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Let $M := (\Omega, \mathcal{F})$ be a measurable space and let $\mathcal{M}$ be the vector space of all $(\mathbb{R},\mathcal{B}(\mathbb{R}))$-valued measurable functions defined on $M$. If $G$ is a sub-$\sigma$-field of $\mathcal{F}$, let $\mathcal{M}_{G}$ be the subspace of $\mathcal{M}$ whose elements are the $G$-measurable functions.

Suppose that $\mathcal{F}_{1},\ldots, \mathcal{F}_{k}$ are sub-$\sigma$-fields of $\mathcal{F}$ and let $\mathfrak{F} := \lor_{i = 1}^{k}\mathcal{F}_{i}$ denote their join.

Is there any relationship between $\mathcal{M}_{\mathfrak{F}}$ and $\mathcal{M}_{\mathcal{F}_{i}}$ besides that the latter is a subset of the former? Something like the former being the sum of the latter?

$\bigoplus_{i = 1}^{k} \mathcal{M}_{\mathcal{F_{i}}} \subseteq \mathcal{M}_{\mathfrak{F}}$ holds (where $\bigoplus_{i = 1}^{k}\mathcal{M}_{\mathcal{F_{i}}} := \{f_{1} + \ldots + f_{k}: f_{i} \in \mathcal{M}_{\mathcal{F}_{i}}, i \in \{1, \ldots, k\}\}$), but I can't see how to prove the reverse inclusion if it is indeed true.

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    $\begingroup$ At least, you have to replace the union by a linear hull, otherwise the set on the left-hand side mail fail to be even a vector space. $\endgroup$
    – gerw
    Nov 11, 2021 at 11:45
  • $\begingroup$ Thank you for your comment. I've edited the question in response. If I misunderstood what you meant, let me know. $\endgroup$
    – MrLJ
    Nov 11, 2021 at 12:16

1 Answer 1

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Consider the measure space $([-1,1],\mathcal{B}_{[-1,1]})$, and let $\mathcal{F}_1=\sigma([0,1])$ and $\mathcal{F}_2=\sigma(f)$, where $f(x)=|x|$. Note that $\mathcal{F}_2$ consists of Borel sets in $[-1,1]$ that are symmetric around $0$, and $\mathcal{F}_1\vee\mathcal{F}_2$ is generated by the sets in $\{A\cap B: A\in\mathcal{F}_1,B\in\mathcal{F}_2\}$. Therefore, $\mathfrak{F}=\mathcal{B}_{[-1,1]}$. Now, the identity function is measurable w.r.t. $\mathfrak{F}$, but it cannot be represented as a sum of $\mathcal{F}_1$ and $\mathcal{F}_2$ measurable functions.

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