Sigma algebra/Borel sigma algebra proof problem In my stochastics class we were given the following problem ($\mathcal{B}(\mathbb{R})$ stands for the Borel $\sigma$-algebra on the real line and $\mathbb{R}$ stands for the real numbers):
Let $f : \Omega → \mathbb{R}$ be a function. Let $F = \{ A \subset \Omega : \text{ there exists } B \in \mathcal{B}(\mathbb{R}) \text{ with } A = f^{−1}(B)\}$. Show that $F$ is a $\sigma$-algebra on $\Omega$.
I'm not sure how I should break this down. Apparently the inverse function $f^{-1}$ has some property where its target space is a $\sigma$-algebra when its starting space is a Borel $\sigma$-algebra... or am I going down the wrong path?
I've been going at this for a good while, any help is appreciated.
 A: This is rather easy and has nothing to do with the fact that the target is the real line with the Borel $\sigma$-algebra.
Consider a function $f: X \to Y$ and let $\Sigma$ be a $\sigma$-algebra on $Y$. Let $f^{\ast}\Sigma = \{ A \subset X \,:\, \exists B \in \Sigma \; \text{such that} \; f^{-1}(B) = A\}$. This is a $\sigma$-algebra:
It is easy to see that $\emptyset, X \in f^{\ast}\Sigma$ and the rest of the axioms (closure under complementation and countable unions) is a formal verification. For instance, if $A_{n} = f^{-1}B_{n}  \in f^{\ast}\Sigma$ for $n \in \mathbb{N}$ then we have that
\[
\bigcup A_{n} = \bigcup f^{-1}B_{n} = f^{-1}\bigcup B_{n} \in f^{\ast}\Sigma
\]
because $\bigcup B_{n} \in \Sigma$. Remember that taking preimages commutes with arbitrary unions: to see this, write $f^{-1}B_{1} = \{ x \in X \,: \, f(x) \in B_{1}\}$ and from this description the claim follows easily.
A: You have to show that $F$ is (i) non-empty, (ii) closed under complementation, and (iii) closed under countable unions.
