# Proof that the pre-image is a sigma algebra

We shall prove that for sets X,Y and a map f: X $\rightarrow$ Y, if B is a $\sigma$-algebra on Y, then

$\{f^{-1}(E) ; E \in B\}$ is a $\sigma$-algebra on X.

I have shown all the properties except for the following one:

i) $B_1,...,B_n \in A \Rightarrow \bigcup \limits_{i=1}^n B_n \in A$

ii) $B_n \in A, n \in \mathbb N \Rightarrow \bigcup \limits^\infty B_n \in A$

I would like to know if for showing i) the following argument would be correct:

For $A := \{f^{-1}(E) ; E \in B\}$, for $a_1,..,a_n \in A, a_1 = f^{-1}(E_1),...,f^{-1}(E_2)$ it holds that: $\bigcup \limits_{i=1}^n a_i = f^{-1}(E_1) \cup...\cup f^{-1}(E_n) = f^{-1}(E_1 \cup ... \cup E_2)\in B$

more precisely I am interested in understand if and why $f^{-1}(E_1) \cup...\cup f^{-1}(E_n) = f^{-1}(E_1 \cup ... \cup E_2)$ holds.

Secondly I would like to know how the generalization to ii) can be done.

To show that two sets are equal, show that each element of the first is an element of the second and vice versa. If $x$ is in the preimage of one of the $E_i$'s then $f(x)$ is in that $E_i$, so $f(x)$ is in the union of the $E_i$'s and therefore $x$ is in the preimage of the union. The converse is similar, and both directions work equally well for finite and infinite unions.
If you can' be bothered to prove that that the preimage of an arbitrary function $f$ commutes with arbitrary unions, you can read proposition 1.4.1 of these notes: http://www.uio.no/studier/emner/matnat/math/MAT2400/v13/mathanalbook.pdf . You can also read some other elementary set theory results here which you might find useful in Measure Theory.