Basic doubt about measurable subspace I was having some doubts related to measure subspace of a given space $(X,\Sigma,\mu)$. I've read that if one defines $\Sigma_E=\{A \cap E: A \in \Sigma\}$, then $\Sigma_E$ is a $\sigma-$algebra of $\mathcal P(E)$. It is clear that $\emptyset=\emptyset \cap E \in \Sigma_E$. I also see that it is closed under countable unions. Now, suppose $A \in \Sigma_E$, then $A=B \cap E$ with $B$ measurable of $X$. Why $A^c=B^c \cup E^c \in \Sigma_E$? I don't see why this is true. I would appreciate if someone could explain this to me.
 A: You have to consider the complements w.r.t. the space $E$, not $X$. 
Let $B \in \Sigma_E$, then $B = A \cap E$ for a $A \in \Sigma$, so 
$$ E \backslash B = E\backslash (A \cap E) = E \backslash A \cup E \backslash E = E \backslash A = X \backslash A \cap E \in \Sigma_E \; ,$$
because $X \backslash A \in \Sigma$.
A: In general, if set $X$ is fixed and $A,E \subset X$ are arbitrary, we may define
$$E \setminus A \;:=\; \{x: x\text{ belongs to } X,\text{ belongs to } E, \text{ but does not belong to } A.\}$$
One instance is $$X \setminus A \;=\; \{x: x\text{ belongs to } X,\text{ belongs to } X, \text{ but does not belong to } A.\}$$
Logically we see that the sets $E \setminus A$ and $(X \setminus A) \cap E$ contain exactly the same elements $x$.
Now, suppose $A \in \Sigma_E$. Thus $A=B \cap E$, and $B \in \Sigma$. We need to show that $E \setminus A \in \Sigma_E$, that is, to show $(X \setminus A) \cap E \in \Sigma_E$.
If we could show that $(X \setminus A) \cap E=(X \setminus B) \cap E$, we would be done, since the form of the latter matches the definition of $\Sigma_E$, and we know $X \setminus B \in \Sigma.$
This we can do by combining $(1)$ the DeMorgan law with $(2)$ the distributivity of intersection over union:
$$X \setminus A = X \setminus (B \cap E) = (X \setminus B) \cup (X \setminus E),\quad\quad\quad\quad (1)$$
$$E \cap (X \setminus A) = [E \cap (X \setminus B)] \cup [E \cap (X \setminus E)].\quad\quad\quad\, (2)$$
Remark: the above work can be shortened if one takes for granted:

For arbitrary $B,E \subset X$, we always have $E \setminus B=E \setminus (B \cap E),$

for we can write $\;\Sigma_E \ni (X \setminus B) \cap E = E \setminus B = E \setminus (B \cap E)=E \setminus A.$
