TITLE: Real Analysis, Folland Proposition 2.1
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.