Question about sigma field My question lies in the opposite inclusion 
Let $F$ denote any $\sigma$-field other than $2^{\Omega}$, and let H $\subset\Omega$ with the property that $H \not\in F$. Then $\sigma(F \cup \{H\})=  \{(H \cap A) \cup (H^c \cap B) : A,B \in F\}$. 
Suppose C = $\{(H \cap A) \cup (H^c \cap B)\}$.
My first argument is to prove that $C \subset \sigma( F \cup H)$ 
case I:
suppose $(H \cap A) \cup (H^c \cap B)$ = $\phi \in \sigma( F \cup H)$ 
case II:
Suppose $(H \cap A) \cup (H^c \cap B) \neq \phi$ $\Longrightarrow$
Either $(H \cap A) \neq \phi$  OR $(H^c \cap B) \neq \phi$. In all cases we get 
$C \subset \sigma(F \cup H)$.
That's my first argument for the first part but I am kind of confused whether its right or not because we are dealing with set that contains sets inside if any kind of critique would be good I am little rusty with dealing with set theory.
For the second inclusion I understand we need to show first that $C$ a sigma field on $H$ and $F$ and involve minimality to prove that $\sigma \subset C$, but what I don't understand how can I do that if any hint would be good. Like I understand I must satisfy all the 3 properties but for example how can I prove $\Omega$ is in $C$ both $H$ and $A,B$ aren't sigma field.
 A: $\sigma(F\cup\{H\})$ is just the smallest $\sigma$-field containing all the sets in $F$ and the set $H$. 
So you don't have to worry about $(H\cap A)\cup(H^c\cap B)$ being empty or not: just use the fact that $\sigma(F\cup\{H\})$ is closed under (countable) set operations!
Call $S:=\sigma(F\cup\{H\})$.
The fact that $C\subseteq S$ is straightforward: if $A,B\in F$, then they belong to $S$ too, as well as $H$, so: 
$$H\cap A\in S,\ \ H^c\in S,\ \ H^c\cap B\in S,\ \ (H\cap A)\cup(H^c\cap B)\in S$$
(we have just used the closure of $S$ under the usual set operations).
The inclusion $S\subseteq C$ is less obvious. To begin, observe that $F\subseteq C$:
if $A\in F$, just take $B:=A$ so that $A=(H\cap A)\cup(H^c\cap B)\in C$. Moreover, $H\in C$: take $A':=\Omega$, $B':=\emptyset$ and you get $H=(H\cap A')\cup(H^c\cap B')\in C$. 
Now we only have to show that $C$ is a $\sigma$-field: then the inclusion will follow because $S$ is the smallest $\sigma$-field containing all the sets in $F$ and $H$ (as we already said). 
Clearly $\emptyset\in C$ (take $A,B:=\emptyset$) and $C$ is closed under countable unions:
if $T_n=(H\cap A_n)\cup(H^c\cap B_n)\in C$ then
$$\bigcup_{n=1}^\infty T_n=\bigcup_{n=1}^\infty (H\cap A_n)\cup\bigcup_{n=1}^\infty(H^c\cap B_n)=\left(H\cap\bigcup_{n=1}^\infty A_n\right)\cup\left(H^c\cap\bigcup_{n=1}^\infty B_n\right)$$
(we have used very standard properties of inclusion and intersection: be sure that you know all of them) and we are done because $A:=\bigcup_{n=1}^\infty A_n\in F$ and $B:=\bigcup_{n=1}^\infty B_n\in F$ and we have just written $\bigcup_{n=1}^\infty T_n$ in the form $(H\cap A)\cup(H^c\cap B)$. 
Finallly the fact that $C$ is closed under complement requires a little more work:
take any $T=(H\cap A)\cup(H^c\cap B)\in C$. Then
$$T^c=(H\cap A)^c\cap(H^c\cap B)^c=(H^c\cup A^c)\cap(H\cup B^c)=(H^c\cap H)\cup (H^c\cap B^c)\cup (A^c\cap H)\cup (A^c\cap B^c)$$
and obviously $H^c\cap H=\emptyset$. The only "bad" term here is $A^c\cap B^c$ (in which $H$ doesn't show up), but we can partition it:
$$A^c\cap B^c=(H\cap(A^c\cap B^c))\cup(H^c\cap(A^c\cap B^c))$$
(why is this true?), so that finally
$$T^c=(H^c\cap B^c)\cup (A^c\cap H)\cup(H\cap(A^c\cap B^c))\cup(H^c\cap(A^c\cap B^c))
=\left(H\cap \left(A^c\cup (A^c\cap B^c)\right)\right)\cup\left(H^c\cap\left(B^c\cup (A^c\cap B^c)\right)\right)$$
so $T^c\in C$ (why?) and we are done.
