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Let $(\Omega, \mathcal{F})$ and be a measurable space and $(X, \mathcal{G})$ be a vector space equipped with some $\sigma$-field.

A simple function $f:\Omega \mapsto X$ is such that $$ f(\omega) = \sum_{i=1}^n x_i \mathbb{1}_{F_i}(\omega)$$

where $F_i \in \mathcal{F}$ are disjoint sets and $x_i \in X$ for all $i = 1,\ldots,n$.

Clasically, one says that $g:\Omega \mapsto X$ is measurable if $$ g^{-1}(G) \in \mathcal{F}, \quad \forall G \in \mathcal{G}$$

Now, for a simple function $f$ this seems to always be the case, since for any $G \in \mathcal{G}$:

\begin{align} f^{-1}(G) & = \{\omega \in \Omega \ | \ f(\omega) \in G \}\\ & = \{\omega \in \Omega \ | \ x_i \in G, \ \omega \in F_i \}\\ & = \bigcup_{i \ \text{such that } x_i \in G} F_i \in \mathcal{F} , \end{align}
since $\mathcal{F}$ is closed under finite (and countable) unions.

Now, this fact is completely independent of the family $\mathcal{G}$, as long as $F_i \in \mathcal{F}$ the resulting simple function $f$ will always be measurable.

I believe there must be a mistake somewhere.

Why? Well, there is a quite wide* family of $\sigma$-fields which satisfies that the point-wise limit of measurable functions is a measurable function

Therefore, if what I showed here is truth, it would imply that any function that is the limit of simple functions, is measurable for all the family of $\sigma$-fields satisfying the point-wise limit property; this feels like a false statement.

*When are pointwise limits of measurable functions measurable?, Limit of measurable functions is measurable?

pancho
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  • Such a theorem about the pointwise limit specifically considers the Borel sigma algebra on the codomain, hence any pointwise limit of simple functions is Borel measurable.. – Berci Jul 26 '19 at 22:32
  • Sure, but that means also that if the topology on the codomain is changed (hence, the Borel sigma algebra also changes) then such functions are still measurable. I.e. pointwise limit of simple functions are always measurable for any Borel sigma-algebra on the codomain? – pancho Jul 29 '19 at 07:25
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    The notion of pointwise limit presumes a topology on $X$. It changes along with the Borel sets. – Berci Jul 29 '19 at 12:17
  • Oh, I see, there it is. Feel free to write this as an answer. – pancho Jul 29 '19 at 12:58

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