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(Modified to be specific enough so that the other post does not answer my question) I have the following conjecture about measurable functions.

Let $\Omega$ be the sample space equipped with $\sigma$-algebra $\mathcal{F}$, and let $\mathcal{H} \subseteq \mathcal{F}$ be a sub-$\sigma$-algebra. For each $\omega \in \Omega$ we can associate a set that consists of answers to the questions "is $\omega \in A$?" for each $A \in \mathcal{H}$. Let this be called the $\mathcal{H}$-set of $\omega$. Concretely, the $\mathcal{H}$-set for $\omega$ is given by $\lbrace(A, 1_A(\omega)):A \in \mathcal{H}\rbrace $. Then, a function $X: \Omega \to \mathbb{R}$ is $\mathcal{H}$-measurable if and only if $X$ is constant across $\omega$'s that have the same $\mathcal{H}$-set.

I was able to prove one direction of this conjecture, namely that if $X$ is $\mathcal{H}$-measurable, then it is constant across $\omega$'s that have the same $\mathcal{H}$-set. However, I do not know how to prove or disprove the other direction -- that if $X$ is constant across $\omega$'s that have the same $\mathcal{H}$-set, then $X$ is $\mathcal{H}$-measurable.

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  • $\begingroup$ Apart from being a duplicate question it is totally unclear what you mean by "knowing whether each $A \in \mathcal{H}$ occurs uniquely". $\endgroup$
    – Kurt G.
    Commented May 17 at 0:46
  • $\begingroup$ @KurtG. The linked post does not answer my question; in particular, I am still uncertain as to whether my conjecture is true, and I do not have a proof. Regarding your second comment, I'm not sure exactly what is unclear about it. Are you familiar with the meaning of "knowing x uniquely determines y"? $\endgroup$ Commented May 17 at 1:12
  • $\begingroup$ After reading Robert Israel's answer I understand that conjecture now. There is an even simpler counter example: $1_A$ and $2\cdot 1_A$ are both measurable w.r.t. the $\sigma$-algebra $\{\emptyset,A,A^c,\Omega\}\,.$ Knowing whether $A$ occurs obviously not determines them uniquely. $\endgroup$
    – Kurt G.
    Commented May 17 at 4:35
  • $\begingroup$ Restricting ourselves to simple functions $X$ that have only finitely many values and are linear combinations of finitely many indicator functions of disjoint sets $A_i$ one can say that the $\sigma(A_1,\dots,A_n)$-measurability of $X$ is equivalent to knowing from the value of $X$ which $A_i$ has occurred. The converse is your conjecture and not true. $\endgroup$
    – Kurt G.
    Commented May 17 at 6:13
  • $\begingroup$ You are correct. My conjecture was not capturing the intuition I meant to convey. Please see the corrected version above. $\endgroup$ Commented May 18 at 0:18

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Your conjecture is false. Let $\mathcal H$ be the $\sigma$-algebra of Lebesgue measurable sets in $\mathbb R$. Knowing whether each half-line $(-\infty, x)$ occurs, i.e. whether $X < x$, uniquely determines $X$. But there are Lebesgue non-measurable functions $X$.

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  • $\begingroup$ My apologies, my conjecture was not capturing what I meant to say. Please see the corrected version above. $\endgroup$ Commented May 18 at 0:17

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