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Let $X$ be a set, let $B(X)$ denote the vector space of bounded real functions on $X$, and let $V\subseteq B(X)$ be a vector subspace containing $1$ and closed under monotone limits.

Let now $\Sigma_V$ be the set of subsets of $X$ such that their indicator function is in $V$, and suppose that $\Sigma_V$ is a $\sigma$-algebra (on the nose). Is it true that all functions of $V$ are $\Sigma_V$-measurable?

If not, do we need to assume any additional conditions?

Edit: This is not the (classic) monotone class theorem. I'm not asking whether the $\Sigma_V$-measurable functions are in $V$, but rather the opposite: I'm asking whether all the functions in $V$ are measurable for $\Sigma_V$.

geodude
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  • This should be a version of the Monotone Class Theorem (https://en.wikipedia.org/wiki/Monotone_class_theorem#Monotone_class_theorem_for_functions). Since $V$ is a linear subspace, 2. is satisfied. Since $V$ is closed under monotone limits, 3. is satisfied. The condition on the indicator functions should give you 1., so you have that $V$ contains all bounded functions that are $\Sigma_V$-measurable. – Syd Amerikaner May 03 '23 at 08:16
  • I don't think this is the monotone class theorem, but a sort of converse. See the edit :) – geodude May 03 '23 at 08:38
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    "Closed under monotone limits" ... I guess this means: If $f_n \in V$ and $f_{n+1} \ge f_n$ for all $n$ and $f :=\lim f_n$ happens to be bounded, then $f \in V$. – GEdgar May 03 '23 at 08:50

1 Answers1

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Try this one: Let $X$ be $[0,1]$ and let $V$ be the set $\{a+bx : a,b\in \mathbb R\}$.

Additional conditions... Maybe it is enough to add
$f,g \in V \Longrightarrow f\vee g \in V$.

GEdgar
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  • Thank you. (If you can help there too, I've asked a related question here, https://math.stackexchange.com/questions/4703733/monotone-class-style-theorem) – geodude May 21 '23 at 15:16