Measurable spaces Proof Explanation First, in the below theorem .  Is proving  $ \mathcal{F'} = \sigma(\mathcal{A'})$  the same as proving that $T$ is $ \mathcal{F} / \sigma(\mathcal{A'})-measurable$? why? $$$$

I have hard time understaning  the highlighted parts 
Could you please explain for each highlighted part why is it true and why do we mention it or why do we use it?
Also, does the final result mean that
$T$ is $ \mathcal{F} / \sigma(\mathcal{A'})-measurable$ ? if Yes why
 A: Concerning the $4$ highlights.
1) If $T$ is measurable then by definition it satisfies $T^{-1}(F)\in\mathcal F$ for every $F\in\mathcal F'$. Here $\mathcal F'=\sigma(\mathcal A')$ implying that $\mathcal A'$ is a subcollection of $\mathcal F'$. So if $T$ is measurable we will have $T^{-1}(F)\in\mathcal F$ for every $F\in\mathcal A'$. This can be restated as: $T$ can only be $\mathcal F/\mathcal F'$-measurable if $T^{-1}(A')\in\mathcal F$ for every $A'\in\mathcal A'$. Again in other words: it is a necessary condition.
2) This is not more than a definition. $\mathcal G'$ denotes the collection of all subsets of $\Omega'$ that have the property that their preimage under $T$ is an element of $\mathcal F$.
3) As we stated in 1) for every $F\in\mathcal A'$ it is true that its preimage under $T$ (i.e. the set $T^{-1}(F)$) is an element of $\mathcal F$. According to the definition of $\mathcal G'$ given in 2) the statement in 1) can be translated into $\mathcal A'\subseteq\mathcal G'$.
4) in i),ii),iii) it is proved that collection $\mathcal G'$ is a $\sigma$-algebra. This with $\mathcal A'\subseteq\mathcal G'$ as said in 3) and from this we are allowed to conclude that also $\mathcal F':=\sigma(\mathcal A')\subseteq\mathcal G'$. So proved is now that every $F\in\mathcal F'$ has the property that its preimage under $T$ is an element of $\mathcal F$. This states exactly that $T:(\Omega,\mathcal F)\to(\Omega',\mathcal F')$ is measurable. Or using other notation that the function $T:\Omega\to\Omega'$ is $\mathcal F/\mathcal F'$-measurable.

edit: 
It can be shown that for every subcollection $\mathcal{A}'\subseteq\wp\left(\Omega'\right)$
we have: $$\sigma\left(T^{-1}\left(\mathcal{A}'\right)\right)=T^{-1}\left(\sigma\left(\mathcal{A}'\right)\right)\tag1$$where $T^{-1}(\mathcal V):=\{T^{-1}(V)\mid V\in\mathcal V\}$ for any $\mathcal V\subseteq\wp(\Omega')$.
For a guide how to prove that see here.
Proving tat $T$ is $\mathcal{F}/\mathcal{F}'$- measurable is the same as proving that: $T^{-1}\left(\mathcal{F}'\right)\subseteq\mathcal{F}$ or
equivalently: $$T^{-1}\left(\sigma\left(\mathcal{A}'\right)\right)\subseteq\mathcal{F}$$
But $\left(1\right)$ tells us that this is the same condition as:
$$\sigma\left(T^{-1}\left(\mathcal{A}'\right)\right)\subseteq\mathcal{F}$$
wich on its turn is equivalent with the condition:$$T^{-1}\left(\mathcal{A}'\right)\subseteq\mathcal{F}$$

I strongly advice you to put $(1)$ in your "math backpack".
