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Let $(\Omega, \mathcal F, \mu)$ be a probability space, $\mathcal{G}$ a sub-$\sigma$-field of $\mathcal{F}$, and $X,Y:\Omega \to \mathbb R$ random variables such that $X=Y$ almost surely. This implies there is $N \in \mathcal F$ such that $\mu(N)=0$ and $X=Y$ on $N^c := \Omega \setminus N$. Assume that $X$ is $\mathcal G$-measurable. Fix $B \in \mathcal B(\mathbb R)$. Then $$ \{Y \in B\} := (\{X \in B\} \cap N^c) \cup (\{Y \in B\} \cap N). $$

Clearly, $\{X \in B\} \in \mathcal G$ because $X$ is $\mathcal G$-measurable. However, it is not necessarily true that $N \in \mathcal G$. As such, $Y$ is not necessarily $\mathcal G$-measurable.

Could you verify if my above understanding is fine?

Akira
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    Yes, it is right. – Christophe Leuridan Jan 15 '23 at 13:59
  • See the notion of "complete" for $\mathcal G$. – GEdgar Jan 15 '23 at 14:58
  • @GEdgar Assume $(\Omega, \mathcal G, \mathbb P)$ is a complete probability space. If $N \in \mathcal G$ then ${Y \in B} \in \mathcal G$ and thus $Y$ is $\mathcal G$-measurable. Could you elaborate on the case $N \notin \mathcal G$? – Akira Jan 15 '23 at 15:03
  • That's very close to this question. Your understanding is right, you have to assume in addition that $\mu$ is complete on $\mathcal G$ to get that $Y$ is $\mathcal G$-measurable. – Matija Jan 16 '23 at 00:39
  • @Matija From your linked question, I got that if $X=Y$ a.s. under $(\Omega, \mathcal G, \mathbb P)$, $X$ is $\mathcal G$-measurable, and $(\Omega, \mathcal G, \mathbb P)$ is complete, then $Y$ is $\mathcal G$-measurable. However, in my case we only have $X=Y$ a.s. under $(\Omega, \mathcal F, \mathbb P)$. – Akira Jan 16 '23 at 00:50
  • @Matija As I'm concerned in my first comment, if $(\Omega, \mathcal G, \mathbb P)$ is complete and $N \notin \mathcal G$, I'm not sure if $Y$ is $\mathcal G$-measurable. – Akira Jan 16 '23 at 00:55
  • Thanks, of course, I was a bit too quick to answer. – Matija Jan 16 '23 at 08:22

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Let $\Omega=\{1,\dots,n\}$ for $n>1$, let $\mathcal F$ be the power set of $\Omega$, and let $X,Y:\Omega\rightarrow\mathbb R$ be given by $X(\omega)=1$, $Y(\omega)=\omega$. Let $\mu$ be the one-point mass on $1$, then we have $X=Y$ almost surely, with $N=\Omega\setminus\{1\}$. Let $\mathcal G=\{\emptyset,\Omega\}$ be trivial, then $X$ is $\mathcal G$-measurable, but $Y$ is not. In particular, $N$ is not $\mathcal G$-measurable, and $\{Y\in B\},\{Y\in B\}\cap N$ is not $\mathcal G$-measurable for any $B\subseteq\mathbb R$ with $B\cap N\neq\emptyset$.

Matija
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