• $(\Omega_i,\mathcal A_i)$ 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 $R_1,\ldots,R_k\in\mathcal R$.

  • $\begingroup$ I am not sure that what you want to show is true. What is true is that if you consider the smallest class containing $\mathcal{E}(...)$ and which is closed under bounded pointwise convergence, then it contains all bounded measurable (w.r.t. the product sigma algebra) functions. This follows from Dynkins multiplicative system theorem, see e.g. coursehero.com/file/p3097l/… $\endgroup$
    – PhoemueX
    Jan 16, 2018 at 7:55
  • $\begingroup$ @PhoemueX Yes, this generalizes the functional monotone class theorem. I've tried to use it, but the problem is to show the closedness under bounded convergence. $\endgroup$
    – 0xbadf00d
    Jan 16, 2018 at 14:21
  • $\begingroup$ I think in general you need to iterate the point-wise limit uncountably many times to reach the sigma algebra generated. But if you have a measure, and you consider a.e. equivalence and a.e. convergence, then the monotone class theorem ensures that few iterations of "point-wise monotone limits" suffice. $\endgroup$ Jan 20, 2018 at 14:00

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.