Let $\phi \in [0,1]^X$ integrable, $A\subseteq B$, and $\phi (X\setminus A)=\{0\}$, is $\frac{\mathbb{E}(\phi |B)}{\mathbb{E}(\phi |A)}=1?$

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I have a conceptual problem with the relation between two integrable functions that are equal a.e.. Here there is a possible setting, to make things more concrete.

Question:

Assumptions:

  • $(X, \Sigma, \mu)$ is a measure space,
  • $A, B \in \Sigma$, with $A \subseteq B$,
  • $\phi \in [0,1]^X$ is a measurable (and integrable) function, with $\phi (X \setminus A) = \{ 0 \}$.

If we focus on the conditional expectation of $\mu$, defined for an arbitrary $Y \in \Sigma$ as $$ \mathbb{E} \ (\ \phi \ | \ Y ) := \frac{1}{\mu (Y)} \int_Y \phi d \mu, $$ can we state – given the previously described setting – that the following is true: $$ \frac{\mathbb{E} \ (\ \phi \ | \ B ) }{\mathbb{E} \ (\ \phi \ | \ A )}= 1 ?$$

To me this make sense, because essentially the two expectations are the same, given the condition $\phi (X \setminus A)$. Still I am not completely sure.

Is the line of reasoning sound?
Is actually correct to talk about equality a.e. in this kind of context?

Thank you for your time.

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Your conditions do not happen in useful situations. Apply your hypothesis to $A=\emptyset$ and $B=X.$ The $\phi$ must be zero on the complement of $A,$ i.e., everywhere.

Now if these conditions apply only for a particular choice of $A$ and $B$ then I have to argue that $E[\phi|A]$ is different from $E[\phi|B]$ as soon as one of these two numbers is nonzero and $\mu(B-A)>0$ (so that $\mu(A)\neq\mu(B)$).