Q:

Sigma algebra and inverses

A theorem in my book is exactly this: Inverse image of $\sigma$-algebra
Then there is another theorem which reads: 

Let $f$ be a function defined on a set $X$ with values in a set $Y$. Let $\mathcal{F}$ be a sigma algebra of subsets of $X$ and let $\mathcal{G}=\{E \subseteq Y : f^{-1}(E) \in \mathcal{F} \}. $ Show that $\mathcal{G}$ is a sigma algebra on $Y$.

The first theorem (the one in the link) essentially says (summarizing the main point of the theorem) "the pre-image of a sigma algebra is a sigma algebra." I want to make sure that the second theorem (the one I wrote out) does not mean that the image of a sigma algebra is a sigma algebra. And could someone connect or show me how these two theorems are related (are they saying the same thing? If not, how would you summarize the main point of the second theorem, the one I wrote out). I am a bit confused with all the pre-image notation.

A:

The two theorems are not the same. They complement each other. One gives  a way of getting a sigma algebra on the domain using a sigma algebra on the range and the second one does the reverse. It starts with a sigma algebra on the domain and produce a sigma algebra on the range. One basic fact that makes these two theorems valid is good behaviour of inverse images w.r.t. unions, complements and interscetions of sets. 

