Let $(\Omega,\mathscr F)$ be a measurable space and $\mathscr G\subseteq\mathscr F$ a $\sigma$-subalgebra of $\mathscr F$. Let $(Y,\mathscr Y)$ be another measurable space. Suppose that $f:\Omega\times Y\to\mathbb R$ is a real-valued function measurable with respect to the product $\sigma$-algebra $\mathscr F\otimes\mathscr Y$ (where $\mathbb R$ is endowed with the Borel $\sigma$-algebra). Finally, suppose that $g:\Omega\to Y$ is a $\mathscr G/\mathscr Y$- measurable function (note that $\mathscr G$ is the $\sigma$-subalgebra here).
Now consider the following nested function mapping from $\Omega$ to $\mathbb R$: \begin{align*} \omega\mapsto f(\omega,g(\omega)) \end{align*}
I am wondering whether this nested function can be approximated by simple functions of sorts of the following form: \begin{align*} f(\omega,g(\omega))\approx\sum_{i=1}^n I_{G_i}(\omega)f(\omega,y_i)\quad\text{for each $\omega\in\Omega$,}\tag{$*$} \end{align*} where
- $n$ is a positive integer;
- $y_1,\ldots,y_n$ are elements of the set $Y$;
- $G_1,\ldots, G_n$ are disjoint sets in $\mathscr G$ whose union is $\Omega$; and
- $I_{G_i}$ is the indicator function of $G_i$ for each $i\in\{1,\ldots,n\}$.
The “$\approx$” in ($*$) is in the usual sense: there exists a sequence of functions of the form on the right-hand side converging pointwise (for each $\omega\in\Omega$) to the left-hand side.
Note that no topological structure on $(Y,\mathscr Y)$ is assumed.
Any suggestion about or reference to a proof or a counterexample would be greatly appreciated.
UPDATE: The original conjecture is false (see below), so I am going to refine it as follows. Let $\mathbb P$ be a probability measure on $(\Omega,\mathscr F)$ and suppose that $f$ is bounded.
Can the integral of $\omega\mapsto f(\omega,g(\omega))$ be approximated by integrals of functions of the form on the right-hand side of ($*$)?
More precisely, let \begin{align*} \mathcal H\equiv\{h:\Omega\to\mathbb R\mid&\bullet h(\omega)=\sum_{i=1}^nI_{G_i}(\omega)f(\omega,y_i)\text{ for each $\omega\in\Omega$},\\ &\bullet n\in\mathbb N,\\ &\bullet y_1,\ldots,y_n\in Y,\\ &\bullet \text{$G_1,\ldots, G_n$ are disjoint sets in $\mathscr G$ whose union is $\Omega$}\}. \end{align*} Does there exist a sequence $(h_m)_{m\in\mathbb N}$ in $\mathcal H$ such that \begin{align*} \lim_{m\to\infty}\left\{\int_{\Omega}h_m(\omega)\,\mathrm d\mathbb P(\omega)\right\}=\int_{\Omega}f(\omega,g(\omega))\,\mathrm d\mathbb P(\omega)? \end{align*}
My hunch (and hope, indeed) is that some kind of countability argument due to the ($\sigma$-)finiteness of $\mathbb P$ may make the approximation for at least integrals work.
To see why pointwise approximation does not work, let $(\Omega,\mathscr F)$ and $(Y,\mathscr Y)$ both be $\mathbb R$ endowed with the discrete $\sigma$-algebra and $\mathscr G=\mathscr F$. Let $g$ be the identity function and $f$ the indicator function of the diagonal $$D\equiv\{(x,x)\,|\,x\in\mathbb R\}.$$ Note that the diagonal is $\mathscr F\otimes\mathscr F$-measurable, since it is of the form $$D=\bigcap_{m\in\mathbb N}\bigcup_{q\in\mathbb Q}\left[q-\frac{1}{m},q+\frac{1}{m}\right]\times\left[q-\frac{1}{m},q+\frac{1}{m}\right].$$ Then, $$f(\omega,g(\omega))=1\quad\text{for each $\omega\in\mathbb R$}.$$ For any sequence $(h_m)_{m\in\mathbb N}$ of functions in $\mathcal H$, let $\widetilde Y$ denote the union of all the $y_i$’s appearing in the definition of the $h_m$’s. Then, $\widetilde Y$ is a countable union of finite sets, so it is countable. Taking $\widetilde\omega\in\mathbb R\setminus\widetilde Y$, one can see that $h_m(\widetilde\omega)=0$ for each $m\in\mathbb N$, so pointwise convergence fails at $\widetilde\omega$.
I feel a bit silly, because even a simpler version of the counterexample for the original conjecture works to refute the modified conjecture.
Let $(\Omega,\mathscr F)$ and $(Y,\mathscr Y)$ both be $[0,1]$ endowed with the Borel $\sigma$-algebra. Let $\mathscr G=\mathscr F$, $\mathbb P$ the Lebesgue measure, $g$ the identity function, and $f$ the indicator function of the diagonal $$D\equiv\{(x,x)\,|\,x\in[0,1]\}.$$ Then, $$\int_{\Omega}f(\omega,g(\omega))\,\mathrm d\mathbb P(\omega)=1.$$ However, if $h\in\mathcal H$, then $$\int_{\Omega}h(\omega)\,\mathrm d\mathbb P(\omega)=\sum_{i=1}^n\int_{G_i}f(\omega,y_i)\,\mathrm d\mathbb P(\omega)=0,$$ because $f(\omega,y_i)=0$ for all $\omega\in G_i\setminus\{y_i\}$ and $i\in\{1,\ldots,n\}$.