Proving $\mathcal{M}$ is a $\sigma$-algebra in X Let $X$ and $Y$ be two sets and $\left(Y,\mathcal{A}\right)$ a measurable space and let
$f:X\rightarrow Y$ be a function. Prove that 
$\mathcal{M} = \{E \subset X : E = f^{-1} (A), A \in \mathcal{A}\}$ is a $\sigma$-algebra in $X$.
Here's what I know the function $f:X\rightarrow Y$ induces the mapping $f^{-1}:P(Y)\rightarrow P(X)$ defined as $f^{-1}(E) = \{x \in X:f(X) \in E\}$. Let $\mathcal{A}$ be a $\sigma$-algebra on $Y$ $\implies$ $\{f^{-1}(E)|E \in \mathcal{A}\}$ 
This is where I'm lost. Any help is always appreciated.
 A: So we are trying to prove that if you have a $\sigma$-algebra $\mathcal{A}$, then if you look at the set of all preimages of subsets of $Y$ that are in $\mathcal{A}$, this set is actually a $\sigma$-algebra of subsets of $X$.
So, we need to show three things:


*

*$X$ is in $\{ f^{-1}(E) \mid E \in \mathcal{A} \}$

*If $A$ is in $\{ f^{-1}(E) \mid E \in \mathcal{A} \}$, then so is $X - A = A^{c}$.

*If $A_{n}$ (for $n = 1,2,3,\dots$) are in $\{ f^{-1}(E) \mid E \in \mathcal{A} \}$, then so is $\bigcup \limits_{n = 1}^{\infty} A_{n}$.


So, essentially, for $1$, you have to prove $X$ can be written as $f^{-1}(E)$ for some $E \in \mathcal{A}$.  Do you have any ideas which $E$ you should pick?
Then, for $2$, you have to prove if $A \subseteq X$ can be written as $f^{-1}(E)$ for some $E \in \mathcal{A}$, then so can $A^{c}$.  Hint: $f^{-1}(E^{c}) = (f^{-1}(E))^{c}$ -- this is an equality you should prove.  Once you prove it, why is this enough to prove $2$?
Then for $3$, you have to prove if $A_{n} \subseteq X$ can be written as $f^{-1}(E_{n})$ for some $E_{n} \in \mathcal{A}$, then we can find $E \in \mathcal{M}$ so that $\bigcup \limits_{n = 1}^{\infty} A_{n} = f^{-1}(E)$.  Hint: $f^{-1}(\bigcup \limits_{n = 1}^{\infty} E_{n}) = \bigcup \limits_{n = 1}^{\infty} f^{-1}(E_{n})$.  You have to prove this equality, and then once you do, you are done.  But why?  Why is proving this equality enough to show $3$?
