ${\bf 1}_\Delta$ can't be reached that way.