Proof that the pre-image is a sigma algebra

Question

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.

Answer 1

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.

Answer 2

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.