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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 $R_1,\ldots,R_k\in\mathcal R$.

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  • $\begingroup$ 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 '18 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 '18 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. $\endgroup$ – Pietro Majer Jan 20 '18 at 14:00
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$\mathcal{H}$ 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.

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