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

So we have: 
$$ \int_B \phi d \mu = \int_{B \setminus A} \phi d \mu + \int_A \phi d \mu= \int_A \phi d \mu$$
because $B \setminus A \subseteq X \setminus A $ and $\phi (X \setminus A) = \{ 0 \}$, so $\phi (B \setminus A) = \{ 0 \}$. 
So we have
$$ \int_B \phi d \mu = \int_A \phi d \mu = c$$ 
for some  $c \in \mathbb{R} $. 
Then, assuming $\mu(A) \neq 0$ we have that $\mu(B) \neq 0$, and we have: 
$$ \mathbb{E} \ (\ \phi \ | \ B ) := \frac{1}{\mu (B)} \int_B \phi d \mu = \frac{c}{\mu (B)} $$
and 
$$ \mathbb{E} \ (\ \phi \ | \ A ) := \frac{1}{\mu (A)} \int_A \phi d \mu = \frac{c}{\mu (A)} $$
Assuming that $\phi$ is NOT null a.e., we have that $c \neq 0$. Assuming $\mu(A)$ is finite, we have $\mathbb{E} \ (\ \phi \ | \ A ) \neq 0$. So we have: 
$$ \frac{\mathbb{E} \ (\ \phi \ | \ B )}{\mathbb{E} \ (\ \phi \ | \ A )} = \frac{\mu (A)}{\mu (B)}$$
Remark: We can conclude that 
$$ \frac{\mathbb{E} \ (\ \phi \ | \ B )}{\mathbb{E} \ (\ \phi \ | \ A )} = 1$$
if and only if 
$$ \mu (A) = \mu (B)$$
Since $A \subseteq B$ and $\mu(A)$ is finite, the last condition is equivalent to $\mu(B \setminus A) =0$.
A: 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 $\mu(A)\neq\mu(B)$).
