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Let's consider the vector space V of bounded scalar functions, which includes the constant function 1. We assume that any uniform limit of a bounded monotonic sequence of functions from V also belongs to this space.

Let D ⊂ V be a subset that is closed under multiplication.

Can we prove that any bounded σ(D) -measurable function belongs to the space V ? Any hints or detailed explanations would be appreciated.

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    $\begingroup$ I made the requested edits. $\endgroup$ Commented Nov 12, 2024 at 21:18
  • $\begingroup$ What is $\sigma(C)$? If this is a minimal sigma-algebra for which all functions from $C$ is measurable, than what if $I=C=C[0,1]$? Not every bounded Borel function is in $I$ $\endgroup$ Commented Nov 13, 2024 at 5:25

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This statement is false in general. E.g., suppose that $C=I=C[0,1]$.

Then the function $1_{[0,1/2]}$ is bounded and $\sigma(C)$-measurable but not in $I$.

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  • $\begingroup$ @NasimMamatkylov : Are you saying that, just because someone sets a task of proving a statement, the statement automatically becomes true? I will then set the task of proving that $0=1$. $\endgroup$ Commented Nov 13, 2024 at 13:47

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