Real Analysis, Folland Proposition 2.1 Background information:
Any mapping $f:X\rightarrow Y$ between two sets induces a mapping $f^{-1}:P(Y)\rightarrow P(X)$, defined by $f^{-1}(E) = \{x\in X:f(x)\in E\}$, which preserves unions, intersections, and complements.
Thus, if $N$ is a $\sigma$-algebra on $Y$, $\{f^{-1}(E):E\in N\}$ is a $\sigma$-algebra on $X$.
If $(X,M)$ and $(Y,N)$ are measurable spaces, a mapping $f:X\rightarrow Y$ is called $(M,N)$-measurable or just measurable when $M$ and $N$ are understood, if $f^{-1}(E)\in M$ for all $E\in N$
Proposition 2.1 - If $N$ is generated by $\varepsilon$, then $f:X\rightarrow Y$ is $(M,N)$-measurable iff $f^{-1}(E)\in M$ for all $E\in \varepsilon$.
Proof $\Rightarrow$ Let $N$ be a $\sigma$-algebra generated by $\varepsilon$ and suppose we have $f:X\rightarrow Y$ that is $(M,N)$- measurable. Then by definition, this mapping induces a mapping $f^{-1}:P(Y)\rightarrow P(X)\}$ defined by $$f^{-1}(E) = \{x\in X: f(x)\in E\}$$ then clearly $f^{-1}(E)\in M$ since $M$ is a $\sigma$-algebra containing $\varepsilon$; it therefore contains $N$. 
I am not sure if this part is true, also I am not sure how to prove the converse any suggestions is greatly appreciated. 
 A: 
Lemma: Let $(X,M)$ be a measurable space and $Y$ be a set. If $f:X\rightarrow Y$ is a function, then 
  $$\Sigma= \{E \subset Y : f^{-1}(E) \in M \}$$
  is a $\sigma$-algebra on $Y$.

Proof: 


*

*Since $f^{-1}(\emptyset)= \emptyset \in M$, we have that $\emptyset \in \Sigma$.

*If $E\in \Sigma$ then $f^{-1}(E) \in M$. So we have $f^{-1}(E^c) = (f^{-1}(E))^c \in M$. So $E\in \Sigma$.

*If $E_n\in \Sigma$, for all $n \in \mathbb{N}$, then $f^{-1}(E_n) \in M$, for all $n \in \mathbb{N}$. So we have 
$$f^{-1}\left ( \bigcup_{n} E_n \right)= \bigcup_{n} f^{-1}( E_n )\in M $$
So $\bigcup_{n} E_n \in \Sigma$ 


So we proved $\Sigma$ is a $\sigma$-algebra (on $Y$).

Proposition 2.1 - If $N$ is generated by $\varepsilon$, then $f:X\rightarrow Y$ is $(M,N)$-measurable iff $f^{-1}(E)\in M$ for all $E\in \varepsilon$.

Proof: 
$(\Rightarrow)$ If $f:X\rightarrow Y$ is $(M,N)$-measurable, then, for all $E\in N$, $f^{-1}(E)\in M$.  Since $\varepsilon \subset N$, we have that, for all $E\in \varepsilon$, $f^{-1}(E)\in M$.
$(\Leftarrow)$ Suppose $f^{-1}(E)\in M$ for all $E\in \varepsilon$.
Let $$\Sigma= \{E \subset Y : f^{-1}(E) \in M \}$$ Then, for all $E\in \varepsilon$, $E\in \Sigma$. It means $\varepsilon \subset \Sigma$. 
From the lemma, we know that $\Sigma$ is a $\sigma$-algebra. 
On the other hand, $N$ is the $\sigma$-algebra generated by $\varepsilon$, so we have that $N \subset \Sigma\:$ (in fact, $N$ is the smallest $\sigma$-algebra containing $\varepsilon$).  
Note that $N \subset \Sigma$ means that, for all $E\in N$, $f^{-1}(E)\in M$, which means that $f:X\rightarrow Y$ is $(M,N)$-measurable.
A: One direction is trivial; if $N$ is generated by $\epsilon$, then by definition, $N$ contains each member of $\epsilon$, and so if $E \in \epsilon$, $E \in N$ also, whence $f^{-1}(E) \in M$, by the definition of measurability.
The trickier direction to prove is the converse. You want to show that if $f^{-1}(E) \in M$ for every $E \in \epsilon$, then $f^{-1}(E) \in M$ for every $E \in N$. 
To show this, we let $K \subseteq \mathcal{P}(Y)$ be the set of all subsets $E$ of $Y$ such that $f^{-1}(E) \in M$. Our goal is to show that $K$ is a $\sigma$-algebra containing $\epsilon$. Then, since $N$ is generated by $\epsilon$, it will follow that $N \subseteq K$, and so we will have that every $E \in N$ satisfies $f^{-1}(E) \in M$, which was what we wanted to prove.
The fact that $K$ contains $\epsilon$ follows immediately from the assumption that $f^{-1}(E) \in M$ for every $E \in \epsilon$, so it remains to show that $K$ is a $\sigma$-algebra. It's clear that $Y \in K$, since $f^{-1}(Y) = X$, and $X \in M$ since $M$ is a $\sigma$-algebra. The fact that $K$ is closed under countable unions and intersections follows immediately from the fact that the preimage mapping $f^{-1}$ preserves unions and intersections. For example, if $E_1, E_2, \dots $ are a countable collection of elements of $K$, then we have that $f^{-1} (\cup_{i=1}^\infty E_i) = \cup_{i=1}^\infty f^{-1} (E_i)$, and since each $f^{-1}(E_i) \in M$ and $M$ is a $\sigma$-algebra, it follows that the union of the sets $f^{-1}(E_i)$ is also in $M$, from which it follows that the union of the $E_i$'s is an element of $K$, and so $K$ is closed under countable unions; a similar argument will establish that $K$ is closed under countable intersections (alternatively, apply De Morgan's laws and use the fact that the preimage map preserves complements). 
This shows that $K$ is a $\sigma$-algebra, and from the deductions in the third paragraph, this concludes the proof.
