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I'm currently going through a book on measure-theoretic probability. In it, we defined conditional expectation:

Let X belong to $M^+(Omega, F)$ and P be a probability measure on F. For each sub-sigma-field G of F, the conditional expectation$ P(X|G)$ is the random variable $X_G$ in $M^+(X,G)$ for which $P(gX) = P(gX_G)$.

However, there is a small disclaimer: be careful to check that $X_G$ is G-measurable. I'm confused about how this can be a problem, since isn't $X_G\in M^+(X,G)$? Doesn't this by definition make it $G-$measurable?

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    $\begingroup$ \Omega produces $\Omega$ $\endgroup$ – parsiad Apr 3 '19 at 0:40
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They are just saying that if you want to prove that $Y$ is $P(X|G)$ you have to verify two facts: $Y \in M^{+}(X,G)$ and the condition $P(gX)=P(gY)$. A common mistake it to ignore the first part.

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  • $\begingroup$ Oh I see.. It says "check that X_G is G-measurable. Otherwise you might be tempted to leap to the conclusion that X_G = X". But we don't know that X is G-measurable, making this equality perhaps false $\endgroup$ – swedishfished Apr 3 '19 at 0:47
  • $\begingroup$ @swedishfished Eaxclty. That is why they are issuing a word of caution. $\endgroup$ – Kavi Rama Murthy Apr 3 '19 at 0:48

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