# Functions measurable with respect to the join of sigma fields

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.

• 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.
– gerw
Nov 11, 2021 at 11:45
• Thank you for your comment. I've edited the question in response. If I misunderstood what you meant, let me know.
– MrLJ
Nov 11, 2021 at 12:16

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.